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NCERT Solutions

Class 12

Maths

Chapter 9 Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

Q: 1. What is a differential equation as stated in Class 12 books?

Differential equations assist you in differentiating any function w.r.t an independent variable. It describes a relationship between the functions and their derivatives. The functions refer to the physical quantities whereas their derivatives refer to the rate at which the function is changing and this relationship is differential equations. A D.E. takes the form dy/dx = g(x). Here, y signifies the function. And the function cited here is f(x).

Q: 2. How do you solve differential equations?

Differential equations can be of different orders. There are various approaches to solve these equations. For solving first-order linear differential equations, we need to perform the substitution. Then, after part factorization, one can perform the separation of variables. Once this step is successfully done, we can substitute the original value. Perform these steps until the solution is obtained for the original equation.

Q: 3. Is Class 12 Maths Chapter 9 tough?

Practice makes Maths easy for any student. Once the student is able to clearly grasp the basic concepts and learns the tricks to solve each question, only then will Maths be easy for that student. Hence, it is important to learn the basics well and practise thoroughly for all types of questions with sincerity. The Class 12 Maths syllabus is a mix of different questions in terms of difficulty. Visit the page NCERT Solutions Class 12 Maths Chapter 9 for the solutions

Q: 4. What is the order and degree of the differential equation? Give an example.

Order simply helps to find out the order of the highest term in any D.E., which is basically the term with the highest exponent value. For instance, in this equation, dy/dx + 4 = 2, the order is 1. The degree of any D.E. is associated with its order. The power raised to which the highest term is characterised is the degree of the D.E. In the above-cited example of the 1st order D.E., the degree of the D.E. is 1. Modules relating to this topic or other topics covered in this chapter can be found on the Vedantu website or on the Vedantu app at free of cost.

Q: 5. What is the best Solution book for NCERT Class 12 Maths Chapter 9?

You may acquire NCERT Class 12 Math Solutions by going to the Vedantu website and searching for Class 12 Maths solutions. Aside from that, you may access a variety of modules that will assist you in achieving high marks in Maths examinations. The exercise solutions are provided on the page NCERT Solutions Class 12 Maths Chapter 9. Click on it to download a PDF of the solutions.

Q: 6. What kind of problems are covered in the NCERT Solutions for Differential Equations Class 12 NCERT Solutions?

The solutions address various problems related to:Identifying the order and degree of differential equations.Forming differential equations from real-world scenarios.Solving differential equations using methods like variable separation, homogeneous equations, and linear equations (for beginners).Applying integrating factors (for slightly advanced problems).Understanding applications of differential equations in areas like population growth, motion, and electrical circuits.

Q: 7. Are there solutions for differential equation class 12 for advanced techniques like Laplace Transform?

The NCERT solutions primarily focus on methods for beginners and intermediate learners. For advanced techniques like Laplace Transform, you might need to refer to other resources like advanced mathematics textbooks or online tutorials dedicated to the topic.

Q: 8. What is a differential equation?

A differential equation is a mathematical equation that relates a function with its derivatives. It describes the rate at which a quantity changes. For example, dy/dx =f(x,y) is a first-order differential equation.

Q: 9. How are exact differential equations solved in differential equations class 12 NCERT solutions?

Exact differential equations are solved by finding a potential function whose partial derivatives match the terms in the differential equation. If the equation is not exact, an integrating factor might be used to make it exact.

Q: 10. What are the applications of differential equations Class 12?

Differential equations are used in various fields such as physics (motion, heat, waves), biology (population models), economics (growth models), and engineering (circuit analysis).

NCERT Maths Chapter 9 Differential Equations Class 12 Solutions - Free PDF Download

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations explores the fundamental concepts, formation, and solutions of differential equations, which are equations involving derivatives of functions. Differential equations are equations that involve derivatives of a function. This chapter introduces students to the concepts of order and degree of a differential equation, the formation of differential equations, and methods of solving them.

Table of Content

Whether you are doing your homework or preparing for exams, these solutions are designed to align with the CBSE marking scheme and guidelines, ensuring you grasp CBSE Class 12 Maths Syllabus thoroughly and perform well in your exams.

Glance on Maths Chapter 9 Class 12 - Differential Equations

Chapter 9 of class 12 differential equations NCERT solutions Maths involves an unknown function and its derivatives.
An equation involving derivatives of the dependent variable to independent variables (variables) is known as a differential equation.
The order of a differential equation is the order of the highest-order derivative occurring in the differential equation.
The degree of a differential equation is defined if it is a polynomial equation in its derivatives.
A function that satisfies the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called a particular solution
To form a differential equation from a given function we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants.
This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 9 - Differential Equations, which you can download as PDFs.
There are five exercises (98 fully solved questions) in class 12th maths chapter 9 Differential Equations.

Access Exercise wise NCERT Solutions for Chapter 9 Maths Class 12

S.No.	Current Syllabus Exercises of Class 12 Maths Chapter 9
1	NCERT Solutions of Class 12 Maths Differential Equations Exercise 9.1
2	NCERT Solutions of Class 12 Maths Differential Equations Exercise 9.2
3	NCERT Solutions of Class 12 Maths Differential Equations Exercise 9.3
4	NCERT Solutions of Class 12 Maths Differential Equations Exercise 9.4
5	NCERT Solutions of Class 12 Maths Differential Equations Exercise 9.5

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Exercises Under NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

The chapter is divided into five exercises and one miscellaneous exercise. Here is a brief summary of each exercise:

Exercise 9.1: This exercise asks students to verify whether a given function is a solution of a given differential equation. A general solution is the one where the independent arbitrary constants of the equation are equal to the order of the equation. So for an equation d2y/dx2 + y = 0, its general solution would be given as y = K Cos x + C sin x, since it has 2 arbitrary constants K and C which are equal to the order of the equation that is 2. To find the particular solution of a differential equation, the arbitrary constants need to be given particular values. So, in the above example, above if we replace K = C = 1, we get the solution y = cos x + sin x which is termed as the particular solution of the differential equation.
Exercise 9.2: This exercise asks students to find the order and degree of given differential equations. In this, you would learn how to formulate differential equations given “n” arbitrary constants and differentiate the equation n times over to get the n + 1 equations. When you eliminate the arbitrary constants from these n + 1 equations, you will get the required differential equation. You would use this method to derive a differential equation that represents a family of curves.
Exercise 9.3: This exercise asks students to solve differential equations of the first order and first degree.
Exercise 9.4: This exercise asks students to solve differential equations of the first order and higher degree.
Exercise 9.5: This exercise asks students to solve differential equations of second order and higher degree.

Overall, this chapter is essential for students who want to pursue higher education in Mathematics and Physics. It provides a solid foundation for students to understand differential equations and their applications in real-world problems.

Access NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations

Exercise 9.1

1. Determine order and degree (if defined) of differential equation $\frac{d^{4} y}{d x^{4}} +sin (y^{'}^{'}^{'}) =0 .$

Ans: Rewrite the equation $\frac{d^{4} y}{d x^{4}} +sin (y^{'}^{'}^{'}) =0$ . as:

$\Rightarrow y^{'}^{'}^{'}^{'} +sin (y^{‴}) = 0$

The highest order between the two terms is of $y^{'}^{'}^{'}^{'}$ which is four.

The differential equation contains a trigonometric derivative term and is not completely polynomial in its derivative, thus degree is not defined.

2. Determine order and degree (if defined) of differential equation $y^{'} +5y=0$ .

Ans: The given differential equation is $y^{'} +5y=0$ .

The highest order term is $y^{'}$ , thus the order is one.

As the derivative is of completely polynomial nature is and highest power of derivative is of $y^{'}$ which is one. Thus degree is one.

3. Determine order and degree (if defined) of differential equation ${(\frac{ds}{dt})}^{4} +3s \frac{d^{2} s}{d t^{2}} =0$ .

Ans: The given differential equation is ${(\frac{ds}{dt})}^{4} +3s \frac{d^{2} s}{d t^{2}} =0$ .

The highest order term is $\frac{d^{2} s}{d t^{2}}$ , thus the order is two.

As the derivative is of completely polynomial nature is and highest power of derivative term $\frac{d^{2} s}{d t^{2}}$ which is one. Thus the degree is one.

4. Determine order and degree (if defined) of differential equation ${(\frac{d^{2} y}{d x^{2}})}^{2} +cos (\frac{dy}{dx}) =0$ .

Ans: The given differential equation is ${(\frac{d^{2} y}{d x^{2}})}^{2} +cos (\frac{dy}{dx}) =0$ .

The highest order term is $\frac{d^{2} y}{d x^{2}}$ , thus the order is two.

The differential equation contains a trigonometric derivative term and is not completely polynomial in its derivative, thus degree is not defined.

5. Determine order and degree (if defined) of differential equation $(\frac{d^{2} y}{d x^{2}}) -cos3x+sin3x=0$ .

Ans: The given differential equation is ${(\frac{d^{2} y}{d x^{2}})}^{2} -cos3x+sin3x=0$ .

The highest order term is $\frac{d^{2} y}{d x^{2}}$ , thus the order is two.

As the derivative is of completely polynomial nature is and highest power of derivative term $\frac{d^{2} y}{d x^{2}}$ which is one. Thus degree is one.

6. Determine order and degree (if defined) of differential equation ${(y^{'}^{'}^{'})}^{2} + {(y^{'}^{'})}^{3} + {(y^{'})}^{4} + y^{5} =0$ .

Ans: The given differential equation is ${(y^{'}^{'}^{'})}^{2} + {(y^{'}^{'})}^{3} + {(y^{'})}^{4} + y^{5} =0$ .

The highest order term is ${(y^{‴})}^{2}$ , thus the order is three.

The differential equation is of the polynomial form and the power of highest order term $y^{'}^{'}^{'}$ is two, thus the degree is two.

7. Determine order and degree (if defined) of differential equation $y^{'}^{'}^{'} +2y^{'}^{'} +y^{'} =0$ .

Ans: The given differential equation is $y^{'}^{'}^{'} +2y^{'}^{'} +y^{'} =0$ .

The highest order derivative in the differential equation is $y^{‴}$ . Thus its order is three.

The differential equation is polynomial with the highest order term $y^{‴}$ having a degree one. Thus the degree is one.

8. Determine order and degree (if defined) of differential equation $y^{'} +y=e^{'}$ .

Ans: The given differential equation is $y^{'} +y=e^{'}$ . Therefore:

$\Rightarrow y^{'} +y-e^{'} =0$

The highest order derivative in the differential equation is $y^{'}$ . Thus its order is one.

The given equation is of polynomial form with the highest order term $y^{'}$ with degree one. Thus the degree is one.

9. Determine order and degree (if defined) of differential equation $y^{'}^{'} + {(y^{'})}^{2} +2y=0$ .

Ans: The given differential equation is $y^{'}^{'} + {(y^{'})}^{2} +2y=0$ .

The highest order derivative in the differential equation is $y^{'}^{'}$ . Thus its order is two.

The given equation is of polynomial form with the highest order term $y^{'}^{'}$ with highest degree one. Thus the degree is one.

10. Determine order and degree (if defined) of differential equation $y^{'}^{'} +2y^{'} +siny=0$ .

Ans: The given differential equation is $y^{'}^{'} +2y^{'} +siny=0$ .

The highest order derivative in the differential equation is $y^{'}^{'}$ . Thus its order is two.

The given equation is of polynomial form with the highest order term $y^{'}^{'}$ with the highest degree one. Thus the degree is one.

11. Find the degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{dy}{dx})}^{2} +sin (\frac{dy}{dx}) +1=0$ .

(A) $3$

(B) $2$

(D)not defined

Ans: The given differential equation is ${(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{dy}{dx})}^{2} +sin (\frac{dy}{dx}) +1=0$ .

The differential equation is not polynomial in its derivative because of the term $sin (\frac{dy}{dx})$ thus its order is not defined.

The correct answer is (D).

12. Find the order of the differential equation $2 x^{2} \frac{d^{2} y}{d x^{2}} -3 \frac{dy}{dx} +y=0$ .

(A) $2$

(B) $1$

(D)not defined

Ans: The given differential equation is $2 x^{2} \frac{d^{2} y}{d x^{2}} -3 \frac{dy}{dx} +y=0$ .

The highest order term of the equation is $\frac{d^{2} y}{d x^{2}}$ , thus the order is two.

The correct answer is (A).

Exercise 9.2

1. Verify the function $y= e^{x} +1$ is solution of differential equation $y^{'}^{'} -y=0$ .

Ans: The given function is $y= e^{x} + 1$ .

Take its derivative:

$\frac{d y}{d x} = \frac{d}{d x} (e^{x} + 1)$

$\Rightarrow y^{'} = e^{x}$ ……(1)

Take the derivative of the above equation:

$\frac{d}{d x} (y^{'}) = \frac{d}{d x} (e^{x})$

$\Rightarrow y^{″} = e^{x}$

Using result from equation (1):

$y^{″} - y^{'} = 0$

Thus the given function is solution of differential equation $y^{'}^{'} -y^{'} =0$ .

2. Verify the function $y= x^{2} +2x+C$ is solution of differential equation $y^{'} -2x-2=0$ .

Ans: The given function is $y= x^{2} +2x+C$ .

Take its derivative:

$\frac{d y}{d x} = \frac{d}{d x} (x^{2} + 2 x + C)$

$\Rightarrow y^{'} = 2 x + 2$

$\Rightarrow y^{'} - 2 x - 2 = 0$

Thus the given function is solution of differential equation $y^{'} -2x-2=0$ .

3. Verify the function $y=cos x +C$ is solution of differential equation $y^{'} +sin x =0$ .

Ans: The given function is $y=cos x + C$ .

Take its derivative:

$\frac{d y}{d x} = \frac{d}{d x} (\cos x + C)$

$\Rightarrow y^{'} = - \sin x$

$\Rightarrow y^{'} + \sin x = 0$

Thus the given function is solution of differential equation $y^{'} + sin x = 0$ .

4. Verify the function $y= \sqrt{1+ x^{2}}$ is solution of differential equation $y^{'} = \frac{xy}{1+ x^{2}}$ .

Ans: The given function is $y = \sqrt{1 + x^{2}}$ .

Take its derivative:

$\frac{d y}{d x} = \frac{d}{d x} (\sqrt{1 + x^{2}})$

$y^{'} = \frac{1}{2 \sqrt{1 + x^{2}}} \times \frac{d}{d x} (1 + x^{2})$

$y^{'} = \frac{2}{2 \sqrt{1 + x^{2}}}$

$\Rightarrow y^{'} = \frac{1}{\sqrt{1 + x^{2}}}$

Multiply numerator and denominator by $\sqrt{1+ x^{2}}$ :

$y^{'} = \frac{1}{\sqrt{1 + x^{2}}} \times \frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}$

Substitute $y = \sqrt{1+ x^{2}}$

$\Rightarrow y^{'} = \frac{x y}{1 + x^{2}}$

Thus the given function is solution of differential equation $y^{'} = \frac{x y}{1 + x^{2}}$ .

5. Verify the function $y=Ax$ is solution of differential equation $xy^{'} =y (x \neq 0)$ .

Ans: The given function is $y = Ax$ .

Take its derivative:

$\frac{d y}{d x} = \frac{d}{d x} (A x)$

$\Rightarrow y^{'} = A$

Multiply by $x$ on both side:

$x y^{'} = A x$

Substitute $y = Ax$ :

$\Rightarrow x y^{'} = y$

Thus the given function is solution of differential equation $x y^{'} = y (x \neq 0)$ .

6. Verify the function $y=xsin x$ is solution of differential equation $xy^{'} =y+x \sqrt{x^{2} - y^{2}} (x \neq 0 and x > y or x < - y)$ .

Ans: The given function is $y = xsin x$ .

Take its derivative:

$\frac{d y}{d x} = \frac{d}{d x} (x \sin x)$

$\Rightarrow y^{'} = \sin x \frac{d}{d x} (x) + x \frac{d}{d x} (\sin x)$

$\Rightarrow y^{'} = \sin x + x \cos x$

Multiply by $x$ on both side:

$x y^{'} = x (\sin x + x \cos x)$

$x y^{'} = x \sin x + x^{2} \cos x$

Substitute $y = xsin x$ :

$\Rightarrow x y^{'} = y + x^{2} \cos x$

Use $sin x = \frac{y}{x}$ and substitute $cos x$ :

$x y^{'} = y + x^{2} \sqrt{1 - \sin^{2} x}$

$x y^{'} = y + x^{2} \sqrt{1 - {(\frac{y}{x})}^{2}}$

$\Rightarrow x y^{'} = y + x \sqrt{y^{2} - x^{2}}$

Thus the given function is solution of differential equation $xy^{'} = y + x \sqrt{y^{2} - x^{2}}$ .

7. Verify the function $xy=log y + C$ is solution of differential equation $y^{'} = \frac{y^{2}}{1-xy} (xy \neq 1)$ .

Ans: The given function is $xy = log y + C$ .

Take derivative on both side:

$\frac{d}{d x} (x y) = \frac{d}{d x} (\log y)$

$\Rightarrow y \frac{d}{d x} (x) + x \frac{d y}{d x} = \frac{1}{y} \frac{d y}{d x}$

$\Rightarrow y + x y^{'} = \frac{1}{y} y^{'}$

$\Rightarrow y^{2} + x y y^{'} = y^{'}$

Shift the $y^{'}$ terms on one side and take it common.

$\Rightarrow (x y - 1) y^{'} = - y^{2}$

$\Rightarrow y^{'} = \frac{y^{2}}{1 - x y}$

Thus the given function is solution of differential equation $y^{'} = \frac{y^{2}}{1-xy}$ .

8. Verify the function $y-cosy=x$ is solution of differential equation $(ysin y + cos y + x) y^{'} = 1$ .

Ans: The given function is $y - cos y = x$ .

Take derivative on both side:

$\frac{d y}{d x} - \frac{d}{d x} (\cos y) = \frac{d}{d x} (x)$

$\Rightarrow y^{'} + y^{'} \sin y = 1$

$\Rightarrow y^{'} (1 + \sin y) = 1$

$\Rightarrow y^{'} = \frac{1}{1 + \sin y}$

Multiply by $(ysin y + cos y + x)$ on both side:

$(ysin y + cos y + x) y^{'} = \frac{(ysin y + cos y + x)}{1 + \sin y}$

Substitute $y = \cos y + x$ in the numerator:

$(ysin y + cos y + x) y^{'} = \frac{(ysin y + y)}{1 + \sin y}$

$(ysin y + cos y + x) y^{'} = \frac{y (sin y + 1)}{1 + \sin y}$

$\Rightarrow (ysin y + cos y + x) y^{'} = y$

Thus the given function is solution of differential equation $(ysin y + cos y + x) y^{'} = y$ .

9. Verify the function $x+y=ta n^{- 1} y$ is solution of differential equation $y^{2} y^{'} + y^{2} +1=0$ .

Ans: The given function is $x + y = ta n^{- 1} y$ .

Take derivative on both side:

$\frac{d}{d x} (x + y) = \frac{d}{d x} (t a n^{- 1} y)$

$1 + y^{'} = (\frac{1}{1 + y^{2}}) y^{'}$

$\Rightarrow y^{'} [\frac{1}{1 + y^{2}} - 1] = 1$

$\Rightarrow y^{'} [\frac{1 - (1 + y^{2})}{1 + y^{2}}] = 1$

$\Rightarrow y^{'} [\frac{- y^{2}}{1 + y^{2}}] = 1$

$\Rightarrow - y^{2} y^{'} = 1 + y^{2}$

$\Rightarrow y^{2} y^{'} + y^{2} + 1 = 0$

Thus the given function is solution of differential equation $y^{2} y^{'} + y^{2} + 1 = 0$ .

10. Verify the function $y= \sqrt{a^{2} - x^{2}} x \in (-a,a)$ is solution of differential equation $x+y \frac{dy}{dx} =0 (y \neq 0)$ .

Ans: The given function is:

$y = \sqrt{a^{2} - x^{2}} x \in (- a, a)$ .

Take derivative on both side:

$\frac{d y}{d x} = \frac{d}{d x} (\sqrt{a^{2} - x^{2}})$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{a^{2} - x^{2}}} \frac{d}{d x} (a^{2} - x^{2})$

$\Rightarrow \frac{d y}{d x} = \frac{- 2 x}{2 \sqrt{a^{2} - x^{2}}}$

$\Rightarrow \frac{d y}{d x} = \frac{- x}{\sqrt{a^{2} - x^{2}}}$

Substitute $y = \sqrt{a^{2} - x^{2}}$

$\Rightarrow \frac{d y}{d x} = \frac{- x}{y}$

$\Rightarrow x + y \frac{d y}{d x} = 0$

Thus the given function is solution of differential equation:

$x + y \frac{d y}{d x} = 0 (y \neq 0)$ .

11. Find the numbers of arbitrary constants in the general solution of a differential equation of fourth order.

(A) $0$

(B) $2$

(D) $4$

Ans: The number of arbitrary constants in the general solution of a differential equation is equal to its order. As the given differential equation is of fourth order, thus it has four arbitrary constants in its solution.

The correct answer is (D).

12. Find the numbers of arbitrary constants in the particular solution of a differential equation of third order.

(A) $3$

(B) $2$

(D) $0$

Ans: The particular solution of any differential equation does not have any arbitrary constants. Thus it has zero constants in its solution.

The correct answer is (D).

Exercise 9.3

1. Find the general solution for $\frac{dy}{dx} = \frac{1- cos x}{1+ cos x}$ .

Ans: The given differential equation is $\frac{dy}{dx} = \frac{1- cos x}{1+ cos x}$ .

Use trigonometric half –angle identities to simplify:
$\frac{d y}{d x} = \frac{1 - \cos x}{1 + \cos x}$

$\Rightarrow \frac{d y}{d x} = \frac{2 \sin^{2} \frac{x}{2}}{2 \cos^{2} \frac{x}{2}}$

$\Rightarrow \frac{d y}{d x} = \tan^{2} \frac{x}{2}$

$\Rightarrow \frac{d y}{d x} = \sec^{2} \frac{x}{2} - 1$

Separate the differentials and integrate:

$\int d y = \int (\sec^{2} \frac{x}{2} - 1) d x$

$\Rightarrow y = \int \sec^{2} \frac{x}{2} d x - \int d x$

$\Rightarrow y = 2 \tan \frac{x}{2} - x + C$

Thus the general solution of given differential equation is $y = 2tan \frac{x}{2} - x + C$ .

2. Find the general solution for $\frac{dy}{dx} = \sqrt{4 - y^{2}} (-2 y 2)$ .

Ans: The given differential equation is $\frac{dy}{dx} = \sqrt{4 - y^{2}} (-2 y 2)$ .

Simplify the expression:

$\frac{d y}{d x} = \sqrt{4 - y^{2}}$

$\Rightarrow \frac{d y}{\sqrt{4 - y^{2}}} = d x$

Use standard integration:

$\int \frac{d y}{\sqrt{4 - y^{2}}} = \int d x$

$\Rightarrow \sin^{- 1} \frac{y}{2} = x + C$

$\Rightarrow \frac{y}{2} = \sin (x + C)$

$\Rightarrow y = 2 \sin (x + C)$

Thus the general solution of given differential equation is $y = 2 sin (x + C)$ .

3. Find the general solution for $\frac{dy}{dx} + y =1 (y \neq 1)$ .

Ans: The given differential equation is $\frac{dy}{dx} + y =1 (y \neq 1)$ .

Simplify the expression:

$\frac{d y}{d x} + y = 1$

$\Rightarrow \frac{d y}{d x} = 1 - y$

$\Rightarrow \frac{d y}{1 - y} = d x$

Use standard integration:

$\int \frac{d y}{1 - y} = \int d x$

$\Rightarrow - \log (1 - y) = x + C$

$\Rightarrow \log (1 - y) = - (x + C)$

$\Rightarrow 1 - y = e^{- (x + C)}$

$y = 1 - A e^{- x} (A = e^{- C})$

Thus the general solution of given differential equation is $y=1- A e^{-x}$ .

4. Find the general solution for $se c^{2} x tany dx+se c^{2} y tanx dy=0$ .

Ans: The given differential equation is:

$\sec^{2} x \tan y d x + \sec^{2} y \tan x d y = 0$ .

Divide both side by $tan x tan y$ :

$\frac{\sec^{2} x \tan y d x + \sec^{2} y \tan x d y}{\tan x \tan y} = 0$

$\Rightarrow \frac{\sec^{2} x}{\tan x} d x + \frac{\sec^{2} y}{\tan y} d y = 0$

Integrate both side:

$\int \frac{\sec^{2} x}{\tan x} d x + \int \frac{\sec^{2} y}{\tan y} d y = 0$

$\Rightarrow \int \frac{\sec^{2} y}{\tan y} d y = - \int \frac{\sec^{2} x}{\tan x} d x$ ……(1)

Use a substitution method for integration. Substitute $tan x = u$ :

For integral on RHS:

$\Rightarrow \tan x = u$

$\Rightarrow \sec^{2} x d x = d u$

$\Rightarrow \int \frac{\sec^{2} x}{\tan x} d x = \int \frac{d u}{u}$

$\Rightarrow \int \frac{\sec^{2} x}{\tan x} d x = \log u$

$\Rightarrow \int \frac{\sec^{2} x}{\tan x} d x = \log (\tan x)$

Thus evaluating result form (1):

$\Rightarrow \log (\tan y) = - \log (\tan x) + \log (C)$

$\Rightarrow \log (\tan y) = \log (\frac{C}{\tan x})$

$\Rightarrow \tan x \tan y = C$

Thus the general solution of given differential equation is $tan x tan y = C$ .

5. Find the general solution for $(e^{x} + e^{-x}) dy- (e^{x} - e^{-x}) dx=0$ .

Ans: The given differential equation is:

$(e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0$ .

Simplify the expression:

$d y = [\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}] d x$

Integrate both side:

$\int d y = \int [\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}] d x$ ……(1)

Use a substitution method for integration. Substitute $e^{x} + e^{-x} =t$ :

For integral on RHS:

$\Rightarrow e^{x} + e^{- x} = t$

$\Rightarrow (e^{x} - e^{- x}) d x = d t$

$\Rightarrow \int [\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}] d x = \int \frac{d t}{t}$ s

$\Rightarrow \int [\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}] d x = \ln t + C$

$\Rightarrow \int [\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}] d x = \log (e^{x} + e^{- x}) + C$

Thus evaluating result form (1):

$y = \log (e^{x} + e^{- x}) + C$

Thus the general solution of given differential equation is $y=log (e^{x} + e^{-x}) +C$ .

6. Find the general solution for $\frac{dy}{dx} = (1+ x^{2}) (1+ y^{2})$ .

Ans: The given differential equation is:

$\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})$ .

Simplify the expression:

$\frac{d y}{1 + y^{2}} = (1 + x^{2}) d x$

Integrate both side:

$\int \frac{d y}{1 + y^{2}} = \int (1 + x^{2}) d x$

Use standard integration:

$\tan^{- 1} y = \int d x + \int x^{2} d x$

$\Rightarrow \tan^{- 1} y = x + \frac{x^{3}}{3} + C$

Thus the general solution of given differential equation is $ta n^{-1} y=x+ \frac{x^{3}}{3} +C$ .

7. Find the general solution for $ylog y dx-xdy=0$ .

Ans: The given differential equation is:

$y \log y d x - x d y = 0$ .

Simplify the expression:

$y \log y d x = x d y$

$\Rightarrow \frac{d x}{x} = \frac{d y}{y \log y}$

Integrate both side:

$\int \frac{d y}{y \log y} = \int \frac{d x}{x}$ ……(1)

Use substitution method for integration on LHS. Substitute $log y=t$ :

$\log y = t$

$\Rightarrow \frac{1}{y} d y = d t$

$\Rightarrow \int \frac{d y}{y \log y} = \int \frac{d t}{t}$

$\Rightarrow \int \frac{d y}{y \log y} = \log t$

$\Rightarrow \int \frac{d y}{y \log y} = \log (\log y)$

Evaluating expression (1):

$\log (\log y) = \log x + \log C$

$\Rightarrow \log (\log y) = \log C x$

$\Rightarrow \log y = C x$

$\Rightarrow y = e^{C x}$

Thus the general solution of given differential equation is $y= e^{Cx}$ .

8. Find the general solution for $x^{5} \frac{dy}{dx} =- y^{5}$ .

Ans: The given differential equation is:

$x^{5} \frac{d y}{d x} = - y^{5}$ .

Simplify the expression:

$\frac{d y}{y^{5}} = - \frac{d x}{x^{5}}$

Integrate both side:

$\int \frac{d y}{y^{5}} = - \int \frac{d x}{x^{- 5}}$

$\Rightarrow \int y^{- 5} d y = - \int x^{- 5} d x$

$\Rightarrow \frac{y^{- 5 + 1}}{- 5 + 1} = - \frac{x^{- 5 + 1}}{- 5 + 1} + C$

$\Rightarrow \frac{y^{- 4}}{- 4} = - \frac{x^{- 4}}{- 4} + C$

$\Rightarrow x^{- 4} + y^{- 4} = - 4 C$

$\Rightarrow x^{- 4} + y^{- 4} = A (A = - 4 C)$

Thus the general solution of given differential equation is $x^{-4} + y^{-4} =A$ .

9. Find the general solution for $\frac{dy}{dx} =si n^{-1} x$ .

Ans: The given differential equation is:

$\frac{d y}{d x} = \sin^{- 1} x$ .

Simplify the expression:

$d y = \sin^{- 1} x d x$

Integrate both side:

$\int d y = \int \sin^{- 1} x d x$

$\Rightarrow y = \int 1 \times \sin^{- 1} x d x$

Use product rule of integration:

$\int \sin^{- 1} x d x = \sin^{- 1} x \int d x - \int (\frac{1}{\sqrt{1 - x^{2}}} \int d x) d x$

$\Rightarrow \int \sin^{- 1} x d x = x \sin^{- 1} x - \int \frac{x}{\sqrt{1 - x^{2}}} d x$

Substitute $1- x^{2} = t^{2}$

$1 - x^{2} = t^{2}$

$\Rightarrow - 2 x d x = 2 t d t$

$\Rightarrow - x d x = t d t$

Evaluating the integral:

$\frac{2 x^{2} + x}{(x + 1) (x^{2} + 1)} = \frac{A}{x + 1} + \frac{B x + C}{x^{2} + 1}$

$\Rightarrow \int \sin^{- 1} x d x = x \sin^{- 1} x + \int \frac{t d t}{\sqrt{t^{2}}}$

$\Rightarrow \int \sin^{- 1} x d x = x \sin^{- 1} x + t + C$

$\Rightarrow \int \sin^{- 1} x d x = x \sin^{- 1} x + \sqrt{1 - x^{2}} + C$

$\Rightarrow y = x \sin^{- 1} x + \sqrt{1 - x^{2}} + C$

Thus the general solution of given differential equation is $y=xsi n^{-1} x+ \sqrt{1- x^{2}} +C$ ,

10. Find the general solution for $e^{x} tanydx+ (1- e^{x}) se c^{2} ydy=0$ .

Ans: The given differential equation is:

$e^{x} \tan y d x + (1 - e^{x}) \sec^{2} y d y = 0$ .

Simplify the expression:

$(1 - e^{x}) \sec^{2} y d y = - e^{x} \tan y d x$

$\Rightarrow \frac{\sec^{2} y}{\tan y} d y = - \frac{e^{x}}{(1 - e^{x})} d x$

Integrate both side:

$\int \frac{\sec^{2} y}{\tan y} d y = - \int \frac{e^{x}}{(1 - e^{x})} d x$ ……(1)

Substitute $tan y=u$

$\tan y = u$

$\Rightarrow \sec^{2} y = d u$

Evaluating the LHS integral of (1):

$\Rightarrow \int \frac{\sec^{2} y}{\tan y} d y = \int \frac{d u}{u}$

$\Rightarrow \int \frac{\sec^{2} y}{\tan y} d y = \log u$

$\Rightarrow \int \frac{\sec^{2} y}{\tan y} d y = \log (\tan y)$

Substitute $1- e^{x} =v$

$1 - e^{x} = v$

$\Rightarrow - e^{x} d x = d v$

Evaluating the RHS integral of (1):

$\Rightarrow - \int \frac{e^{x}}{(1 - e^{x})} d x = \int \frac{d v}{v}$

$\Rightarrow - \int \frac{e^{x}}{(1 - e^{x})} d x = \log v$

$\Rightarrow - \int \frac{e^{x}}{(1 - e^{x})} d x = \log (1 - e^{x})$

Therefore the integral (1) will be:

$\log (\tan y) = \log (1 - e^{x}) + \log C$

$\Rightarrow \log (\tan y) = \log C (1 - e^{x})$

$\Rightarrow \tan y = C (1 - e^{x})$

Thus the general solution of given differential equation is $tany=C (1- e^{x})$ ,

11. Find the particular solution of $(x^{3} + x^{2} +x+1) \frac{dy}{dx} =2 x^{2} +x; y=1, x=0$ to satisfy the given condition.

Ans: The given differential equation is:

$(x^{3} + x^{2} + x + 1) \frac{d y}{d x} = 2 x^{2} + x; y = 1, x = 0$ .

Simplify the expression:

$\frac{d y}{d x} = \frac{2 x^{2} + x}{(x^{3} + x^{2} + x + 1)}$

$\Rightarrow \frac{d y}{d x} = \frac{2 x^{2} + x}{(x^{3} + x + x^{2} + 1)}$

$\Rightarrow \frac{d y}{d x} = \frac{2 x^{2} + x}{(x + 1) (x^{2} + 1)}$

$\Rightarrow d y = \frac{2 x^{2} + x}{(x + 1) (x^{2} + 1)} d x$

Integrate both side:

$\int d y = \int \frac{2 x^{2} + x}{(x + 1) (x^{2} + 1)} d x$ ……(1)

Use partial fraction method to simplify the RHS:

$\frac{2 x^{2} + x}{(x + 1) (x^{2} + 1)} = \frac{A}{x + 1} + \frac{B x + C}{x^{2} + 1}$

$\Rightarrow 2 x^{2} + x = A (x^{2} + 1) + (B x + C) (x + 1)$

$\Rightarrow 2 x^{2} + x = (A + B) x^{2} + (B + C) x + (A + C)$

By comparing coefficients:

$A + B = 2$

$B + C = 1$

$A + C = 0$

Solving this we get:

$\frac{2 x^{2} + x}{(x + 1) (x^{2} + 1)} = \frac{(\frac{1}{2})}{x + 1} + \frac{(\frac{3}{2}) x + (- \frac{1}{2})}{x^{2} + 1}$

$\Rightarrow \frac{2 x^{2} + x}{(x + 1) (x^{2} + 1)} = \frac{1}{2} (\frac{1}{x + 1} + \frac{3 x - 1}{x^{2} + 1})$

Rewriting the integral(1):

$y = \frac{1}{2} \int (\frac{1}{x + 1} + \frac{3 x - 1}{x^{2} + 1}) d x$

$\Rightarrow y = \frac{1}{2} \int \frac{1}{x + 1} d x + \frac{1}{2} \int \frac{3 x - 1}{x^{2} + 1} d x$

$\Rightarrow y = \frac{1}{2} \log (x + 1) + \frac{3}{2} \int \frac{x}{x^{2} + 1} d x - \frac{1}{2} \int \frac{1}{x^{2} + 1} d x$

$\Rightarrow y = \frac{1}{2} \log (x + 1) + \frac{3}{4} \int \frac{2 x}{x^{2} + 1} d x - \frac{1}{2} \tan^{- 1} x$

$\Rightarrow y = \frac{1}{2} \log (x + 1) + \frac{3}{4} \log (x^{2} + 1) - \frac{1}{2} \tan^{- 1} x + C$

For $y = 1$ when $x=0$ .

$1 = \frac{1}{2} \log (0 + 1) + \frac{3}{4} \log (0 + 1) - \frac{1}{2} \tan^{- 1} 0 + C$

$\Rightarrow C = 1$

Thus the required particular solution is :

$\Rightarrow y = \frac{1}{2} \log (x + 1) + \frac{3}{4} \log (x^{2} + 1) - \frac{1}{2} \tan^{- 1} x + 1$ .

12. Find the particular solution of $x (x^{2} -1) \frac{dy}{dx} =1$ ; $y=0$ when $x=2$ to satisfy the given condition.

Ans: The given differential equation is:

$x (x^{2} - 1) \frac{d y}{d x} = 1$ ; $y = 0$ when $x = 2$

Simplify the expression:

$x (x^{2} - 1) \frac{d y}{d x} = 1$

$\Rightarrow d y = \frac{d x}{x (x^{2} - 1)}$

$\Rightarrow d y = \frac{d x}{x (x - 1) (x + 1)}$

Integrate both side:

$\int d y = \int \frac{d x}{x (x - 1) (x + 1)}$ ……(1)

Use partial fraction method to simplify the RHS:

$\frac{1}{x (x - 1) (x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}$

$\Rightarrow 1 = A (x^{2} - 1) + B x (x + 1) + C x (x - 1)$

$\Rightarrow 1 = (A + B + C) x^{2} + (B - C) x - A$

By comparing coefficients:

$A + B + C = 0$

$B - C = 0$

$- A = 1$

Solving this we get:

$\frac{1}{x (x - 1) (x + 1)} = \frac{(- 1)}{x} + \frac{(\frac{1}{2})}{x - 1} + \frac{(\frac{1}{2})}{x + 1}$

$\Rightarrow \frac{1}{x (x - 1) (x + 1)} = - \frac{1}{x} + \frac{1}{2} (\frac{1}{x - 1} + \frac{1}{x + 1})$

Rewriting the integral(1):

$y = \int (- \frac{1}{x} + \frac{1}{2} (\frac{1}{x - 1} + \frac{1}{x + 1})) d x$

$\Rightarrow y = - \int \frac{1}{x} d x + \frac{1}{2} \int \frac{1}{x - 1} d x + \frac{1}{2} \int \frac{1}{x + 1} d x$

$\Rightarrow y = - \log x + \frac{1}{2} \log (x - 1) + \frac{1}{2} \log (x + 1) + \log C$

$\Rightarrow y = - \frac{2}{2} \log x + \frac{1}{2} \log (x - 1) + \frac{1}{2} \log (x + 1) + \frac{2}{2} \log C$

$\Rightarrow y = \frac{1}{2} (- \log x^{2} + \log (x - 1) + \log (x + 1) + \log C^{2})$

$\Rightarrow y = \frac{1}{2} \log [\frac{C^{2} (x^{2} - 1)}{x^{2}}]$

For $y = 0$ when $x=2$ .

$0 = \frac{1}{2} \log [\frac{C^{2} (2^{2} - 1)}{2^{2}}]$

$\Rightarrow 0 = \log [\frac{3 C^{2}}{4}]$

$\Rightarrow \frac{3 C^{2}}{4} = 1$

$\Rightarrow C^{2} = \frac{4}{3}$

Thus the required particular solution is :

$y = \frac{1}{2} \log [\frac{4 (x^{2} - 1)}{3 x^{2}}]$ .

13. Find the particular solution of $cos (\frac{dy}{dx}) =a (a \in R)$ ; $y=1$ when $x=0$ to satisfy the given condition.

Ans: The given differential equation is:

$\cos (\frac{d y}{d x}) = a (a \in R)$ ; $y = 1$ when $x = 0$

Simplify the expression:

$\cos (\frac{d y}{d x}) = a$

$\Rightarrow \frac{d y}{d x} = \cos^{- 1} a$

$\Rightarrow d y = \cos^{- 1} a d x$

Integrate both side:

$\int d y = \int \cos^{- 1} a d x$

$\Rightarrow y = \cos^{- 1} a \int d x$

$\Rightarrow y = x \cos^{- 1} a + C$

For $y = 1$ when $x=0$ .

$1 = 0 \cos^{- 1} a + C$

$\Rightarrow C = 1$

Thus the required particular solution is:

$y = x \cos^{- 1} a + 1$ .

$\Rightarrow \frac{y - 1}{x} = \cos^{- 1} a$

$\Rightarrow \cos (\frac{y - 1}{x}) = a$

14. Find the particular solution of $\frac{dy}{dx} =ytanx$ ; $y=1$ when $x=0$ to satisfy the given condition.

Ans: The given differential equation is:

$\frac{d y}{d x} = y \tan x$ ; $y = 1$ when $x = 0$

Simplify the expression:

$\frac{d y}{d x} = y \tan x$

$\Rightarrow \frac{d y}{y} = \tan x d x$

Integrate both side:

$\int \frac{d y}{y} = \int \tan x d x$

$\Rightarrow \log y = \log (\sec x) + \log C$

$\Rightarrow \log y = \log (C \sec x)$

$\Rightarrow y = C \sec x$

For $y=1$ when $x=0$ .

$1 = C \sec 0$

$\Rightarrow C = 1$

Thus the required particular solution is :

$y = \sec x$ .

15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is $y^{'} = e^{x} sin x$ .

Ans: The given differential equation is:

$y^{'} = e^{x} \sin x$

The curve passes through $(0, 0)$ .

Simplify the expression:

$\Rightarrow \frac{d y}{d x} = e^{x} \sin x$

$\Rightarrow d y = e^{x} \sin x d x$

Integrate both side:

$\int d y = \int e^{x} \sin x d x$

Use product rules for integration of RHS. Let:

$I = \int e^{x} \sin x d x$

$\Rightarrow I = \sin x \int e^{x} d x - \int (\cos x \int e^{x} d x) d x$

$\Rightarrow I = e^{x} \sin x - \int e^{x} \cos x d x$

$\Rightarrow I = e^{x} \sin x - (\cos x \int e^{x} d x + \int (\sin x \int e^{x} d x) d x)$

$\Rightarrow I = e^{x} \sin x - e^{x} \cos x - \int (e^{x} \sin x) d x$

$\Rightarrow I = e^{x} \sin x - e^{x} \cos x - I$

$\Rightarrow I = \frac{e^{x}}{2} (\sin x - \cos x)$

Thus integral will be:

$y = \frac{e^{x}}{2} (\sin x - \cos x) + C$

Thus as the curve passes through $(0,0)$

$0 = \frac{e^{0}}{2} (\sin 0 - \cos 0) + C$

$\Rightarrow 0 = \frac{1}{2} (0 - 1) + C$

$\Rightarrow C = \frac{1}{2}$

Thus the equation of the curve will be:

$y = \frac{e^{x}}{2} (\sin x - \cos x) + \frac{1}{2}$

$\Rightarrow y = \frac{e^{x}}{2} (\sin x - \cos x + 1)$

16. For the differential equation $xy \frac{dy}{dx} = (x+2) (y+2)$ find the solution curve passing through the point $(1,-1)$ .

Ans: The given differential equation is:

$x y \frac{d y}{d x} = (x + 2) (y + 2)$

The curve passes through $(1, - 1)$ .

Simplify the expression:

$\Rightarrow (\frac{y}{y + 2}) d y = \frac{(x + 2)}{x} d x$

$\Rightarrow (1 - \frac{2}{y + 2}) d y = \frac{(x + 2)}{x} d x$

Integrate both side:

$\int (1 - \frac{2}{y + 2}) d y = \int \frac{(x + 2)}{x} d x$

$\Rightarrow \int d y - 2 \int \frac{1}{y + 2} d y = \int \frac{x}{x} d x + \int \frac{2}{x} d x$

$\Rightarrow y - 2 \log (y + 2) = x + 2 \log x + C$

$\Rightarrow y - x = 2 \log (y + 2) + 2 \log x + C$

$\Rightarrow\Rightarrow y - x = 2 \log [x (y + 2)] + C$

$\Rightarrow y - x = \log [x^{2} {(y + 2)}^{2}] + C$

Thus as the curve passes through $(1,-1)$

$\Rightarrow - 1 - 1 = \log [{(1)}^{2} {(- 1 + 2)}^{2}] + C$

$\Rightarrow - 2 = \log 1 + C$

$\Rightarrow C = - 2$

Thus the equation of the curve will be:

$y - x = \log [x^{2} {(y + 2)}^{2}] - 2$

$\Rightarrow y - x + 2 = \log (x^{2} {(y + 2)}^{2})$

17. Find the equation of a curve passing through the point $(0,-2)$ given that at any point $(x,y)$ on the curve, the product of the slope of its tangent and $y$ -coordinate of the point is equal to the $x$ -coordinate of the point.

Ans: According to , the equation is given by:

$y \frac{d y}{d x} = x$

The curve passes through $(0, - 2)$ .

Simplify the expression:

$\Rightarrow y d y = x d x$

Integrate both side:

$\int y d y = \int x d x$

$\Rightarrow \frac{y^{2}}{2} = \frac{x^{2}}{2} + C$

$\Rightarrow y^{2} - x^{2} = 2 C$

Thus as the curve passes through $(0,-2)$

$\Rightarrow {(- 2)}^{2} - 0^{2} = 2 C$

$\Rightarrow 4 = 2 C$

$\Rightarrow C = 2$

Thus the equation of the curve will be:

$y^{2} - x^{2} = 2 (2)$

$\Rightarrow y^{2} - x^{2} = 4$ .

18. At any point $(x,y)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point $(-4,-3)$ . Find the equation of the curve given that it passes through $(-2,1)$ .

Ans: Let the point of contact of the tangent be $(x,y)$ .Then the slope of the segment joining point of contact and $(-4,-3)$ :

$m = \frac{y + 3}{x + 4}$

According to the for the slope of tangent $\frac{dy}{dx}$ it follows:

$\frac{d y}{d x} = 2 m$

$\Rightarrow \frac{d y}{d x} = 2 (\frac{y + 3}{x + 4})$

Simplify the expression:

$\frac{d y}{d x} = 2 (\frac{y + 3}{x + 4})$

$\frac{d y}{y + 3} = \frac{2}{x + 4} d x$

Integrate both side:

$\int \frac{d y}{y + 3} = \int \frac{2}{x + 4} d x$

$\Rightarrow \log (y + 3) = 2 \log (x + 4) + \log C$

$\Rightarrow \log (y + 3) = \log {(x + 4)}^{2} + \log C$

$\Rightarrow \log (y + 3) = \log C {(x + 4)}^{2}$

$\Rightarrow y + 3 = C {(x + 4)}^{2}$

Thus as the curve passes through $(-2,1)$

$1 + 3 = C {(- 2 + 4)}^{2}$

$\Rightarrow 4 = 4 C$

$\Rightarrow C = 1$

Thus the equation of the curve will be:

$y + 3 = {(x + 4)}^{2}$ .

19. The volume of spherical balloons being inflated changes at a constant rate. If initially its radius is $3$ units and after $3$ seconds it is $6$ units. Find the radius of balloon after $t$ seconds.

Ans: Let the volume of spherical balloon be $V$ and its radius $r$ . Let the rate of change of volume be $k$ .

$\frac{d V}{d t} = k$

$\Rightarrow \frac{d}{d t} (\frac{4}{3} π r^{3}) = k$

$\Rightarrow \frac{4}{3} π \frac{d}{d t} (r^{3}) = k$

$\Rightarrow \frac{4}{3} π (3 r^{2}) \frac{d r}{d t} = k$

$\Rightarrow 4 π r^{2} \frac{d r}{d t} = k$

$\Rightarrow 4 π r^{2} d r = k d t$

Integrate both side:
$\int 4 π r^{2} d r = \int k d t$

$\Rightarrow 4 π \int r^{2} d r = k t + C$

$\Rightarrow \frac{4}{3} π r^{3} = k t + C$

At initial time, $t=0$ and $r=3$ :

$\frac{4}{3} π 3^{3} = k (0) + C$

$\Rightarrow C = 36 π$

At $t=3$ the radius $r=6$ :

$\frac{4}{3} π (6^{3}) = k (3) + 36 π$

$\Rightarrow 3 k = 288 π - 36 π$

$\Rightarrow k = 84 π$

Thus the radius-time relation can be given by:

$\frac{4}{3} π r^{3} = 84 π t + 36 π$

$\Rightarrow r^{3} = 63 t + 27$

$\Rightarrow r = {(63 t + 27)}^{\frac{1}{3}}$

The radius of balloon after $t$ seconds is given by: $r = {(63 t + 27)}^{\frac{1}{3}}$ .

20. In a bank, principal increases continuously at the rate of $r$ % per year. Find the value of $r$ if Rs. $100$ doubles itself in $10$ years $(lo g_{e} 2=0 .6931)$ .

Ans: Let the principal be $p$ , according to :

$\frac{d p}{d t} = (\frac{r}{100}) p$

Simplify the expression:

$\frac{d p}{p} = (\frac{r}{100}) d t$

Integrate both side:
$\int \frac{d p}{p} = \int (\frac{r}{100}) d t$

$\Rightarrow \log p = \frac{r t}{100} + c$

$\Rightarrow p = e^{\frac{r t}{100} + c}$

$\Rightarrow p = A e^{\frac{r t}{100}} (A = e^{c})$

At $t=0$ , $p=100$ :

$100 = A e^{\frac{r (0)}{100}}$

$\Rightarrow A = 100$

Thus the principle and rate of interest relation:

$p = 100 e^{\frac{r t}{100}}$

At $t=10$ , $p=2 \times 100=200$ :

$200 = 100 e^{\frac{r (10)}{100}}$

$\Rightarrow 2 = e^{\frac{r}{10}}$

Take logarithm on both side:

$\log (e^{\frac{r}{10}}) = \log (2)$

$\Rightarrow \frac{r}{10} = 0.6931$

$\Rightarrow r = 6.931$

Thus the rate of interest $r=6 .931$ .

21. In a bank, principal increases continuously at the rate of $5$ % per year. An amount of Rs $1000$ is deposited with this bank, how much will it worth after $10$ years $(e^{0 .5} =1 .648)$ .

Ans: Let the principal be $p$ , according to principle increases at the rate of $5$ % per year.

$\frac{d p}{d t} = (\frac{5}{100}) p$

Simplify the expression:

$\frac{d p}{d t} = \frac{p}{20}$

$\Rightarrow \frac{d p}{p} = \frac{1}{20} d t$

Integrate both side:

$\int \frac{d p}{p} = \int \frac{1}{20} d t$

$\Rightarrow \log p = \frac{t}{20} + C$

$\Rightarrow p = e^{\frac{t}{20} + C}$

$\Rightarrow p = A e^{\frac{t}{20}} (A = e^{C})$
At $t=0$ , $p=1000$ :

$1000 = A e^{\frac{0}{20}}$

$\Rightarrow A = 1000$

Thus the relation of principal and time relation:

$\Rightarrow p = 1000 e^{\frac{t}{20}}$

At $t=10$ :

$p = 1000 e^{\frac{10}{20}}$

$\Rightarrow p = 1000 e^{0.5}$

$\Rightarrow p = 1000 \times 1.648$

$\Rightarrow p = 1648$

Thus after $10$ this year the amount will become Rs. $1648$ .

22. In a culture, the bacteria count is $100000$ . The number is increased by $10$ in $2$ hours. In how many hours will the count reach $200000$ , if the rate of growth of bacteria is proportional to the number present?

Ans: Let the number of bacteria be $y$ at time $t$ . According to :

$\frac{d y}{d t} \propto y$

$\frac{d y}{d t} = c y$

Here $c$ is constant.

Simplify the expression:
$\frac{d y}{y} = c d t$

Integrate both side:
$\int \frac{d y}{y} = \int c d t$

$\Rightarrow \log y = c t + D$

$\Rightarrow y = e^{c t + D}$

$\Rightarrow y = A e^{c t} (A = e^{D})$

At $t=0$ , $y=100000$ :

$100000 = A e^{c (0)}$

$\Rightarrow A = 100000$

At $t=2$ , $y= \frac{11}{10} (100000) = 110000$ :

$y = 100000 e^{c t}$

$\Rightarrow 110000 = 100000 e^{c (2)}$

$\Rightarrow e^{2 c} = \frac{11}{10}$

$\Rightarrow 2 c = \log (\frac{11}{10})$

$\Rightarrow c = \frac{1}{2} \log (\frac{11}{10})$ ……(1)

For $y=200000$ :

$200000 = 100000 e^{c t}$

$\Rightarrow e^{c t} = 2$

$\Rightarrow c t = \log 2$

$\Rightarrow t = \frac{\log 2}{c}$

Back substituting using expression (1):
$t = \frac{\log 2}{\frac{1}{2} \log (\frac{11}{10})}$

$\Rightarrow t = \frac{2 \log 2}{\log (\frac{11}{10})}$

Thus time required for bacteria to reach $200000$ is $t= \frac{log2}{\frac{1}{2} log (\frac{11}{10})}$ hrs.

23. Find the general solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ .

(A) $e^{x} + e^{-y} =C$

(B) $e^{x} + e^{y} =C$

(D) $e^{-x} + e^{-y} =C$

Ans: The given differential equation is $\frac{dy}{dx} = e^{x+y}$ . Simplify the expression:

$\frac{d y}{d x} = e^{x} e^{y}$

$\Rightarrow \frac{d y}{e^{y}} = e^{x} d x$

$\Rightarrow e^{- y} d y = e^{x} d x$

Integrate both side:

$\int e^{- y} d y = \int e^{x} d x$

$\Rightarrow - e^{- y} = e^{x} + D$

$\Rightarrow e^{x} + e^{- y} = - D$

$\Rightarrow e^{x} + e^{- y} = C (C = - D)$

Thus the general solution of given differential equation is $e^{x} + e^{- y} = C$

Thus the correct option is (A).

Exercise 9.4

1. Show that, differential equation $(x^{2} +xy) dy= (x^{2} + y^{2}) dx$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{d y}{d x} = \frac{x^{2} + y^{2}}{x^{2} + x y}$

Checking for homogeneity:

$F (x, y) = \frac{x^{2} + y^{2}}{x^{2} + x y}$

$\Rightarrow F (λ x, λ y) = \frac{{(λ x)}^{2} + {(λ y)}^{2}}{{(λ x)}^{2} + (λ x) (λ y)}$

$\Rightarrow F (λ x, λ y) = \frac{λ^{2} (x^{2} + y^{2})}{λ^{2} (x^{2} + x y)}$

$\Rightarrow F (λ x, λ y) = \frac{x^{2} + y^{2}}{x^{2} + x y}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = \frac{x^{2} + {(v x)}^{2}}{x^{2} + x (v x)}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = \frac{x^{2} (1 + v^{2})}{x^{2} (1 + v)}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{1 + v^{2}}{1 + v}$

$\Rightarrow x \frac{d v}{d x} = \frac{1 + v^{2}}{1 + v} - v$

$\Rightarrow x \frac{d v}{d x} = \frac{1 + v^{2} - v - v^{2}}{1 + v}$

$\Rightarrow x \frac{d v}{d x} = \frac{1 - v}{1 + v}$

Separate the differentials:

$\frac{1 + v}{1 - v} d v = \frac{d x}{x}$

Integrate both side:

$\int \frac{1 + v}{1 - v} d v = \int \frac{d x}{x}$

$\Rightarrow \int \frac{2 - (1 - v)}{1 - v} d v = \log x - \log k$

$\Rightarrow \int \frac{2}{1 - v} d v - \int \frac{1 - v}{1 - v} d v = \log \frac{x}{k}$

$\Rightarrow - 2 \log (1 - v) - \int d v = \log \frac{x}{k}$

$\Rightarrow - 2 \log (1 - v) - v = \log \frac{x}{k}$

$\Rightarrow v = - \log \frac{x}{k} - 2 \log (1 - v)$

$\Rightarrow v = \log (\frac{k}{x {(1 - v)}^{2}})$

Back substitute $v= \frac{y}{x}$ :

$\Rightarrow \frac{y}{x} = \log (\frac{k}{x {(1 - \frac{y}{x})}^{2}})$

$\Rightarrow \frac{y}{x} = \log (\frac{k x}{{(x - y)}^{2}})$

$\Rightarrow e^{\frac{y}{x}} = \frac{k x}{{(x - y)}^{2}}$

$\Rightarrow {(x - y)}^{2} = k x e^{- \frac{y}{x}}$

The solution of the given differential equation ${(x-y)}^{2} =kx e^{- \frac{y}{x}}$ .

2. Show that, differential equation $y^{'} = \frac{x+y}{x}$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{d y}{d x} = \frac{x + y}{x}$

Checking for homogeneity:

. $F (x, y) = \frac{x + y}{x}$

$\Rightarrow F (λ x, λ y) = \frac{λ x + λ y}{λ x}$

$\Rightarrow F (λ x, λ y) = \frac{λ (x + y)}{λ (x)}$

$\Rightarrow F (λ x, λ y) = \frac{x + y}{x}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = \frac{x + (v x)}{x}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = \frac{x (1 + v)}{x}$

$\Rightarrow v + x \frac{d v}{d x} = 1 + v$

$\Rightarrow x \frac{d v}{d x} = 1$

Separate the differentials:

$d v = \frac{d x}{x}$

Integrate both side:

$\int d v = \int \frac{d x}{x}$

$\Rightarrow \int d v = \log x + \log k$

$\Rightarrow v = \log k x$

Back substitute $v= \frac{y}{x}$ :

$\Rightarrow \frac{y}{x} = \log k x$

$\Rightarrow y = x \log k x$

The solution of the given differential equation $y=x log kx$

3. Show that, differential equation $(x-y) dy- (x+y) dx=0$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{d y}{d x} = \frac{x + y}{x - y}$

Checking for homogeneity:

$F (x, y) = \frac{x + y}{x - y}$

$\Rightarrow F (λ x, λ y) = \frac{λ x + λ y}{λ x - λ y}$

$\Rightarrow F (λ x, λ y) = \frac{λ (x + y)}{λ (x - y)}$

$\Rightarrow F (λ x, λ y) = \frac{x + y}{x - y}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = \frac{x + (v x)}{x - v x}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = \frac{x (1 + v)}{x (1 - v)}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{1 + v}{1 - v}$

$\Rightarrow x \frac{d v}{d x} = \frac{1 + v}{1 - v} - v$

$\Rightarrow x \frac{d v}{d x} = \frac{1 + v - v + v^{2}}{1 - v}$

$\Rightarrow x \frac{d v}{d x} = \frac{1 + v^{2}}{1 - v}$

Separate the differentials:

$\frac{1 - v}{1 + v^{2}} d v = \frac{d x}{x}$

Integrate both side:

$\int \frac{1 - v}{1 + v^{2}} d v = \int \frac{d x}{x}$

$\Rightarrow \int \frac{1 - v}{1 + v^{2}} d v = \int \frac{d x}{x}$

$\Rightarrow \int \frac{1}{1 + v^{2}} d v - \int \frac{v}{1 + v^{2}} d v = \log x + C$

$\Rightarrow \tan^{- 1} v - \frac{1}{2} \int \frac{2 v}{1 + v^{2}} d v = \log x + C$

$\Rightarrow \tan^{- 1} v - \frac{1}{2} \log (1 + v^{2}) = \log x + C$

$\Rightarrow \tan^{- 1} v = \log x + \frac{1}{2} \log (1 + v^{2}) + C$

$\Rightarrow \tan^{- 1} v = \frac{1}{2} \log x^{2} + \frac{1}{2} \log (1 + v^{2}) + C$

$\Rightarrow \tan^{- 1} v = \frac{1}{2} \log [x^{2} (1 + v^{2})] + C$

Back substitute $v= \frac{y}{x}$ :

$\Rightarrow \tan^{- 1} \frac{y}{x} = \frac{1}{2} \log [x^{2} (1 + \frac{y^{2}}{x^{2}})] + C$

$\Rightarrow \tan^{- 1} \frac{y}{x} = \frac{1}{2} \log (x^{2} + y^{2}) + C$

The solution of the given differential equation $ta n^{-1} \frac{y}{x} = \frac{1}{2} log (x^{2} + y^{2}) +C$ .

4. Show that, differential equation $(x^{2} - y^{2}) dx+2xy dy=0$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{d y}{d x} = - \frac{x^{2} - y^{2}}{2 x y}$

Checking for homogeneity:

$F (x, y) = - \frac{x^{2} - y^{2}}{2 x y}$

$\Rightarrow F (λ x, λ y) = - \frac{{(λ x)}^{2} - {(λ y)}^{2}}{2 (λ x) (λ y)}$

$\Rightarrow F (λ x, λ y) = - \frac{λ^{2} x^{2} - λ^{2} y^{2}}{2 λ^{2} x y}$

$\Rightarrow F (λ x, λ y) = - \frac{λ^{2} (x^{2} - y^{2})}{λ^{2} (2 x y)}$

$\Rightarrow F (λ x, λ y) = - \frac{x^{2} - y^{2}}{2 x y}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = - \frac{x^{2} - {(v x)}^{2}}{2 x (v x)}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = \frac{x^{2} (v^{2} - 1)}{x^{2} (2 v)}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{v^{2} - 1}{2 v}$

$\Rightarrow x \frac{d v}{d x} = \frac{v^{2} - 1}{2 v} - v$

$\Rightarrow x \frac{d v}{d x} = \frac{v^{2} - 1 - 2 v^{2}}{2 v}$

$\Rightarrow x \frac{d v}{d x} = - \frac{1 + v^{2}}{2 v}$

Separate the differentials:

$\frac{2 v}{1 + v^{2}} d v = - \frac{d x}{x}$

Integrate both side:

$\int \frac{2 v}{1 + v^{2}} d v = - \int \frac{d x}{x}$

$\Rightarrow \log (1 + v^{2}) = - \log x + C$

$\Rightarrow \log (1 + v^{2}) + \log x = C$

$\Rightarrow \log [x (1 + v^{2})] = C$

Back substitute $v= \frac{y}{x}$ :

$\Rightarrow \log [x (1 + \frac{y^{2}}{x^{2}})] = C$

$\Rightarrow (\frac{x^{2} + y^{2}}{x}) = k k = e^{C}$

$\Rightarrow x^{2} + y^{2} = k x$

The solution of the given differential equation $x^{2} + y^{2} =kx$ .

5. Show that, differential equation $x^{2} \frac{dy}{dx} = x^{2} -2 y^{2} +xy$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$\frac{d y}{d x} = \frac{x^{2} - 2 y^{2} + x y}{x^{2}}$

Checking for homogeneity:

$F (x, y) = \frac{x^{2} - 2 y^{2} + x y}{x^{2}}$

$\Rightarrow F (λ x, λ y) = \frac{λ^{2} x^{2} - 2 λ^{2} y^{2} + (λ x) (λ y)}{λ^{2} x^{2}}$

$\Rightarrow F (λ x, λ y) = \frac{λ^{2} (x^{2} - 2 y^{2} + x y)}{λ^{2} (x^{2})}$

$\Rightarrow F (λ x, λ y) = \frac{x^{2} - 2 y^{2} + x y}{x^{2}}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = \frac{x^{2} - 2 {(v x)}^{2} + x (v x)}{x^{2}}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = \frac{x^{2} (1 - 2 v^{2} + v)}{x^{2}}$

$\Rightarrow v + x \frac{d v}{d x} = 1 - 2 v^{2} + v$

$\Rightarrow x \frac{d v}{d x} = 1 - 2 v^{2}$

Separate the differentials:

$\frac{1}{1 - 2 v^{2}} d v = \frac{d x}{x}$

Integrate both side:

$\int \frac{1}{1 - 2 v^{2}} d v = \int \frac{d x}{x}$

$\Rightarrow \frac{1}{2} \int \frac{1}{\frac{1}{2} - v^{2}} d v = \log x + C$

$\Rightarrow \frac{1}{2} \int \frac{1}{{(\frac{1}{\sqrt{2}})}^{2} - v^{2}} d v = \log x + C$

$\Rightarrow \frac{1}{2} (\frac{1}{2 \times \frac{1}{\sqrt{2}}}) \log \frac{| \frac{1}{\sqrt{2}} + v |}{| \frac{1}{\sqrt{2}} - v |} = \log x + C$

$\Rightarrow \frac{1}{2 \sqrt{2}} \log \frac{| 1 + \sqrt{2} v |}{| 1 - \sqrt{2} v |} = \log x + C$

Back substitute $v= \frac{y}{x}$ :s

$\Rightarrow \frac{1}{2 \sqrt{2}} \log \frac{| 1 + \sqrt{2} (\frac{y}{x}) |}{| 1 - \sqrt{2} (\frac{y}{x}) |} = \log x + C$

$\Rightarrow \frac{1}{2 \sqrt{2}} \log \frac{| x + \sqrt{2} y |}{| x - \sqrt{2} y |} = \log x + C$

The solution of the given differential equation $\frac{1}{2 \sqrt{2}} log \frac{| x+ \sqrt{2} y |}{| x- \sqrt{2} y |} =logx+C$ .

6. Show that, differential equation $xdy-ydx= \sqrt{x^{2} + y^{2}} dx$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$x d y = \sqrt{x^{2} + y^{2}} d x + y d x$

$\Rightarrow \frac{d y}{d x} = \frac{\sqrt{x^{2} + y^{2}} + y}{x}$

Checking for homogeneity:

$F (x, y) = \frac{\sqrt{x^{2} + y^{2}} + y}{x}$

$\Rightarrow F (λ x, λ y) = \frac{\sqrt{λ^{2} x^{2} + λ^{2} y^{2}} + λ y}{λ x}$

$\Rightarrow F (λ x, λ y) = \frac{\sqrt{λ^{2} (x^{2} + y^{2})} + λ y}{λ x}$

$\Rightarrow F (λ x, λ y) = \frac{λ (\sqrt{x^{2} + y^{2}} + y)}{λ (x)}$

$\Rightarrow F (λ x, λ y) = \frac{\sqrt{x^{2} + y^{2}} + y}{x}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = \frac{\sqrt{x^{2} + {(v x)}^{2}} + v x}{x}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = \frac{x (\sqrt{1 + v^{2}} + v)}{x}$

$\Rightarrow v + x \frac{d v}{d x} = \sqrt{1 + v^{2}} + v$

$\Rightarrow x \frac{d v}{d x} = \sqrt{1 + v^{2}}$

Separate the differentials:

$\frac{1}{\sqrt{1 + v^{2}}} d v = \frac{d x}{x}$

Integrate both side:

$\int \frac{1}{\sqrt{1 + v^{2}}} d v = \int \frac{d x}{x}$

$\Rightarrow \log | v + \sqrt{1 + v^{2}} | = \log x + \log C$

Back substitute $v= \frac{y}{x}$ :s

$\Rightarrow \log | \frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}} | = \log C x$

$\Rightarrow \log | \frac{y + \sqrt{y^{2} + x^{2}}}{x} | = \log C x$

$y + \sqrt{y^{2} + x^{2}} = C x^{2}$

7. Show that, differential equation ${xcos (\frac{y}{x}) +ysin (\frac{y}{x})} ydx= {ysin (\frac{y}{x}) -xcos (\frac{y}{x})} xdy$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

${x \cos (\frac{y}{x}) + y \sin (\frac{y}{x})} y d x = {y \sin (\frac{y}{x}) - x \cos (\frac{y}{x})} x d y$

$\frac{d y}{d x} = \frac{{x \cos (\frac{y}{x}) + y \sin (\frac{y}{x})} y}{{y \sin (\frac{y}{x}) - x \cos (\frac{y}{x})} x}$

Checking for homogeneity:

$F (x, y) = \frac{{x \cos (\frac{y}{x}) + y \sin (\frac{y}{x})} y}{{y \sin (\frac{y}{x}) - x \cos (\frac{y}{x})} x}$

$\Rightarrow F (λ x, λ y) = \frac{{λ x \cos (\frac{λ y}{λ x}) + y \sin (\frac{λ y}{λ x})} λ y}{{λ y \sin (\frac{λ y}{λ x}) - λ x \cos (\frac{λ y}{λ x})} λ x}$

$\Rightarrow F (λ x, λ y) = \frac{λ^{2} {x \cos (\frac{y}{x}) + y \sin (\frac{y}{x})} y}{λ^{2} {y \sin (\frac{y}{x}) - x \cos (\frac{y}{x})} x}$

$\Rightarrow F (λ x, λ y) = \frac{{x \cos (\frac{y}{x}) + y \sin (\frac{y}{x})} y}{{y \sin (\frac{y}{x}) - x \cos (\frac{y}{x})} x}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = \frac{{x \cos (\frac{x v}{x}) + v x \sin (\frac{v x}{x})} v x}{{v x \sin (\frac{v x}{x}) - x \cos (\frac{v x}{x})} x}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = \frac{x^{2} {\cos v + v \sin v} v}{x^{2} {v \sin v - \cos v}}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{v \cos v + v^{2} \sin v}{v \sin v - \cos v}$

$\Rightarrow x \frac{d v}{d x} = \frac{v \cos v + v^{2} \sin v}{v \sin v - \cos v} - v$

$\Rightarrow x \frac{d v}{d x} = \frac{v \cos v + v^{2} \sin v - v^{2} \sin v + v \cos v}{v \sin v - \cos v}$

$\Rightarrow x \frac{d v}{d x} = \frac{2 v \cos v}{v \sin v - \cos v}$

Separate the differentials:

$\frac{v \sin v - \cos v}{v \cos v} d v = 2 \frac{d x}{x}$

Integrate both side:

$\int \frac{v \sin v - \cos v}{v \cos v} d v = 2 \int \frac{d x}{x}$

$\Rightarrow \int \frac{v \sin v}{v \cos v} d v - \int \frac{\cos v}{v \cos v} d v = 2 \int \frac{d x}{x}$

$\Rightarrow \int \tan v d v - \int \frac{1}{v} d v = 2 \log | x | + \log C$

$\Rightarrow \log | \sec v | - \log | v | = \log C {| x |}^{2}$

$\Rightarrow \log \frac{| \sec v |}{| v |} = \log C {| x |}^{2}$

$\Rightarrow \sec v = C v x^{2}$

Back substitute $v= \frac{y}{x}$ :s

$\Rightarrow \sec \frac{y}{x} = C (\frac{y}{x}) x^{2}$

$\Rightarrow \cos \frac{y}{x} = \frac{k}{x y} k = \frac{1}{C}$

The solution of the given differential equation $cos \frac{y}{x} = \frac{k}{xy}$ .

8. Show that, differential equation $x \frac{dy}{dx} -y+xsin (\frac{y}{x}) =0$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$x \frac{d y}{d x} - y + x \sin (\frac{y}{x}) = 0$

$\frac{d y}{d x} = \frac{y - x \sin (\frac{y}{x})}{x}$

Checking for homogeneity:

$F (x, y) = \frac{y - x \sin (\frac{y}{x})}{x}$

$\Rightarrow F (λ x, λ y) = \frac{λ y - λ x \sin (\frac{λ y}{λ x})}{λ x}$

$\Rightarrow F (λ x, λ y) = \frac{λ (y - x \sin (\frac{y}{x}))}{λ x}$

$\Rightarrow F (λ x, λ y) = \frac{y - x \sin (\frac{y}{x})}{x}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = \frac{v x - x \sin (\frac{v x}{x})}{x}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = \frac{x (v - \sin v)}{x}$

$\Rightarrow v + x \frac{d v}{d x} = v - \sin v$

$\Rightarrow x \frac{d v}{d x} = - \sin v$

Separate the differentials:

$\frac{1}{\sin v} d v = - \frac{d x}{x}$

Integrate both side:

$\int cosec v d v = - \int \frac{d x}{x}$

$\Rightarrow \log | cosec v - \cot v | = - \log x + \log C$

$\Rightarrow \log | cosec v - \cot v | = \log \frac{C}{x}$

$\Rightarrow cosec v - \cot v = \frac{C}{x}$

$\Rightarrow \frac{1}{\sin v} - \frac{\cos v}{\sin v} = \frac{C}{x}$

$\Rightarrow 1 - \cos v = \frac{C}{x} \sin v$

Back substitute $v= \frac{y}{x}$ :

$\Rightarrow 1 - \cos \frac{y}{x} = \frac{C}{x} \sin \frac{y}{x}$

$\Rightarrow x (1 - \cos \frac{y}{x}) = C \sin (\frac{y}{x})$

The solution of the given differential equation $x (1-cos \frac{y}{x}) =Csin (\frac{y}{x})$ .

9. Show that, differential equation $ydx+xlog (\frac{y}{x}) dy-2xdy=0$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$y d x = 2 x d y - x \log (\frac{y}{x}) d y$

$\frac{d y}{d x} = \frac{y}{2 x - x \log (\frac{y}{x})}$

Checking for homogeneity:

$F (x, y) = \frac{y}{2 x - x \log (\frac{y}{x})}$

$\Rightarrow F (λ x, λ y) = \frac{λ y}{2 λ x - λ x \log (\frac{λ y}{λ x})}$

$\Rightarrow F (λ x, λ y) = \frac{λ y}{λ (2 x - x \log (\frac{y}{x}))}$

$\Rightarrow F (λ x, λ y) = \frac{y}{2 x - x \log (\frac{y}{x})}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = \frac{v x}{2 x - x \log (\frac{v x}{x})}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = \frac{x v}{x (2 - \log (v))}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{v}{2 - \log (v)}$

$\Rightarrow x \frac{d v}{d x} = \frac{v}{2 - \log v} - v$

$\Rightarrow x \frac{d v}{d x} = \frac{v - 2 v + v \log v}{2 - \log v}$

$\Rightarrow x \frac{d v}{d x} = \frac{v \log v - v}{2 - \log v}$

Separate the differentials:

$\frac{2 - \log v}{v \log v - v} d v = \frac{d x}{x}$

Integrate both side:

$\int \frac{2 - \log v}{v \log v - v} d v = \int \frac{d x}{x}$

$\int \frac{1 + 1 - \log v}{v (\log v - v)} d v = \log x + \log C$

$\Rightarrow \int \frac{1}{v (\log v - 1)} d v + \int \frac{1 - \log v}{v (\log v - 1)} d v = \log x + \log C$

$\Rightarrow \int \frac{1}{v (\log v - 1)} d v - \int \frac{1}{v} d v = \log C x$ ……(1)

Solving :

$\int \frac{1}{v (\log v - 1)} d v$

Substitute $log v-1=t$ :

$\log v - 1 = t$

$\Rightarrow \frac{1}{v} d v = d t$

Thus the integral will be:

$\Rightarrow \int \frac{1}{v (\log v - 1)} d v = \int \frac{d t}{t}$

$\Rightarrow \int \frac{1}{v (\log v - 1)} d v = \log t$

$\Rightarrow \int \frac{1}{v (\log v - 1)} d v = \log (\log v - 1)$

Using above result for solving (1):

$\Rightarrow \log (\log v - 1) - \log v = \log C x$

$\Rightarrow \log \frac{\log v - 1}{v} = \log C x$

$\Rightarrow \frac{\log v - 1}{v} = C x$

Back substitute $v= \frac{y}{x}$ :

$\Rightarrow \frac{\log \frac{y}{x} - 1}{\frac{y}{x}} = C x$

$\Rightarrow \log \frac{y}{x} - 1 = C x (\frac{y}{x})$

$\Rightarrow \log \frac{y}{x} - 1 = C y$

The solution of the given differential equation $log \frac{y}{x} -1=Cy$ .

10. Show that, differential equation $(1+ e^{\frac{x}{y}}) dx+ e^{\frac{x}{y}} (1- \frac{x}{y}) dy=0$ is homogenous and solves it.

Ans: Rewrite the equation in standard form:

$(1 + e^{\frac{x}{y}}) d x = - e^{\frac{x}{y}} (1 - \frac{x}{y}) d y$

$\frac{d x}{d y} = - \frac{e^{\frac{x}{y}} (1 - \frac{x}{y})}{(1 + e^{\frac{x}{y}})}$

Checking for homogeneity:

$F (x, y) = - \frac{e^{\frac{x}{y}} (1 - \frac{x}{y})}{(1 + e^{\frac{x}{y}})}$

$\Rightarrow F (λ x, λ y) = - \frac{e^{\frac{λ x}{λ y}} (1 - \frac{λ x}{λ y})}{(1 + e^{\frac{λ x}{λ y}})}$

$\Rightarrow F (λ x, λ y) = - \frac{e^{\frac{x}{y}} (1 - \frac{x}{y})}{(1 + e^{\frac{x}{y}})}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $x=vy$ :

$\frac{d (v y)}{d y} = - \frac{e^{\frac{v y}{y}} (1 - \frac{v y}{y})}{1 + e^{\frac{v y}{y}}}$

$\Rightarrow v \frac{d y}{d y} + y \frac{d v}{d y} = - \frac{e^{v} (1 - v)}{1 + e^{v}}$

$\Rightarrow v + y \frac{d v}{d y} = - \frac{e^{v} (1 - v)}{1 + e^{v}}$

$\Rightarrow y \frac{d v}{d y} = - \frac{e^{v} (1 - v)}{1 + e^{v}} - v$

$\Rightarrow y \frac{d v}{d y} = \frac{- e^{v} + v e^{v} - v - v e^{v}}{1 + e^{v}}$

$\Rightarrow y \frac{d v}{d y} = \frac{- (e^{v} + v)}{1 + e^{v}}$

Separate the differentials:

$\frac{1 + e^{v}}{e^{v} + v} d v = - \frac{d y}{y}$

Integrate both side:

$\int \frac{1 + e^{v}}{e^{v} + v} d v = - \int \frac{d y}{y}$

$\int \frac{e^{v} + 1}{e^{v} + v} d v = - \log y + \log C$ ……(1)

Solving the LHS integral. Substitute $e^{v} +v=t$ :

$e^{v} + v = t$

$\Rightarrow (e^{v} + 1) d v = d t$

Solving the expression (1):

$\Rightarrow \int \frac{1}{t} d t = \log \frac{C}{y}$

$\Rightarrow \log (t) = \log \frac{C}{y}$

$\Rightarrow \log (e^{v} + v) = \log \frac{C}{y}$

$\Rightarrow e^{v} + v = \frac{C}{y}$

Back substitute $v= \frac{x}{y}$ :

$\Rightarrow e^{\frac{x}{y}} + \frac{x}{y} = \frac{C}{y}$

$\Rightarrow y e^{\frac{x}{y}} + x = C$

The solution of the given differential equation $y e^{\frac{x}{y}} +x=C$ .

11. For the differential equation $(x+y) dy+ (x-y) dx=0$ .Find the particular solution for the condition $y=1$ when $x=1$ .

Ans: Given differential equation is:

$(x + y) d y + (x - y) d x = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{x - y}{x + y}$

Checking for homogeneity:

$F (x, y) = - \frac{x - y}{x + y}$

$\Rightarrow F (λ x, λ y) = - \frac{λ x - λ y}{λ x + λ y}$

$\Rightarrow F (λ x, λ y) = - \frac{λ (x - y)}{λ (x + y)}$

$\Rightarrow F (λ x, λ y) = - \frac{x - y}{x + y}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = - \frac{x - (v x)}{x + (v x)}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = \frac{x (v - 1)}{x (v + 1)}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{v - 1}{v + 1}$

$\Rightarrow x \frac{d v}{d x} = \frac{v - 1}{v + 1} - v$

$\Rightarrow x \frac{d v}{d x} = \frac{v - 1 - v^{2} - v}{v + 1}$

$\Rightarrow x \frac{d v}{d x} = - \frac{1 + v^{2}}{v + 1}$

Separate the differentials:

$\frac{v + 1}{1 + v^{2}} d v = - \frac{d x}{x}$

Integrate both side:

$\int \frac{v + 1}{1 + v^{2}} d v = - \int \frac{d x}{x}$

$\Rightarrow \int \frac{v}{1 + v^{2}} d v + \int \frac{1}{1 + v^{2}} d v = - \log x + k$

$\Rightarrow \frac{1}{2} \log (1 + v^{2}) + \tan^{- 1} v + \log x = k$

$\Rightarrow \frac{1}{2} \log [x (1 + v^{2})] + \tan^{- 1} v = k$

Back substitute $v= \frac{y}{x}$ :

$\Rightarrow \frac{1}{2} \log [x (1 + \frac{y^{2}}{x^{2}})] + \tan^{- 1} \frac{y}{x} = k$

$\Rightarrow \frac{1}{2} \log [\frac{x^{2} + y^{2}}{x}] + \tan^{- 1} \frac{y}{x} = k$

Now $y=1$ and $x=1$ :

$\Rightarrow \frac{1}{2} \log [\frac{1^{2} + 1^{2}}{1}] + \tan^{- 1} \frac{1}{1} = k$

$k = \frac{1}{2} \log 2 + \frac{π}{4}$

The required particular solution:

$\Rightarrow \frac{1}{2} \log [\frac{x^{2} + y^{2}}{x}] + \tan^{- 1} \frac{y}{x} = \frac{1}{2} \log 2 + \frac{π}{4}$

12. For the differential equation $x^{2} dy+ (xy+ y^{2}) dx=0$ .Find the particular solution for the condition $y=1$ when $x=1$ .

Ans: Given differential equation is:

$x^{2} d y + (x y + y^{2}) d x = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{x y + y^{2}}{x^{2}}$

Checking for homogeneity:

$F (x, y) = - \frac{x y + y^{2}}{x^{2}}$

$\Rightarrow F (λ x, λ y) = - \frac{(λ x) (λ y) + λ^{2} y^{2}}{λ^{2} x^{2}}$

$\Rightarrow F (λ x, λ y) = - \frac{λ^{2} (x y + y^{2})}{λ^{2} x^{2}}$

$\Rightarrow F (λ x, λ y) = - \frac{x y + y^{2}}{x^{2}}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = - \frac{x (v x) + {(v x)}^{2}}{x^{2}}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = - \frac{v x^{2} + v^{2} x^{2}}{x^{2}}$

$\Rightarrow v + x \frac{d v}{d x} = - \frac{x^{2} (v + v^{2})}{x^{2}}$

$\Rightarrow v + x \frac{d v}{d x} = - v - v^{2}$

$\Rightarrow x \frac{d v}{d x} = - v^{2} - 2 v$

Separate the differentials:

$\frac{1}{v^{2} + 2 v} d v = - \frac{d x}{x}$

Integrate both side:

$\int \frac{1}{v^{2} + 2 v} d v = - \int \frac{d x}{x}$

$\Rightarrow \frac{1}{2} \int \frac{v + 2 - v}{v (v + 2)} d v = - \log x + \log C$

$\Rightarrow \frac{1}{2} \int \frac{v + 2}{v (v + 2)} d v - \frac{1}{2} \int \frac{v}{v (v + 2)} d v = - \log x + \log C$

$\Rightarrow \frac{1}{2} \int \frac{1}{v} d v - \frac{1}{2} \int \frac{1}{v + 2} d v = - \log x + \log C$

$\Rightarrow \frac{1}{2} \log v - \frac{1}{2} \log (v + 2) = \log \frac{C}{x}$

$\Rightarrow \frac{1}{2} \log \frac{v}{v + 2} = \log \frac{C}{x}$

$\Rightarrow \frac{v}{v + 2} = {(\frac{C}{x})}^{2}$

$\Rightarrow \frac{v}{v + 2} = \frac{C^{2}}{x^{2}}$

Back substitute $v= \frac{y}{x}$ :

$\Rightarrow \frac{\frac{y}{x}}{\frac{y}{x} + 2} = \frac{C^{2}}{x^{2}}$

$\Rightarrow \frac{x^{2} y}{y + 2 x} = C^{2}$

Now $y=1$ and $x=1$ :

$\Rightarrow \frac{1^{2} (1)}{1 + 2 (1)} = C^{2}$

$\Rightarrow C^{2} = \frac{1}{3}$

The required particular solution:

$\Rightarrow \frac{x^{2} y}{y + 2 x} = \frac{1}{3}$

$\Rightarrow y + 2 x = 3 x^{2} y$

13. For the differential equation $[xsi n^{2} (\frac{x}{y}) -y] dx+xdy=0$ .Find the particular solution for the condition $y= \frac{π}{4}$ when $x=1$ .

Ans: Given differential equation is:

$[x \sin^{2} (\frac{x}{y}) - y] d x + x d y = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{[x \sin^{2} (\frac{x}{y}) - y]}{x}$

Checking for homogeneity:

$F (x, y) = - \frac{x \sin^{2} (\frac{x}{y}) - y}{x}$

$\Rightarrow F (λ x, λ y) = - \frac{λ x \sin^{2} (\frac{λ x}{λ y}) - λ y}{λ x}$

$\Rightarrow F (λ x, λ y) = - \frac{λ (x \sin^{2} (\frac{y}{x}) - y)}{λ x}$

$\Rightarrow F (λ x, λ y) = - \frac{x \sin^{2} (\frac{y}{x}) - y}{x}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = - \frac{x \sin^{2} (\frac{v x}{x}) - v x}{x}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = \frac{- x \sin^{2} (v) + v x}{x}$

$\Rightarrow v + x \frac{d v}{d x} = - \sin^{2} v + v$

$\Rightarrow x \frac{d v}{d x} = - \sin^{2} v$

Separate the differentials:

$cose c^{2} v d v = - \frac{d x}{x}$

Integrate both side:

$\int cose c^{2} v d v = - \int \frac{d x}{x}$

$\Rightarrow - \cot v = - \log x - \log C$

$\Rightarrow \cot v = \log C x$

Back substitute $v= \frac{y}{x}$ :

$\Rightarrow \cot \frac{y}{x} = \log C x$

Now $y= \frac{π}{4}$ and $x=1$ :

$\cot \frac{\frac{π}{4}}{1} = \log C (1)$

$\Rightarrow \log C = \cot \frac{π}{4}$

$\Rightarrow \log C = 1$

$\Rightarrow C = e$

The required particular solution:

$\Rightarrow \cot \frac{y}{x} = \log | e x |$

14. For the differential equation $\frac{dy}{dx} - \frac{y}{x} +cosec (\frac{y}{x}) =0$ .Find the particular solution for the condition $y=0$ when $x=1$ .

Ans: Given differential equation is:

$\frac{d y}{d x} - \frac{y}{x} + cosec (\frac{y}{x}) = 0$

$\Rightarrow \frac{d y}{d x} = \frac{y}{x} - cosec (\frac{y}{x})$

Checking for homogeneity:

$F (x, y) = \frac{y}{x} - cosec (\frac{y}{x})$

$\Rightarrow F (λ x, λ y) = \frac{λ y}{λ x} - cosec (\frac{λ y}{λ x})$

$\Rightarrow F (λ x, λ y) = \frac{y}{x} - cosec (\frac{y}{x})$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = v - cosec (v)$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = v - cosec (v)$

$\Rightarrow v + x \frac{d v}{d x} = v - cosec (v)$

$\Rightarrow x \frac{d v}{d x} = - cosec (v)$

Separate the differentials:

$\sin v d v = - \frac{d x}{x}$

Integrate both side:

$\int \sin v d v = - \int \frac{d x}{x}$

$\Rightarrow - \cos v = - \log | x | - \log C$

$\cos v = \log | C x |$

Back substitute $v= \frac{y}{x}$ :

$\Rightarrow \cos \frac{y}{x} = \log | C x |$

Now $y=0$ and $x=1$ :

$\Rightarrow \cos \frac{0}{1} = \log | C 1 |$

$\Rightarrow \log C = 1$

$\Rightarrow C = e$

The required particular solution:

$\Rightarrow \cos \frac{y}{x} = \log | e x |$ .

15. For the differential equation $2xy+ y^{2} -2 x^{2} \frac{dy}{dx} =0$ .Find the particular solution for the condition $y=2$ when $x=1$ .

Ans: Given differential equation is:

$2 x y + y^{2} - 2 x^{2} \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = \frac{2 x y + y^{2}}{2 x^{2}}$

Checking for homogeneity:

$F (x, y) = \frac{2 x y + y^{2}}{2 x^{2}}$

$\Rightarrow F (λ x, λ y) = \frac{2 (λ x) (λ y) + λ^{2} y^{2}}{2 λ^{2} x^{2}}$

$\Rightarrow F (λ x, λ y) = \frac{2 x y + y^{2}}{2 x^{2}}$

$\Rightarrow F (x, y) = F (λ x, λ y)$

Thus it is an homogenous equation.

Let $y=vx$ :

$\frac{d (v x)}{d x} = \frac{2 x (v x) + {(v x)}^{2}}{2 x^{2}}$

$\Rightarrow v \frac{d x}{d x} + x \frac{d v}{d x} = \frac{2 v + v^{2}}{2}$

$\Rightarrow v + x \frac{d v}{d x} = v + \frac{v^{2}}{2}$

$\Rightarrow x \frac{d v}{d x} = \frac{v^{2}}{2}$

Separate the differentials:

$\frac{d v}{v^{2}} = \frac{1}{2} (\frac{d x}{x})$

Integrate both side:

$2 \int \frac{d v}{v^{2}} = \int (\frac{d x}{x})$

$\Rightarrow \frac{v^{- 2 + 1}}{- 2 + 1} = \log | x | + C$

$- \frac{2}{v} = \log | x | + C$

Back substitute $v= \frac{y}{x}$ :

$\Rightarrow - \frac{2 x}{y} = \log | x | + C$

Now $y=2$ and $x=1$ :

$\Rightarrow - \frac{2 (1)}{2} = \log | 1 | + C$

$\Rightarrow C = - 1$

The required particular solution:

$\Rightarrow - \frac{2 x}{y} = \log | x | - 1$ .

$\Rightarrow y = \frac{2 x}{1 - \log | x |} (x \neq 0, e)$

16. What substitution should be used for solving homogeneous differential equation $\frac{dx}{dy} =h (\frac{x}{y})$ .

A. $y=vx$

B. $v=yx$

C. $x=vy$

D. $x=v$

Ans: The required substitution will be:

$\frac{x}{y} = v$

$\Rightarrow x = v y$

The correct answer is (C).

17. Which of the following equation is homogeneous:

$(4x+6y+5) dy- (3y+2x+4) dx=0$
$(xy) dx- (x^{3} + y^{3}) dy=0$
$(x^{3} +2 y^{2}) dx+2xydy=0$
$y^{2} dx+ (x^{2} -xy- y^{2}) dy=0$

Ans: For option (A):

$F (x, y) = \frac{3 y + 2 x + 4}{4 x + 6 y + 5}$

$F (λ x, λ y) = \frac{3 λ y + 2 λ x + 4}{4 λ x + 6 λ y + 5}$

$F (λ x, λ y) \neq F (x, y)$

For option (B):

$F (x, y) = \frac{x y}{x^{3} + y^{3}}$

$F (λ x, λ y) = \frac{(λ x) (λ y)}{{(λ x)}^{3} + {(λ y)}^{3}}$

$F (λ x, λ y) = \frac{x y}{λ (x + y)}$

$F (λ x, λ y) \neq F (x, y)$

For option (C):

$F (x, y) = - \frac{x^{3} + 2 y^{2}}{2 x y}$

$F (λ x, λ y) = - \frac{λ^{3} x^{3} + 2 λ^{2} y^{2}}{2 (λ x) (λ y)}$

$F (λ x, λ y) = - \frac{λ x^{3} + 2 y^{2}}{2 x y}$

$F (λ x, λ y) \neq F (x, y)$

For option (D):

$F (x, y) = - \frac{y^{2}}{x^{2} - x y - y^{2}}$

$F (λ x, λ y) = - \frac{λ^{2} y^{2}}{λ^{2} x^{2} - (λ x) (λ y) - λ^{2} y^{2}}$

$F (λ x, λ y) = - \frac{y^{2}}{x^{2} - x y - y^{2}}$

$F (λ x, λ y) = F (x, y)$

Thus the correct answer is option (D).

Exercise 9.5

1. Find the general solution for the differential equation $\frac{dy}{dx} +2y=sinx$ .

Ans: The given differential equation is:

$\frac{d y}{d x} + 2 y = \sin x$

It is a linear differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = 2$

$Q = \sin x$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{\int 2 d x}$

$\Rightarrow I . F = e^{2 x}$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$y e^{2 x} = \int \sin x (e^{2 x}) d x + C$

$\Rightarrow y e^{2 x} = I + C (I = \int \sin x (e^{2 x}) d x)$ ……(1)

$I = \int \sin x (e^{2 x}) d x$

$\Rightarrow I = (\sin x) \int e^{2 x} d x - \int ({(\sin x)}^{'} \int e^{2 x} d x) d x$

$\Rightarrow I = \frac{e^{2 x}}{2} \sin x - \int (\cos x (\frac{e^{2 x}}{2})) d x$

$\Rightarrow I = \frac{e^{2 x}}{2} \sin x - \frac{1}{2} [\cos x \int e^{2 x} d x - \int ({(\cos x)}^{'} (\int e^{2 x} d x)) d x]$

$\Rightarrow I = \frac{e^{2 x}}{2} \sin x - \frac{1}{2} [\frac{e^{2 x}}{2} \cos x + \frac{1}{2} \int e^{2 x} (\sin x) d x]$

$\Rightarrow I = \frac{e^{2 x}}{2} \sin x - \frac{1}{2} [\frac{e^{2 x}}{2} \cos x + \frac{1}{2} I]$

$\Rightarrow I = \frac{e^{2 x}}{2} \sin x - \frac{e^{2 x}}{4} \cos x - \frac{1}{4} I$

$\Rightarrow \frac{5}{4} I = \frac{e^{2 x}}{2} \sin x - \frac{e^{2 x}}{4} \cos x$

$\Rightarrow I = \frac{2 e^{2 x}}{5} \sin x - \frac{e^{2 x}}{5} \cos x$

$\Rightarrow I = \frac{e^{2 x}}{5} [2 \sin x - \cos x]$

Back substituting $I$ in expression (1):

$\Rightarrow y e^{2 x} = \frac{e^{2 x}}{5} [2 \sin x - \cos x] + C$

$\Rightarrow y = \frac{1}{5} (2 \sin x - \cos x) + C e^{- 2 x}$

The general solution for given differential equation is $y= \frac{1}{5} (2sinx-cosx) +C e^{-2x}$ .

2. Find the general solution for the differential equation $\frac{dy}{dx} +3y= e^{-2x}$ .

Ans: The given differential equation is:

$\frac{d y}{d x} + 3 y = e^{- 2 x}$

It is a linear differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = 3$

$Q = e^{- 2 x}$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{\int 3 d x}$

$\Rightarrow I . F = e^{3 x}$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y e^{3 x} = \int e^{- 2 x} (e^{3 x}) d x + C$

$\Rightarrow y e^{3 x} = \int e^{- 2 x + 3 x} d x + C$

$\Rightarrow y e^{3 x} = \int e^{x} d x + C$

$\Rightarrow y e^{3 x} = e^{x} + C$

$\Rightarrow y = e^{- 2 x} + C e^{- 3 x}$

The general solution for given differential equation is $y= e^{-2x} +C e^{-3x}$ .

3. Find the general solution for the differential equation $\frac{dy}{dx} + \frac{y}{x} = x^{2}$ .

Ans: The given differential equation is:

$\frac{d y}{d x} + \frac{y}{x} = x^{2}$

It is a linear differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = \frac{1}{x}$

$Q = x^{2}$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{\int \frac{1}{x} d x}$

$\Rightarrow I . F = e^{\log x}$

$\Rightarrow I . F = x$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y x = \int x^{2} (x) d x + C$

$\Rightarrow x y = \int x^{3} d x + C$

$\Rightarrow x y = \frac{x^{3 + 1}}{3 + 1} + C$

$\Rightarrow x y = \frac{x^{4}}{4} + C$

The general solution for given differential equation is $xy= \frac{x^{4}}{4} +C$ .

4. Find the general solution for the differential equation $\frac{dy}{dx} + (secx) y=tanx (0 \leq x \leq \frac{π}{2})$ .

Ans: The given differential equation is:

$\frac{d y}{d x} + (\sec x) y = \tan x$

It is a linear differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = \sec x$

$Q = \tan x$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{\int \sec x d x}$

$\Rightarrow I . F = e^{\log (\sec x + \tan x)}$

$\Rightarrow I . F = (\sec x + \tan x)$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y (\sec x + \tan x) = \int \tan x (\sec x + \tan x) d x + C$

$\Rightarrow y (\sec x + \tan x) = \int \tan x \sec x d x + \int \tan^{2} x d x + C$

$\Rightarrow y (\sec x + \tan x) = \sec x + \int (\sec^{2} x - 1) d x + C$

$\Rightarrow y (\sec x + \tan x) = \sec x + \int \sec^{2} x d x - \int d x + C$

$\Rightarrow y (\sec x + \tan x) = \sec x + \tan x - x + C$

The general solution for given differential equation is $y (secx+tanx) =secx+tanx-x+C$ .

5. Find the general solution for the differential equation $co s^{2} x \frac{dy}{dx} +y=tanx (0 \leq x \leq \frac{π}{2})$ .

Ans: The given differential equation is:

$\cos^{2} x \frac{d y}{d x} + y = \tan x$

$\frac{d y}{d x} + (\sec^{2} x) y = \sec^{2} x \tan x$

It is a linear differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = \sec^{2} x$

$Q = \sec^{2} x \tan x$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{\int \sec^{2} x d x}$

$\Rightarrow I . F = e^{\tan x}$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y e^{\tan x} = \int e^{\tan x} (\sec^{2} x \tan x) d x + C$

$\Rightarrow y e^{\tan x} = I + C (I = \int e^{\tan x} (\sec^{2} x \tan x) d x)$ ……(1)

Solving the integral $I$ :

$I = \int e^{\tan x} (\sec^{2} x \tan x) d x$

Substitute $tanx=t$ :
$\tan x = t$

$\Rightarrow \sec^{2} x d x = d t$

$\Rightarrow I = \int e^{t} t d t$

$\Rightarrow I = t \int e^{t} d t - \int ({(t)}^{'} \int e^{t} d t) d t$

$\Rightarrow I = t e^{t} - \int (e^{t}) d t$

$\Rightarrow I = t e^{t} - e^{t}$

Back substitute $t$ :

$I = \tan x e^{\tan x} - e^{\tan x}$

Back substitute $I$ in expression (1):

$\Rightarrow y e^{\tan x} = I + C$

$\Rightarrow y e^{\tan x} = \tan x e^{\tan x} - e^{\tan x} + C$

The general solution for given differential equation is $y e^{tanx} =tanx e^{tanx} - e^{tanx} +C$ .

6. Find the general solution for the differential equation $x \frac{dy}{dx} +2y= x^{2} logx$ .

Ans: The given differential equation is:

$x \frac{d y}{d x} + 2 y = x^{2} \log x$

$\frac{d y}{d x} + \frac{2}{x} y = x \log x$

It is a linear differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = \frac{2}{x}$

$Q = x \log x$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{\int \frac{2}{x} d x}$

$\Rightarrow I . F = e^{2 \log x}$

$\Rightarrow I . F = e^{\log x^{2}}$

$\Rightarrow I . F = x^{2}$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y x^{2} = \int x^{2} (x \log x) d x + C$

$\Rightarrow y x^{2} = \int x^{3} \log x d x + C$

$\Rightarrow y x^{2} = I + C (I = \int x^{3} \log x d x)$ ……(1)

Solving the integral $I$ :

$I = \int x^{3} \log x d x$

$\Rightarrow I = \log x \int x^{3} d x - \int ({(\log x)}^{'} \int x^{3} d x) d x$

$\Rightarrow I = \frac{x^{4}}{4} \log x - \int (\frac{1}{x} (\frac{x^{4}}{4})) d x$

$\Rightarrow I = \frac{x^{4}}{4} \log x - \frac{1}{4} \int x^{3} d x$

$\Rightarrow I = \frac{x^{4}}{4} \log x - \frac{1}{4} (\frac{x^{4}}{4})$

$\Rightarrow I = \frac{x^{4}}{4} \log x - \frac{x^{4}}{16}$

Back substitute $I$ in expression (1):

$\Rightarrow y x^{2} = I + C$

$\Rightarrow y x^{2} = \frac{x^{4}}{4} \log x - \frac{x^{4}}{16} + C$

$\Rightarrow y = \frac{x^{2}}{16} (4 \log x - 1) + C x^{- 2}$

The general solution for given differential equation is $y= \frac{x^{2}}{16} (4logx-1) +C x^{-2}$ .

7. Find the general solution for the differential equation $xlogx \frac{dy}{dx} +y= \frac{2}{x} logx$ .

Ans: The given differential equation is:

$x \log x \frac{d y}{d x} + y = \frac{2}{x} \log x$

$\frac{d y}{d x} + \frac{y}{x \log x} = \frac{2}{x^{2}}$

It is a linear differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = \frac{1}{x \log x}$

$Q = \frac{2}{x^{2}}$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{\int \frac{1}{x \log x} d x}$

Substitute $logx=t$ :
$\log x = t$

$\Rightarrow \frac{1}{x} d x = d t$

$\Rightarrow I . F = e^{\int \frac{1}{t} d t}$

$\Rightarrow I . F = e^{\log t}$

$\Rightarrow I . F = t$

$\Rightarrow I . F = \log x$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y \log x = \int \frac{2}{x^{2}} (\log x) d x + C$

$\Rightarrow y \log x = I + C (I = \int \frac{2}{x^{2}} (\log x) d x)$ ……(1)

Solving the integral $I$ :

$I = \int \frac{2}{x^{2}} (\log x) d x$

$I = 2 [\log x \int \frac{1}{x^{2}} d x - \int ({(\log x)}^{'} \int \frac{1}{x^{2}} d x) d x]$

$I = 2 [\log x (\frac{- 1}{x}) - \int (\frac{1}{x} (\frac{- 1}{x})) d x]$

$I = 2 [- \frac{\log x}{x} + \int (\frac{1}{x^{2}}) d x]$

$I = 2 [- \frac{\log x}{x} - \frac{1}{x}]$

Back substitute $I$ in expression (1):

$y \log x = I + C$

$\Rightarrow y \log x = 2 [- \frac{\log x}{x} - \frac{1}{x}] + C$

$\Rightarrow y \log x = - \frac{2}{x} (\log x + 1) + C$

The general solution for given differential equation is $ylogx=- \frac{2}{x} (logx+1) +C$ .

8. Find the general solution for the differential equation $(1+ x^{2}) dy+2xydx=cotxdx (x \neq 0)$ .

Ans: The given differential equation is:

$(1 + x^{2}) d y + 2 x y d x = \cot x d x (x \neq 0)$

$\Rightarrow \frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{\cot x}{1 + x^{2}}$

It is a linear differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = \frac{2 x}{1 + x^{2}}$

$Q = \frac{\cot x}{1 + x^{2}}$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{\int \frac{2 x}{1 + x^{2}} d x}$

Substitute $logx=t$ :
$1 + x^{2} = t$

$\Rightarrow 2 x d x = d t$

$\Rightarrow I . F = e^{\int \frac{1}{t} d t}$

$\Rightarrow I . F = e^{\log t}$

$\Rightarrow I . F = t$

$\Rightarrow I . F = 1 + x^{2}$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y (1 + x^{2}) = \int \frac{\cot x}{1 + x^{2}} (1 + x^{2}) d x + C$

$\Rightarrow y (1 + x^{2}) = \int \cot x d x + C$

$\Rightarrow y (1 + x^{2}) = \log | \sin x | + C$

The general solution for given differential equation is $y (1+ x^{2}) =log | sinx | +C$ .

9. Find the general solution for the differential equation $x \frac{dy}{dx} +y-x+xycotx=0 (x \neq 0)$ .

Ans: The given differential equation is:

$x \frac{d y}{d x} + y - x + x y \cot x = 0$

$\Rightarrow \frac{d y}{d x} + \frac{y}{x} - 1 + y \cot x = 0$

$\Rightarrow \frac{d y}{d x} + (\frac{1}{x} + \cot x) y = 1$

It is a linear differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = (\frac{1}{x} + \cot x)$

$Q = 1$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{\int (\frac{1}{x} + \cot x) d x}$

$\Rightarrow I . F = e^{\int \frac{1}{x} d x + \int \cot x d x}$

$\Rightarrow I . F = e^{\log x + \log (\sin x)}$

$\Rightarrow I . F = e^{\log (x \sin x)}$

$\Rightarrow I . F = x \sin x$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y (x \sin x) = \int (1) (x \sin x) d x + C$

$\Rightarrow x y \sin x = I + C (I = \int x \sin x d x)$ ……(1)

Solving the integral $I$ :

$I = \int x \sin x d x$

$\Rightarrow I = x \int \sin x d x - \int ({(x)}^{'} \int \sin x d x) d x$

$\Rightarrow I = x (- \cos x) + \int (\cos x) d x$

$\Rightarrow I = - x \cos x + \sin x$

Back substitute $I$ in expression (1):

$x y \sin x = I + C$

$\Rightarrow x y \sin x = - x \cos x + \sin x + C$

$\Rightarrow y = - \cot x + \frac{1}{x} + \frac{C}{x \sin x}$

The general solution for given differential equation is $y=-cotx+ \frac{1}{x} + \frac{C}{xsinx}$ .

10. Find the general solution for the differential equation $(x+y) \frac{dy}{dx} = 1$ .

Ans: The given differential equation is:

$(x + y) \frac{d y}{d x} = 1$

$\Rightarrow \frac{d y}{d x} = \frac{1}{x + y}$

$\Rightarrow \frac{d x}{d y} = x + y$

$\Rightarrow \frac{d x}{d y} - x = y$

It is differential equation of the form $\frac{dx}{dy} +px=Q$ , with:

$p = - 1$

$Q = y$

Calculate the integrating factor:

$I . F = e^{\int p d y}$

$\Rightarrow I . F = e^{\int - 1 d y}$

$\Rightarrow I . F = e^{- y}$

General solution is of the form:

$x (I . F) = \int (Q \times I . F) d y + C$

$\Rightarrow x (e^{- y}) = \int (y) (e^{- y}) d y + C$

$\Rightarrow x e^{- y} = I + C (I = \int y e^{- y} d y)$ ……(1)

Solving the integral $I$ :

$I = \int y e^{- y} d y$

$\Rightarrow I = y \int e^{- y} d y - \int ({(y)}^{'} \int e^{- y} d y) d y$

$\Rightarrow I = - y e^{- y} + \int ((1) e^{- y}) d y$

$\Rightarrow I = - y e^{- y} + \int e^{- y} d y$

$\Rightarrow I = - y e^{- y} - e^{- y}$

Back substitute $I$ in expression (1):

$x e^{- y} = I + C$

$\Rightarrow x e^{- y} = - y e^{- y} - e^{- y} + C$

$\Rightarrow x = - y - 1 + C e^{y}$

$\Rightarrow x + y + 1 = C e^{y}$

The general solution for given differential equation is $x+y+1=C e^{y}$ .

11. Find the general solution for the differential equation $ydx+ (x- y^{2}) dy=0$ .

Ans: The given differential equation is:

$y d x + (x - y^{2}) d y = 0$

$\Rightarrow \frac{d x}{d y} = \frac{y^{2} - x}{y}$

$\Rightarrow \frac{d x}{d y} + (\frac{1}{y}) x = y$

It is differential equation of the form $\frac{dx}{dy} +px=Q$ , with:

$p = \frac{1}{y}$

$Q = y$

Calculate the integrating factor:

$I . F = e^{\int p d y}$

$\Rightarrow I . F = e^{\int \frac{1}{y} d y}$

$\Rightarrow I . F = e^{\log y}$

$\Rightarrow I . F = y$

General solution is of the form:

$x (I . F) = \int (Q \times I . F) d y + C$

$\Rightarrow x (y) = \int (y) (y) d y + C$

$\Rightarrow x y = \int y^{2} d y + C$

$\Rightarrow x y = \frac{y^{3}}{3} + C$

$\Rightarrow x = \frac{y^{2}}{3} + \frac{C}{y}$

The general solution for given differential equation is $x= \frac{y^{2}}{3} + \frac{C}{y}$ .

12. Find the general solution for the differential equation $(x+3 y^{2}) \frac{dy}{dx} =y (y0)$ .

Ans: The given differential equation is:

$(x + 3 y^{2}) \frac{d y}{d x} = y$

$\Rightarrow \frac{d x}{d y} = \frac{x + 3 y^{2}}{y}$

$\Rightarrow \frac{d x}{d y} - (\frac{1}{y}) x = 3 y$

It is differential equation of the form $\frac{dx}{dy} +px=Q$ , with:

$p = - \frac{1}{y}$

$Q = 3 y$

Calculate the integrating factor:

$I . F = e^{\int p d y}$

$\Rightarrow I . F = e^{- \int \frac{1}{y} d y}$

$\Rightarrow I . F = e^{- \log y}$

$\Rightarrow I . F = e^{\log y^{- 1}}$

$\Rightarrow I . F = \frac{1}{y}$

General solution is of the form:

$x (I . F) = \int (Q \times I . F) d y + C$

$\Rightarrow x (\frac{1}{y}) = \int (3 y) (\frac{1}{y}) d y + C$

$\Rightarrow \frac{x}{y} = 3 \int d y + C$

$\Rightarrow \frac{x}{y} = 3 y + C$

$\Rightarrow x = 3 y^{2} + C y$

The general solution for given differential equation is $x=3 y^{2} +Cy$ .

13. Find particular solution for $\frac{dy}{dx} +2ytanx=sinx$ satisfying $y=0$ when $x= \frac{π}{3}$ .

Ans: The given differential equation is:

$\frac{d y}{d x} + 2 y \tan x = \sin x$

$\frac{d y}{d x} + (2 \tan x) y = \sin x$

It is differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = 2 \tan x$

$Q = \sin x$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{2 \int \tan x d x}$

$\Rightarrow I . F = e^{2 \log | \sec x |}$

$\Rightarrow I . F = e^{\log {(\sec x)}^{2}}$

$\Rightarrow I . F = \sec^{2} x$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y \sec^{2} x = \int (\sin x) (\sec^{2} x) d x + C$

$\Rightarrow y \sec^{2} x = \int \tan x \sec x d x + C$

$\Rightarrow y \sec^{2} x = \sec x + C$

$\Rightarrow y = \cos x + C \cos^{2} x$

Given $y=0$ when $x= \frac{π}{3}$ :

$0 = \cos (\frac{π}{3}) + C \cos^{2} (\frac{π}{3})$

$\Rightarrow 0 = \frac{1}{2} + C {(\frac{1}{2})}^{2}$

$\Rightarrow C = - 2$

Therefore the particular solution will be:

$\Rightarrow y = \cos x - 2 \cos^{2} x$

The particular solution for given differential equation satisfying the given conditions is $y=cosx-2co s^{2} x$ .

14. Find particular solution for $(1+ x^{2}) \frac{dy}{dx} +2xy= \frac{1}{1+ x^{2}}$ satisfying $y=0$ when $x=1$ .

Ans: The given differential equation is:

$(1 + x^{2}) \frac{d y}{d x} + 2 x y = \frac{1}{1 + x^{2}}$

$\frac{d y}{d x} + (\frac{2 x}{1 + x^{2}}) y = \frac{1}{{(1 + x^{2})}^{2}}$

It is differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = \frac{2 x}{1 + x^{2}}$

$Q = \frac{1}{{(1 + x^{2})}^{2}}$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{\int \frac{2 x}{1 + x^{2}} d x}$

$\Rightarrow I . F = e^{2 \log | \sec x |}$

$\Rightarrow I . F = e^{\log (1 + x^{2})}$

$\Rightarrow I . F = 1 + x^{2}$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y (1 + x^{2}) = \int (\frac{1}{{(1 + x^{2})}^{2}}) (1 + x^{2}) d x + C$

$\Rightarrow y (1 + x^{2}) = \int \frac{1}{1 + x^{2}} d x + C$

$\Rightarrow y (1 + x^{2}) = \tan^{- 1} x + C$

Given $y=0$ when $x=1$ :

$0 (1 + 1) = \tan^{- 1} (1) + C$

$\Rightarrow C + \frac{π}{4} = 0$

$\Rightarrow C = - \frac{π}{4}$

Therefore the particular solution will be:

$\Rightarrow y (1 + x^{2}) = \tan^{- 1} x - \frac{π}{4}$

The particular solution for given differential equation satisfying the given conditions is $y (1+ x^{2}) =ta n^{-1} x- \frac{π}{4}$ .

15. Find particular solution for $\frac{dy}{dx} -3ycotx=sin2x$ satisfying $y=2$ when $x= \frac{π}{2}$ .

Ans: The given differential equation is:

$\frac{d y}{d x} - 3 y \cot x = \sin 2 x$

$\frac{d y}{d x} + (- 3 \cot x) y = \sin 2 x$

It is differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = - 3 \cot x$

$Q = \sin 2 x$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{- 3 \int \cot x d x}$

$\Rightarrow I . F = e^{- 3 \log | \sin x |}$

$\Rightarrow I . F = e^{\log (\frac{1}{\sin^{3} x})}$

$\Rightarrow I . F = \frac{1}{\sin^{3} x}$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y (\frac{1}{\sin^{3} x}) = \int (\sin 2 x) (\frac{1}{\sin^{3} x}) d x + C$

$\Rightarrow y (\frac{1}{\sin^{3} x}) = 2 \int (\sin x \cos x) (\frac{1}{\sin^{3} x}) d x + C$

$\Rightarrow \frac{y}{\sin^{3} x} = 2 \int (\frac{\cos x}{\sin^{2} x}) d x + C$

$\Rightarrow \frac{y}{\sin^{3} x} = 2 \int \cot x cosec x d x + C$

$\Rightarrow \frac{y}{\sin^{3} x} = - 2 cosec x + C$

$\Rightarrow y = - 2 \sin^{2} x + C \sin^{3} x$

Given $y=2$ when $x= \frac{π}{2}$ :

$2 = - 2 \sin^{2} (\frac{π}{2}) + C \sin^{3} (\frac{π}{2})$

$\Rightarrow C - 2 = 2$

$\Rightarrow C = 4$

Therefore the particular solution will be:

$\Rightarrow y = - 2 \sin^{2} x + 4 \sin^{3} x$

The particular solution for given differential equation satisfying the given conditions is $y=-2si n^{2} x+4si n^{3} x$ .

16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point $(x,y)$ is equal to the sum of the coordinates of the point.

Ans: According to the slope of tangent $\frac{dy}{dx}$ is equal to sum of the coordinates:

$\frac{d y}{d x} = x + y$

The given differential equation is:

$\frac{d y}{d x} = x + y$

$\frac{d y}{d x} + (- 1) y = x$

It is differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = - 1$

$Q = x$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{- 1 \int d x}$

$\Rightarrow I . F = e^{- x}$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y (e^{- x}) = \int x (e^{- x}) d x + C$

$\Rightarrow y e^{- x} = x \int e^{- x} d x - \int ({(x)}^{'} \int e^{- x} d x) d x + C$

$\Rightarrow y e^{- x} = - x e^{- x} + \int (e^{- x}) d x + C$

$\Rightarrow y e^{- x} = - x e^{- x} - e^{- x} + C$

$\Rightarrow y = - x - 1 + C e^{x}$

$\Rightarrow y + x + 1 = C e^{x}$

Given $y=0$ when $x=0$ as it passes through origin:

$0 + 0 + 1 = C e^{0}$

$\Rightarrow C = 1$

Therefore the equation of the required curve is $y+x+1= e^{x}$ .

17. Find the equation of a curve passing through the point $(0,2)$ given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by $5$ .

Ans: Let the slope of tangent be $\frac{dy}{dx}$ .

According to :

$x + y = \frac{d y}{d x} + 5$

The given differential equation is:

$x + y = \frac{d y}{d x} + 5$

$\frac{d y}{d x} + (- 1) y = x - 5$

It is differential equation of the form $\frac{dy}{dx} +py=Q$ , with:

$p = - 1$

$Q = x - 5$

Calculate the integrating factor:

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{- 1 \int d x}$

$\Rightarrow I . F = e^{- x}$

General solution is of the form:

$y (I . F) = \int (Q \times I . F) d x + C$

$\Rightarrow y (e^{- x}) = \int (x - 5) (e^{- x}) d x + C$

$\Rightarrow y (e^{- x}) = \int x (e^{- x}) d x - 5 \int e^{- x} d x + C$

$\Rightarrow y e^{- x} = x \int e^{- x} d x - \int ({(x)}^{'} \int e^{- x} d x) d x - 5 \int e^{- x} d x C$

$\Rightarrow y e^{- x} = - x e^{- x} + \int (e^{- x}) d x + 5 e^{- x} + C$

$\Rightarrow y e^{- x} = - x e^{- x} - e^{- x} + 5 e^{- x} + C$

$\Rightarrow y e^{- x} = - x e^{- x} + 4 e^{- x} + C$

$\Rightarrow y = - x + 4 + C e^{x}$

$\Rightarrow y + x - 4 = C e^{x}$

Given as it passes through $(0,2)$ :

$2 + 0 - 4 = C e^{0}$

$\Rightarrow C = - 2$

Therefore the equation of the required curve is:

$y + x - 4 = - 2 e^{x}$ .

18. Find the integrating factor of the differential equation $x \frac{dy}{dx} -y=2 x^{2}$ .

A. $e^{- x}$

B. $e^{- y}$

C. $\frac{1}{x}$

D. $x$

Ans: Given differential equation is:

$x \frac{d y}{d x} - y = 2 x^{2}$

$\Rightarrow \frac{d y}{d x} - (\frac{1}{x}) y = 2 x^{2}$

Thus it is a linear differential equation of the form $\frac{dy}{dx} +py=Q$ :

$p = - \frac{1}{x}$

$I . F = e^{\int p d x}$

$\Rightarrow I . F = e^{- \int \frac{1}{x} d x}$

$\Rightarrow I . F = e^{- \log | x |}$

$\Rightarrow I . F = e^{\log x^{- 1}}$

$\Rightarrow I . F = \frac{1}{x}$

Therefore integrating factor is $\frac{1}{x}$ . Thus the correct option is (C).

19. Find the integrating factor of the differential equation $(1- y^{2}) \frac{dx}{dy} +yx=ay (-1y1)$ .

A. $\frac{1}{y^{2} - 1}$

B. $\frac{1}{\sqrt{y^{2} - 1}}$

C. $\frac{1}{1 - y^{2}}$

D. $\frac{1}{\sqrt{1 - y^{2}}}$

Ans: Given differential equation is:

$(1 - y^{2}) \frac{d x}{d y} + y x = a y$

$\Rightarrow \frac{d x}{d y} + (\frac{y}{1 - y^{2}}) x = \frac{a y}{(1 - y^{2})}$

Thus it is a linear differential equation of the form $\frac{dx}{dy} +px=Q$ :

$p = \frac{y}{1 - y^{2}}$

$I . F = e^{\int p d y}$

$\Rightarrow I . F = e^{\int \frac{y}{1 - y^{2}} d y}$

$\Rightarrow I . F = e^{\frac{1}{- 2} \int \frac{- 2 y}{1 - y^{2}} d y}$

$\Rightarrow I . F = e^{\frac{1}{- 2} \log (1 - y^{2})}$

$\Rightarrow I . F = e^{\log {(1 - y^{2})}^{- \frac{1}{2}}}$

$\Rightarrow I . F = \frac{1}{\sqrt{1 - y^{2}}}$

Therefore integrating factor is $\frac{1}{\sqrt{1- y^{2}}}$ . Thus the correct option is (D).

Miscellaneous Exercise

1. For each of the differential equations given below, indicate its order and degree (if defined).

(i) $\frac{d^{2} y}{d x^{2}} +5x {(\frac{dy}{dx})}^{2} -6y=logx$

Ans: The given differential equation is:

$\frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y - \log x = 0$

The highest order derivative in the equation is of the term $\frac{d^{2} y}{d x^{2}}$ , thus the order of the equation is $2$ and its highest power is $1$ . Therefore its degree is $1$ .

(ii) ${(\frac{dy}{dx})}^{3} - 4 {(\frac{dy}{dx})}^{2} +7y=sin x$

Ans: The given differential equation is:

${(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y - \sin x = 0$

The highest order derivative in the equation is of the term ${(\frac{dy}{dx})}^{3}$ , thus the order of the equation is $1$ and its highest power is $3$ . Therefore its degree is $3$ .

(iii) $\frac{d^{4} y}{d x^{4}} -sin (\frac{d^{3} y}{d x^{3}}) =0$

Ans: The given differential equation is:

$\frac{d^{4} y}{d x^{4}} - \sin (\frac{d^{3} y}{d x^{3}}) = 0$

The highest order derivative in the equation is of the term $\frac{d^{4} y}{d x^{4}}$ , thus the order of the equation is $4$ .

As the differential equation is not polynomial in its derivative, therefore its degree is not defined.

2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

$xy=a e^{x} +b e^{-x} + x^{2} :x \frac{d^{2} y}{d x^{2}} +2 \frac{dy}{dx} -xy+ x^{2} -2=0$

Ans: The given function is:
$x y = a e^{x} + b e^{- x} + x^{2}$

Take derivative on both side:

$\Rightarrow y + x \frac{d y}{d x} = a e^{x} - b e^{- x} + 2 x$

Take derivative on both side:

$\Rightarrow \frac{d y}{d x} + x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = a e^{x} + b e^{- x} + 2$

$\Rightarrow x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} = a e^{x} + b e^{- x} + 2$ ……(1)

The given differential equation is:
$x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0$

Solving LHS:

Substitute $x \frac{d^{2} y}{d x^{2}} +2 \frac{dy}{dx}$ from the result (1) and $xy$ :

$\Rightarrow (x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x}) - x y + x^{2} - 2$

$\Rightarrow (a e^{x} + b e^{- x} + 2) - (a e^{x} + b e^{- x} + x^{2}) + x^{2} - 2$

$\Rightarrow 2 - x^{2} + x^{2} - 2$

$\Rightarrow 0$

Thus LHS=RHS, the given function is the solution of the given differential equation.

$y= e^{x} (acosx+bsinx) : \frac{d^{2} y}{d x^{2}} -2 \frac{dy}{dx} +2y=0$

Ans: The given function is:
$y = e^{x} (a \cos x + b \sin x)$

Take derivative on both side:

$\Rightarrow \frac{d y}{d x} = e^{x} (a \cos x + b \sin x) + e^{x} (- a \sin x + b \cos x)$

$\Rightarrow \frac{d y}{d x} = e^{x} ((a + b) \cos x + (b - a) \sin x)$

Take derivative on both side:

$\Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x} ((a + b) \cos x + (b - a) \sin x) + e^{x} (- (a + b) \sin x + (b - a) \cos x)$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x} ((a + b + b - a) \cos x + (b - a - a - b) \sin x)$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x} (2 b \cos x - 2 a \sin x)$

The given differential equation is:
$\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0$

Solving LHS:

$\Rightarrow e^{x} (2 b \cos x - 2 a \sin x) - 2 e^{x} ((a + b) \cos x + (b - a) \sin x) + 2 y$

$\Rightarrow e^{x} ((2 b - 2 a - 2 b) \cos x + (- 2 a - 2 b + 2 a) \sin x) - 2 y$

$\Rightarrow e^{x} (- 2 a \cos x - 2 b \sin x) - 2 y$

$\Rightarrow - 2 e^{x} (a \cos x + b \sin x) - 2 y$

$\Rightarrow 0$

Thus LHS=RHS, the given function is the solution of the given differential equation.

$y=x sin3x : \frac{d^{2} y}{d x^{2}} +9y-6cos3x=0$

Ans: The given function is:
$y = x \sin 3 x$

Take derivative on both side:

$\Rightarrow \frac{d y}{d x} = \sin 3 x + 3 x \cos 3 x$

Take derivative on both side:

$\Rightarrow \frac{d^{2} y}{d x^{2}} = 3 \cos 3 x + 3 (\cos 3 x + x (- 3 \sin 3 x))$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = 3 \cos 3 x + 3 \cos 3 x - 9 x \sin 3 x$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = 6 \cos 3 x - 9 x \sin 3 x$

The given differential equation is:
$\frac{d^{2} y}{d x^{2}} + 9 y - 6 \cos 3 x = 0$

Solving LHS:

$\Rightarrow \frac{d^{2} y}{d x^{2}} + 9 y - 6 \cos 3 x$

$\Rightarrow (6 \cos 3 x - 9 x \sin 3 x) + 9 (x \sin 3 x) - 6 \cos 3 x$

$\Rightarrow 6 \cos 3 x - 9 x \sin 3 x + 9 x \sin 3 x - 6 \cos 3 x$

$\Rightarrow 0$

Thus LHS=RHS, the given function is the solution of the given differential equation.

$x^{2} =2 y^{2} log y : (x^{2} + y^{2}) \frac{dy}{dx} -xy=0$

Ans: The given function is:
$x^{2} = 2 y^{2} \log y$

Take derivative on both side:

$\Rightarrow 2 x = 2 (2 y \log y + y^{2} (\frac{1}{y})) \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x} = \frac{x}{(2 y \log y + y)}$

Multiply numerator and denominator by $y$ :

$\Rightarrow \frac{d y}{d x} = \frac{x y}{(2 y^{2} \log y + y^{2})}$

$\Rightarrow \frac{d y}{d x} = \frac{x y}{(x^{2} + y^{2})}$

The given differential equation is:
$(x^{2} + y^{2}) \frac{d y}{d x} - x y = 0$

Solving LHS:

$\Rightarrow (x^{2} + y^{2}) \frac{d y}{d x} - x y$

$\Rightarrow (x^{2} + y^{2}) (\frac{x y}{x^{2} + y^{2}}) - x y$

$\Rightarrow x y - x y$

$\Rightarrow 0$

Thus LHS=RHS, the given function is the solution of the given differential equation.

3. Prove that $x^{2} - y^{2} =c {(x^{2} + y^{2})}^{2}$ is the general solution of differential equation $(x^{3} -3x y^{2}) dx= (y^{3} -3 x^{2} y) dy$ , where $c$ is a parameter.

Ans: Given differential equation:

$(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$

$\Rightarrow \frac{d y}{d x} = \frac{x^{3} - 3 x y^{2}}{y^{3} - 3 x^{2} y}$

As it can be seen that this is an homogenous equation. Substitute $y=vx$ :

$\Rightarrow \frac{d (v x)}{d x} = \frac{x^{3} - 3 x {(v x)}^{2}}{{(v x)}^{3} - 3 x^{2} (v x)}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{x^{3} (1 - 3 v^{2})}{x^{3} (v^{3} - 3 v)}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v}$

$\Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v} - v$

$\Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2} - v^{4} + 3 v^{2}}{v^{3} - 3 v}$

$\Rightarrow x \frac{d v}{d x} = \frac{1 - v^{4}}{v^{3} - 3 v}$

Separate the differentials:

$\frac{v^{3} - 3 v}{1 - v^{4}} d v = \frac{d x}{x}$

Integrate both side:

$\int \frac{v^{3} - 3 v}{1 - v^{4}} d v = \int \frac{d x}{x}$

$\Rightarrow \int \frac{v^{3} - 3 v}{1 - v^{4}} d v = \log x + \log C$

$\Rightarrow I = \log x + \log C (I = \int \frac{v^{3} - 3 v}{1 - v^{4}} d v)$ ……(1)

Solving integral $I$ :

$\Rightarrow I = \int \frac{v^{3} - 3 v}{1 - v^{4}} d v$

$\Rightarrow \frac{v^{3} - 3 v}{1 - v^{4}} = \frac{v^{3} - 3 v}{(1 - v^{2}) (1 + v^{2})}$

$\Rightarrow \frac{v^{3} - 3 v}{1 - v^{4}} = \frac{v^{3} - 3 v}{(1 - v) (1 + v) (1 + v^{2})}$

Using partial fraction:

$\Rightarrow \frac{v^{3} - 3 v}{1 - v^{4}} = \frac{A}{1 - v} + \frac{B}{1 + v} + \frac{C v + D}{1 + v^{2}}$

Solving for $A,B,C and D$ :

$A = - \frac{1}{2}$

$B = \frac{1}{2}$

$C = - 2$

$D = 0$

$\Rightarrow \frac{v^{3} - 3 v}{1 - v^{4}} = \frac{- \frac{1}{2}}{1 - v} + \frac{\frac{1}{2}}{1 + v} + \frac{- 2 v + 0}{1 + v^{2}}$

$I = - \frac{1}{2} \int \frac{1}{1 - v} d v + \frac{1}{2} \int \frac{1}{1 + v} d v - \int \frac{2 v}{1 + v^{2}} d v$

$I = - \frac{1}{2} (- \log (1 - v)) + \frac{1}{2} (\log (1 + v)) - \log (1 + v^{2})$

$I = \frac{1}{2} (\log (1 - v^{2})) - \frac{2}{2} \log (1 + v^{2})$

$I = \frac{1}{2} (\log \frac{(1 - v^{2})}{{(1 + v^{2})}^{2}})$

$\Rightarrow I = \frac{1}{2} (\log \frac{(1 - \frac{y^{2}}{x^{2}})}{{(1 + \frac{y^{2}}{x^{2}})}^{2}})$

$\Rightarrow I = \frac{1}{2} (\log \frac{x^{2} (x^{2} - y^{2})}{{(x^{2} + y^{2})}^{2}})$

$\Rightarrow I = \frac{1}{2} (\log \frac{(x^{2} - y^{2})}{{(x^{2} + y^{2})}^{2}}) + \frac{1}{2} \log x^{2}$

$\Rightarrow I = \frac{1}{2} (\log \frac{(x^{2} - y^{2})}{{(x^{2} + y^{2})}^{2}}) + \log x$

Back substitute $I$ in expression (1):

$\Rightarrow I = \log x + \log C$

$\Rightarrow \frac{1}{2} (\log \frac{(x^{2} - y^{2})}{{(x^{2} + y^{2})}^{2}}) + \log x = \log x + \log C$

$\Rightarrow \log \frac{(x^{2} - y^{2})}{{(x^{2} + y^{2})}^{2}} = 2 \log C$

$\Rightarrow \frac{x^{2} - y^{2}}{{(x^{2} + y^{2})}^{2}} = C^{2}$

$\Rightarrow x^{2} - y^{2} = c {(x^{2} + y^{2})}^{2} (c = C^{2})$

Thus for given differential equation, its general solution is $x^{2} - y^{2} =c {(x^{2} + y^{2})}^{2}$ .

4. Find the general solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1- y^{2}}{1- x^{2}}} =0$ .

Ans: The given differential equation is:

$\frac{d y}{d x} + \sqrt{\frac{1 - y^{2}}{1 - x^{2}}} = 0$

$\frac{d y}{d x} = - \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$

$\Rightarrow \frac{d y}{\sqrt{1 - y^{2}}} = - \frac{d x}{\sqrt{1 - x^{2}}}$

Integrate both side:

$\Rightarrow \int \frac{d y}{\sqrt{1 - y^{2}}} = - \int \frac{d x}{\sqrt{1 - x^{2}}}$

$\Rightarrow \sin^{- 1} y = - \sin^{- 1} x + C$

$\Rightarrow \sin^{- 1} y + \sin^{- 1} x = C$

Thus the general solution of given differential equation is $si n^{-1} y+si n^{-1} x=C$ .

5. Show that the general solution of the differential equation $\frac{dy}{dx} + \frac{y^{2} +y+1}{x^{2} +x+1} =0$ is given by $(x+y+1) =A (1-x-y-2xy)$ where $A$ is a parameter.

Ans: The given differential equation is:

$\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{y^{2} + y + 1}{x^{2} + x + 1}$

$\Rightarrow \frac{d y}{d x} = - \frac{y^{2} + 2 (\frac{1}{2}) y + \frac{1}{4} - \frac{1}{4} + 1}{x^{2} + 2 (\frac{1}{2}) x + \frac{1}{4} - \frac{1}{4} + 1}$

$\Rightarrow \frac{d y}{d x} = - \frac{{(y + \frac{1}{2})}^{2} + \frac{3}{4}}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}$

$\Rightarrow \frac{d y}{{(y + \frac{1}{2})}^{2} + \frac{3}{4}} = - \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}$

Integrate both side:

$\Rightarrow \int \frac{d y}{{(y + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = - \int \frac{d x}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}$

$\Rightarrow \frac{1}{(\frac{\sqrt{3}}{2})} \tan^{- 1} [\frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}}] = - \frac{1}{(\frac{\sqrt{3}}{2})} \tan^{- 1} [\frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}}] + C$

$\Rightarrow \frac{2}{\sqrt{3}} (\tan^{- 1} [\frac{2 y + 1}{\sqrt{3}}] + \tan^{- 1} [\frac{2 x + 1}{\sqrt{3}}]) = C$

$\Rightarrow \tan^{- 1} [\frac{2 y + 1}{\sqrt{3}}] + \tan^{- 1} [\frac{2 x + 1}{\sqrt{3}}] = \frac{\sqrt{3}}{2} C$

Thus the general solution for given differential equation is $ta n^{-1} [\frac{2y+1}{\sqrt{3}}] +ta n^{-1} [\frac{2x+1}{\sqrt{3}}] = \frac{\sqrt{3}}{2} C$ .

6. Find the equation of the curve passing through the point $(0, \frac{π}{4})$ whose differential equation is $sin x cos ydx+cos x sin ydy=0$ .

Ans: Given differential equation is:

$\sin x \cos y d x + \cos x \sin y d y = 0$

$\Rightarrow \sin x \cos y d x + \cos x \sin y d y = 0$

Divide both side by $cos x cos y$ :

$\Rightarrow \frac{\sin x \cos y d x + \cos x \sin y d y}{\cos x \cos y} = 0$

$\Rightarrow \tan x d x + \tan y d y = 0$

$\Rightarrow \tan y d y = - \tan x d x$

Integrate both side:

$\Rightarrow \int \tan y d y = - \int \tan x d x$

$\Rightarrow \log (\sec y) = - \log (\sec x) + C$

$\Rightarrow \log (\sec y) + \log (\sec x) = C$

$\Rightarrow \log (\sec x \sec y) = C$

$\Rightarrow \sec x \sec y = k (k = e^{C})$

As curve passes through $(0, \frac{π}{4})$ :

$\sec 0 \sec (\frac{π}{4}) = k$

$\Rightarrow k = \sqrt{2}$

$\Rightarrow \sec x \sec y = \sqrt{2}$

Thus the equation of required curve is $sec x sec y= \sqrt{2}$ .

7. Find the particular solution of the differential equation $(1+ e^{2x}) dy+ (1+ y^{2}) e^{x} dx=0$ given that $y=1$ when $x=0$ .

Ans: The given differential equation is:

$(1 + e^{2 x}) d y + (1 + y^{2}) e^{x} d x = 0$

Divide both side $(1+ e^{2x}) (1+ y^{2})$ :

$\frac{d y}{(1 + y^{2})} + \frac{e^{x}}{(1 + e^{2 x})} d x = 0$

$\int \frac{d y}{(1 + y^{2})} = - \int \frac{e^{x}}{(1 + e^{2 x})} d x$

$\tan^{- 1} y = - \int \frac{e^{x}}{(1 + {(e^{x})}^{2})} d x$

Substitute $t = e^{x}$ :

$d t = e^{x} d x$

$\Rightarrow \tan^{- 1} y = - \int \frac{1}{(1 + t^{2})} d t$

$\Rightarrow \tan^{- 1} y = - \tan^{- 1} t + C$

$\Rightarrow \tan^{- 1} y = - \tan^{- 1} e^{x} + C$

$\Rightarrow \tan^{- 1} y + \tan^{- 1} e^{x} = C$

As $y=1$ when $x=0$ :

$\tan^{- 1} (1) + \tan^{- 1} (e^{0}) = C$

$\Rightarrow \frac{π}{4} + \frac{π}{4} = C$

$\Rightarrow C = \frac{π}{2}$

$\Rightarrow \tan^{- 1} y + \tan^{- 1} e^{x} = \frac{π}{2}$

Thus the required particular solution is $ta n^{-1} y+ta n^{-1} e^{x} = \frac{π}{2}$ .

8. Solve the differential equation $y e^{\frac{x}{y}} dx= (x e^{\frac{x}{y}} + y^{2}) dy (y \neq 0)$ .

Ans: The given differential equation is:

$y e^{\frac{x}{y}} d x = (x e^{\frac{x}{y}} + y^{2}) d y$

$y e^{\frac{x}{y}} \frac{d x}{d y} = x e^{\frac{x}{y}} + y^{2}$

$\Rightarrow y e^{\frac{x}{y}} \frac{d x}{d y} - x e^{\frac{x}{y}} = y^{2}$

$\Rightarrow e^{\frac{x}{y}} \frac{[y \frac{d x}{d y} - x]}{y^{2}} = 1$

Substitute $z= e^{\frac{x}{y}}$ :

$z = e^{\frac{x}{y}}$

$\frac{d}{d y} z = \frac{d}{d y} e^{\frac{x}{y}}$

$\Rightarrow \frac{d z}{d y} = \frac{d}{d y} (e^{\frac{x}{y}})$

$\Rightarrow \frac{d z}{d y} = e^{\frac{x}{y}} \frac{d}{d y} (\frac{x}{y})$

$\Rightarrow \frac{d z}{d y} = e^{\frac{x}{y}} [(\frac{1}{y}) \frac{d x}{d y} - \frac{x}{y^{2}}]$

$\Rightarrow \frac{d z}{d y} = e^{\frac{x}{y}} [\frac{y \frac{d x}{d y} - x}{y^{2}}]$

$\Rightarrow \frac{d z}{d y} = 1$

$\Rightarrow d z = d y$

$\Rightarrow \int d z = \int d y$

$\Rightarrow z = y + C$

$\Rightarrow e^{\frac{x}{y}} = y + C$

Thus the required general solution is $e^{\frac{x}{y}} =y+C$ .

9. Find a particular solution of the differential equation $(x-y) (dx+dy) =dx-dy$ given that $y=-1$ when $x=0$ . Hint (put $x-y=t$ ).

Ans: Given differential equation is:

$(x - y) (d x + d y) = d x - d y$

$\Rightarrow (x - y) d x - d x = (y - x) d y - d y$

$\Rightarrow (x - y + 1) d y = (1 - x + y) d x$

$\Rightarrow \frac{d y}{d x} = \frac{1 - x + y}{x - y + 1}$

Put $x-y=t$ :

$x - y = t$

$\Rightarrow 1 - \frac{d y}{d x} = \frac{d t}{d x}$

$\Rightarrow 1 - \frac{d t}{d x} = \frac{d y}{d x}$

$\Rightarrow 1 - \frac{d t}{d x} = \frac{1 - t}{1 + t}$

$\Rightarrow \frac{d t}{d x} = 1 - \frac{1 - t}{1 + t}$

$\Rightarrow \frac{d t}{d x} = \frac{1 + t - 1 + t}{1 + t}$

$\Rightarrow \frac{d t}{d x} = \frac{2 t}{1 + t}$

$\Rightarrow \frac{1 + t}{t} d t = 2 d x$

Integrate both side:

$\Rightarrow \int \frac{1 + t}{t} d t = 2 \int d x$

$\Rightarrow \int \frac{1}{t} d t + \int d t = 2 x + C$

$\Rightarrow \log | t | + t = 2 x + C$

$\Rightarrow \log | x - y | + x - y = 2 x + C$

$\Rightarrow \log | x - y | - y = x + C$

As $y=-1$ when $x=0$ :

$\Rightarrow \log | 0 - (- 1) | - (- 1) = 0 + C$

$\Rightarrow \log 1 + 1 = C$

$\Rightarrow C = 1$

Thus the required particular solution is:

$\log | x - y | - y = x + 1$ .

10. Solve the differential equation $[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}] \frac{dx}{dy} =1 (x \neq 0)$ .

Ans: Given differential equation is:
$[\frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}] \frac{d x}{d y} = 1$

$\Rightarrow \frac{d y}{d x} = \frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}$

$\Rightarrow \frac{d y}{d x} + \frac{y}{\sqrt{x}} = \frac{e^{- 2 \sqrt{x}}}{\sqrt{x}}$

It is linear differential equation of the form $\frac{dy}{dx} +py=Q$ :

$p = \frac{1}{\sqrt{x}}$

$Q = \frac{e^{- 2 \sqrt{x}}}{\sqrt{x}}$

Calculating integrating factor:

$I . F = e^{\int p d x}$

$I . F = e^{\int \frac{1}{\sqrt{x}} d x}$

$I . F = e^{2 \sqrt{x}}$

The general solution is given by:

$y \times I . F = \int (Q \times I . F) d x + C$

$y \times (e^{2 \sqrt{x}}) = \int (\frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} \times e^{2 \sqrt{x}}) d x + C$

$\Rightarrow y e^{2 \sqrt{x}} = \int (\frac{e^{- 2 \sqrt{x} + 2 \sqrt{x}}}{\sqrt{x}}) d x + C$

$\Rightarrow y e^{2 \sqrt{x}} = \int \frac{1}{\sqrt{x}} d x + C$

$\Rightarrow y e^{2 \sqrt{x}} = 2 \sqrt{x} + C$

Thus the general solution for the given differential equation is

$y e^{2 \sqrt{x}} = 2 \sqrt{x} + C$ .

11. Find a particular solution of the differential equation $\frac{dy}{dx} +ycotx=4x cosec x (x \neq 0)$ given that $y=0$ when $x= \frac{π}{2}$ .

Ans:
The given differential equation is:

$\frac{d y}{d x} + y \cot x = 4 x cosec x$

It is linear differential equation of the form $\frac{dy}{dx} +py=Q$ :

$p = \cot x$

$Q = 4 x cosec x$

Calculating integrating factor:

$I . F = e^{\int p d x}$

$I . F = e^{\int \cot x d x}$

$I . F = e^{\log | \sin x |}$

$I . F = \sin x$

The general solution is given by:

$y \times I . F = \int (Q \times I . F) d x + C$

$\Rightarrow y \times \sin x = \int (4 x \cos ec x) \sin x d x + C$

$\Rightarrow y \sin x = 4 \int x d x + C$

$\Rightarrow y \sin x = 4 (\frac{x^{2}}{2}) + C$

$\Rightarrow y \sin x = 2 x^{2} + C$

As $y=0$ when $x= \frac{π}{2}$ :

$0 \times \sin (\frac{π}{2}) = 2 {(\frac{π}{2})}^{2} + C$

$C = - 2 (\frac{π^{2}}{4})$

$C = - \frac{π^{2}}{2}$

Thus the required particular solution is:

$y \sin x = 2 x^{2} - \frac{π^{2}}{2}$

12. Find a particular solution of the differential equation $(x+1) \frac{dy}{dx} =2 e^{-y} -1$ given that $y=0$ when $x=0$ .

Ans:
The given differential equation is:

$(x + 1) \frac{d y}{d x} = 2 e^{- y} - 1$

$\Rightarrow \frac{d y}{2 e^{- y} - 1} = \frac{d x}{x + 1}$

Integrate both side:

$\Rightarrow \int \frac{d y}{2 e^{- y} - 1} = \int \frac{d x}{x + 1}$

$\Rightarrow \int \frac{d y}{2 e^{- y} - 1} = \log (x + 1) + \log C$ ……(1)

Evaluating LHS integral:

$\Rightarrow \int \frac{d y}{2 e^{- y} - 1} = \int \frac{e^{y} d y}{2 - e^{y}}$

Put $t=2- e^{y}$ :

$t = 2 - e^{y}$

$d t = - e^{y} d y$

$\Rightarrow \int \frac{d y}{2 e^{- y} - 1} = - \int \frac{d t}{t}$

$\Rightarrow \int \frac{d y}{2 e^{- y} - 1} = - \log (t)$

$\Rightarrow \int \frac{d y}{2 e^{- y} - 1} = \log \frac{1}{t}$

$\Rightarrow \int \frac{d y}{2 e^{- y} - 1} = \log \frac{1}{2 - e^{y}}$

Back substituting in expression (1):

$\Rightarrow \int \frac{d y}{2 e^{- y} - 1} = \log (x + 1) + \log C$

$\Rightarrow \log (\frac{1}{2 - e^{y}}) = \log C (x + 1)$

$\Rightarrow 2 - e^{y} = \frac{1}{C (x + 1)}$

As $y=0$ when $x=0$ :

$\Rightarrow 2 - e^{0} = \frac{1}{C (0 + 1)}$

$\Rightarrow 2 - 1 = \frac{1}{C}$

$\Rightarrow C = 1$

Thus the required particular solution is:

$\Rightarrow 2 - e^{y} = \frac{1}{(x + 1)}$

$\Rightarrow e^{y} = 2 - \frac{1}{(x + 1)}$

$\Rightarrow e^{y} = \frac{2 x + 2 - 1}{(x + 1)}$

$\Rightarrow e^{y} = \frac{2 x + 1}{x + 1}$

$\Rightarrow y = \log (\frac{2 x + 1}{x + 1})$

Thus for given conditions the particular solution is $y=log (\frac{2x+1}{x+1})$ .

13. The general solution of the differential equation $\frac{ydx-xdy}{y} =0$ .

A. $x y = C$

B. $x = C y^{2}$

C. $y = C x$

D. $y = C x^{2}$

Ans: Given differential equation:

$\frac{y d x - x d y}{y} = 0$

Divide both side by $x$ :

$\Rightarrow \frac{y d x - x d y}{x y} = 0$

$\Rightarrow \frac{d x}{x} - \frac{d y}{y} = 0$

Integrate both side:

$\Rightarrow \int \frac{d x}{x} - \int \frac{d y}{y} = 0$

$\Rightarrow \log | x | - \log | y | = \log k$

$\Rightarrow \log | \frac{x}{y} | = \log k$

$\Rightarrow \frac{x}{y} = k$

$\Rightarrow y = C x (C = \frac{1}{k})$

Thus the correct option is (C)

14. Find the general solution of a differential equation of the type $\frac{dx}{dy} + P_{1} x= Q_{1}$ .

A. $y e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C$

B. $y e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d y + C$

C. $x e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C$

D. $x e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d x}) d y + C$

Ans: The given differential equation is:

$\frac{d x}{d y} + P_{1} x = Q_{1}$

It is a linear differential equation and its general solution is:

$x e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C$

With integrating factor $I . F = e^{\int P_{1} d y}$ .

Thus the correct option is (C).

15. Find the general solution of the differential equation $e^{x} dy+ (y e^{x} +2x) dx=0$ .

A. $x e^{y} + x^{2} =C$

B. $x e^{y} + y^{2} =C$

C. $y e^{x} + x^{2} =C$

D. $y e^{y} + x^{2} =C$

Ans: The given differential equation is:
$e^{x} d y + (y e^{x} + 2 x) d x = 0$

$\Rightarrow e^{x} \frac{d y}{d x} + y e^{x} = - 2 x$

$\Rightarrow \frac{d y}{d x} + y = - 2 x e^{- x}$

The given differential equation is of the form:

$\frac{d y}{d x} + P y = Q$

$\Rightarrow P = 1$

$\Rightarrow Q = - 2 x e^{- x}$

Calculating integrating factor:
$I . F = e^{\int P d x}$

$\Rightarrow I . F = e^{\int d x}$

$\Rightarrow I . F = e^{x}$

It is a linear differential equation and its general solution is:

$y (I . F) = \int (Q \times I . F) d x + C$

$y (e^{x}) = \int (- 2 x e^{- x} \times e^{x}) d y + C$

$\Rightarrow y e^{x} = - 2 \int x d x + C$

$\Rightarrow y e^{x} = - 2 (\frac{x^{2}}{2}) + C$

$\Rightarrow y e^{x} + x^{2} = C$

Thus the correct answer is option (C).

NCERT Solutions for Class 12 Maths Chapter 9 Important Points

Differential Equation: A differential equation has an independent variable, a dependent variable, derivatives of the dependent variable with respect to the independent variable, and a constant.
Ordinary Differential Equation: An ordinary differential equation is one that involves derivatives of the dependent variable with respect to only one independent variable.
Order of a Differential Equation: The order of a differential equation is defined as the highest order derivative of the dependent variable with respect to the independent variable.
Degree of a Differential Equation: The degree of a differential equation is the highest exponent of the highest order derivative if the exponent of each derivative is a non-negative integer and the unknown variable in the differential equation is a non-negative integer.
General solution: The general solution of a differential equation is one that contains as many arbitrary constants as the order of the differential equation, i.e., if the solution of a differential equation of order n has n arbitrary constants, it is the general solution.
Particular Solution: The particular solution is a solution obtained by giving particular values to arbitrary constants in the general solution of a differential equation.

Overview of Deleted Syllabus for CBSE Class 12 Maths Differential Equations

Chapter	Dropped Topics
Differential Equations	9.4 Formation of Differential Equations Whose General Solution is Given
	Page 415-416 Example 25
	Ques. 3, 5 and 15 (Miscellaneous Exercise)
	Point Six of the Summary

Class 12 Maths Chapter 9: Exercises Breakdown

Chapter 10 - Circles Exercises in PDF Format
Exercise 9.1	12 Questions (10 Short Questions, 2 MCQs)
Exercise 9.2	12 Questions (10 Short Questions, 2 MCQs)
Exercise 9.3	23 Questions (10 Short Questions, 12 Long Questions, 1 MCQs)
Exercise 9.4	17 Questions (15 Short Questions, 2 MCQs)
Exercise 9.5	19 Questions (2 Long Questions, 15 Short Questions, 2 MCQs)

Conclusion

The Solutions for Maths Chapter 9 class 12 differential equations NCERT solutions, provided by Vedantu, are designed to help students understand the ability to analyse and solve problems involving rates of change and relationships between variables - a valuable skill in various scientific disciplines. This chapter provides tools to analyse and solve equations, which have numerous applications in various scientific fields.

From previous year's question papers, around 8 questions are typically asked from this chapter. Understanding and practising these solutions will help students score well on their exams.

Other Study Material for CBSE Class 12 Maths Chapter 9

S.No.	Important Links for Chapter 9 Differential Equations
1	Class 12 Differential Equations Important Questions
2	Class 12 Differential Equations Revision Notes
3	Class 12 Differential Equations Formulas
4	Class 12 Differential Equations NCERT Exemplar Solutions
5	Class 12 Differential Equations RD Sharma Solutions
6	Class 12 Differential Equations RS Aggarwal Solutions

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

S.No.	NCERT Solutions Class 12 Maths Chapter-wise List
1	Chapter 1 - Relations and Functions Solutions
2	Chapter 2 - Inverse Trigonometric Functions Solutions
3	Chapter 3 - Matrices Solutions
4	Chapter 4 - Determinants Solutions
5	Chapter 5 - Continuity and Differentiability Solutions
6	Chapter 6 - Application of Derivatives Solutions
7	Chapter 7 - Integrals Solutions
8	Chapter 8 - Application of Integrals Solutions
9	Chapter 10 - Vector Algebra Solutions
10	Chapter 11 - Three Dimensional Geometry Solutions
11	Chapter 12 - Linear Programming Solutions
12	Chapter 13 - Probability Solutions

Explore these essential links for NCERT Class 12 Maths in Hindi, providing detailed solutions, explanations, and study resources to help students excel in their mathematics exams.

S.No.	Related NCERT Solutions for Class 12 Maths
1	NCERT Book for Class 12 Maths in Hindi
2	NCERT Solutions for Class 12 Maths in hindi

S.No	Important Resources Links for Class 12 Maths
1	CBSE Class 12 Maths Revision Notes
2	CBSE Syllabus for Class 12 Maths
3	CBSE Class 12 Maths Important Questions
4	CBSE Class 12 Maths Previous Year Question Paper
5	CBSE Class 12 Maths Sample Paper
6	NCERT Books for Class 12 Maths
7	NCERT Exemplar for Class 12 Maths
8	CBSE Class 12 Maths Formulas
9	RS Aggarwal Class 12 Maths Solutions
10	RD Sharma Class 12 Maths Solutions

FAQs on NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

1. What is a differential equation as stated in Class 12 books?

Differential equations assist you in differentiating any function w.r.t an independent variable. It describes a relationship between the functions and their derivatives. The functions refer to the physical quantities whereas their derivatives refer to the rate at which the function is changing and this relationship is differential equations. A D.E. takes the form dy/dx = g(x). Here, y signifies the function. And the function cited here is f(x).

2. How do you solve differential equations?

Differential equations can be of different orders. There are various approaches to solve these equations. For solving first-order linear differential equations, we need to perform the substitution. Then, after part factorization, one can perform the separation of variables. Once this step is successfully done, we can substitute the original value. Perform these steps until the solution is obtained for the original equation.

3. Is Class 12 Maths Chapter 9 tough?

Practice makes Maths easy for any student. Once the student is able to clearly grasp the basic concepts and learns the tricks to solve each question, only then will Maths be easy for that student. Hence, it is important to learn the basics well and practise thoroughly for all types of questions with sincerity. The Class 12 Maths syllabus is a mix of different questions in terms of difficulty. Visit the page NCERT Solutions Class 12 Maths Chapter 9 for the solutions

4. What is the order and degree of the differential equation? Give an example.

Order simply helps to find out the order of the highest term in any D.E., which is basically the term with the highest exponent value. For instance, in this equation, dy/dx + 4 = 2, the order is 1. The degree of any D.E. is associated with its order. The power raised to which the highest term is characterised is the degree of the D.E. In the above-cited example of the 1st order D.E., the degree of the D.E. is 1. Modules relating to this topic or other topics covered in this chapter can be found on the Vedantu website or on the Vedantu app at free of cost.

5. What is the best Solution book for NCERT Class 12 Maths Chapter 9?

You may acquire NCERT Class 12 Math Solutions by going to the Vedantu website and searching for Class 12 Maths solutions. Aside from that, you may access a variety of modules that will assist you in achieving high marks in Maths examinations. The exercise solutions are provided on the page NCERT Solutions Class 12 Maths Chapter 9. Click on it to download a PDF of the solutions.

6. What kind of problems are covered in the NCERT Solutions for Differential Equations Class 12 NCERT Solutions?

The solutions address various problems related to:

Identifying the order and degree of differential equations.
Forming differential equations from real-world scenarios.

Solving differential equations using methods like variable separation, homogeneous equations, and linear equations (for beginners).
Applying integrating factors (for slightly advanced problems).
Understanding applications of differential equations in areas like population growth, motion, and electrical circuits.

7. Are there solutions for differential equation class 12 for advanced techniques like Laplace Transform?

The NCERT solutions primarily focus on methods for beginners and intermediate learners. For advanced techniques like Laplace Transform, you might need to refer to other resources like advanced mathematics textbooks or online tutorials dedicated to the topic.

8. What is a differential equation?

A differential equation is a mathematical equation that relates a function with its derivatives. It describes the rate at which a quantity changes. For example, dy/dx =f(x,y) is a first-order differential equation.

9. How are exact differential equations solved in differential equations class 12 NCERT solutions?

Exact differential equations are solved by finding a potential function whose partial derivatives match the terms in the differential equation. If the equation is not exact, an integrating factor might be used to make it exact.

10. What are the applications of differential equations Class 12?

Differential equations are used in various fields such as physics (motion, heat, waves), biology (population models), economics (growth models), and engineering (circuit analysis).

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

NCERT Maths Chapter 9 Differential Equations Class 12 Solutions - Free PDF Download

Glance on Maths Chapter 9 Class 12 - Differential Equations

Access Exercise wise NCERT Solutions for Chapter 9 Maths Class 12

Exercises Under NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations

Access NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations

Exercise 9.1

Exercise 9.2

Exercise 9.3

Exercise 9.4

Exercise 9.5

Miscellaneous Exercise

NCERT Solutions for Class 12 Maths Chapter 9 Important Points

Overview of Deleted Syllabus for CBSE Class 12 Maths Differential Equations

Class 12 Maths Chapter 9: Exercises Breakdown

Conclusion

Other Study Material for CBSE Class 12 Maths Chapter 9

NCERT Solutions for Class 12 Maths | Chapter-wise List

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FAQs on NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations