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Class 12 Important Question

Maths

Chapter 9: Differential Equations

CBSE Class 12 Maths Important Questions - Chapter 9 Differential Equations

Important Questions for CBSE Class 12 Maths Chapter 9 Differential Equations FREE PDF Download

Looking for important questions from Chapter 9, Differential Equations, for CBSE Class 12 Maths? You’re in the right place! We’ve put together a collection of key questions according to the Latest Class 12 Maths Syllabus to help you prepare for your exams. This FREE PDF includes important topics like solving differential equations, applications in real-life problems, and methods like separation of variables and integrating factors. Download Class 12 Maths important Questions now to boost your preparation and practise effectively!

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Access Important Question for Class 12 Mathematics Chapter 9 - Differential Equations

Very Short Answer Type Questions (1 Mark)

1. Write the order and degree of the following differential equations.

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

Ans: $\frac{d y}{d x} + \cos y = 0$

$y^{'} + \cos x = 0$

Highest order of derivative $= 1$

$∴$ Order $= 1$

Degree = Power of $y^{'}$

Degree $= 1$

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

Ans:

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

Highest order of derivative = 2

Order = 2

Degree = Power of y

Degree = 1

(iii) $\frac{\partial^{4} y}{\partial x^{4}} + \sin x = (\frac{\partial^{2} y}{\partial x^{2}})^{5} .$

Ans:

$\frac{\partial^{4} y}{\partial x^{4}} + \sin x = (\frac{\partial^{2} y}{\partial x^{2}})^{5}$

Highest order of derivative = 4

Order = 4

Degree = Power of y

Degree = 1

(iv) $\frac{\partial^{5} y}{\partial x^{5}} + \log (\frac{d y}{d x}) = 0$

Ans:

$\frac{\partial^{5} y}{\partial x^{5}} + \log (\frac{d y}{d x}) = 0$

Highest order of derivative = 5

Order = 5

Degree = Power of y

Degree = not defined

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

Ans:

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

Highest order of derivative = 2

Order = 2

Degree = Power of y

Degree = 2

(vi) $[1 + (\frac{d y}{d x})^{2}]^{\frac{3}{2}} = k \frac{\partial^{2} y}{\partial x^{2}}$

Ans:

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

Squaring on both sides

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

Highest order of derivative = 2

Order = 2

Degree = Power of y

Degree = 2

(vii) $(\frac{\partial^{3} y}{\partial x^{3}})^{2} + (\frac{\partial^{2} y}{\partial x^{2}})^{3} = \sin x$

Ans:

$(\frac{\partial^{3} y}{\partial x^{3}})^{2} + (\frac{\partial^{2} y}{\partial x^{2}})^{3} = \sin x$

Highest order of derivative = 3

Order = 3

Degree = Power of y

Degree = 2

(viii) $\frac{d y}{d x} + \tan (\frac{d y}{d x}) = 0$

Ans:

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

Highest order of derivative = 1

Order = 1

Degree = Power of y

Degree = Not defined

2 Write the general solution of following differential equations

(i) $\frac{d y}{d x} = x^{5} + x^{2} - \frac{2}{x}$

Ans:

$\frac{d y}{d x} = x^{5} + x^{2} - \frac{2}{x}$

Integrating on both side

$\int \frac{d y}{d x} d x = \int x^{5} d x + \int x^{2} d x - \int \frac{2}{x} d x$

$y = \frac{x^{6}}{6} + \frac{x^{3}}{3} - 2 \log | x | + c$

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

Ans:

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

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

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

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

Integrating both sides.

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

$y = \int \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} dx$

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

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

$dx = \frac{d t}{e^{x} - e^{- x}}$

Putting value of $t$ and $dt$ in (1)

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

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

$y = \log | t | + c$

Putting back $t = e^{x} - e^{- x}$

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

(iii) $\frac{d y}{d x} = x^{3} + e^{x} + x^{e}$

Ans:

$\frac{d y}{d x} = x^{3} + e^{x} + x^{e}$

$d y = (x^{3} + e^{x} + x^{e}) d x$

Integrating on both sides

$\int d y = \int (x^{3} + e^{x} + x^{e}) d x$

$y = \frac{x^{4}}{4} + e^{x} + \int x^{e} d x$

$y = \frac{x^{4}}{4} + e^{x} + \frac{x^{e + 1}}{e + 1} + c$

(iv) $\frac{d y}{d x} = 5^{x + y}$

Ans:

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

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

$\frac{1}{5^{y}} d y = 5^{x} d x$

Integrating on Both sides

$\int \frac{1}{5^{y}} d y = \int 5^{x} d x$

$5^{- y} = 5^{x} + c$

$5^{x} + 5^{- y} = c$

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

Ans:

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

We know that

$\cos 2 x = 2 \cos^{2} x - 1$

Putting $x = \frac{x}{2}$

$\cos \frac{2 x}{2} = 2 \cos^{2} \frac{x}{2} - 1$

$\cos x = 2 \cos^{2} \frac{x}{2} - 1$

$1 + \cos x = 2 \cos^{2} \frac{x}{2}$

We know

$\cos 2 x = 1 - 2 \sin^{2} x$

Putting $x = \frac{x}{2}$

$\cos^{2} \frac{2 x}{2} = 1 - 2 \sin^{2} \frac{x}{2}$

$\cos x = 1 - 2 \sin^{2} \frac{x}{2}$

$1 - \cos x = 2 \sin^{2} \frac{x}{2}$

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

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

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

Integrating on both sides

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

$y - \frac{y}{2} + \frac{\sin 2 y}{4} = \frac{x}{2} - \frac{\sin 2 x}{4} + c$

$\frac{y}{2} + \frac{\sin 2 y}{4} = \frac{x}{2} - \frac{\sin 2 x}{4} + c$

By equating

$2 y + \sin 2 y = 2 x - \sin 2 y + c$

$2 y - 2 x + \sin 2 y - \sin 2 x = c 2 (y - x) + \sin 2 y - \sin 2 x = c$

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

Ans:

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

$2 y + \sin 2 y = 2 x - \sin 2 y + c$

$2 y - 2 x + \sin 2 y - \sin 2 x = c 2 (y - x) + \sin 2 y - \sin 2 x = c$

Integrating on both sides

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

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

$2 \log (3 x + 1) + 3 \log (1 - 2 y) = c$

3. Write integrating factor of the following differential equations.

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

Ans:

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

$d y / d x + y = \cos x - \sin x$ is a linear differential equation of the type $\frac{d y}{d x} + p y = Q$

Here $I . F . = e^{\int 1 . d x} = e^{x}$

Its solution is given by $\Rightarrow y e^{x} = \int e^{x} (\cos x - \sin x) d x$

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

Integrate by parts $\Rightarrow y e^{x} = e^{x} \cos x - \int - \sin x e^{x} d x - \int e^{x} \sin d x$

$∴ y e^{x} = e^{x} \cos x + C$

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

(II) $\frac{d y}{d x} + y \sec^{2} x = \sec x + \tan x$

Ans:

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

$\frac{d y}{d x} + P y = Q$ .

$\int e^{\tan x} (\frac{d y}{d x} + y \sec^{'} x) d x = \int e^{\tan x} \cdot \tan x \sec x$

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

$I = \int e^{\tan x} \tan x \sec^{1} x \cdot d x$

$t = \tan x$

$d t = \sec^{2} u d n$

$\int_{\frac{1}{I}} e^{t} t d t$

$\Rightarrow \int e^{t} t - \int e^{t} d t$

$tar x e^{t} \tan x - e^{t} - e^{\tan x} + c$

$y e^{\tan x} = \tan x \cdot e^{\tan x} - e^{\tan x} + C$

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

Ans:

$\frac{d y}{d x} = \frac{y}{x} [\log \frac{y}{x} + 1]$

$y = V x$

$\int \frac{d V}{V \log V} = \int \frac{d x}{x}$

$\log V = x c$

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

(IV) $x \frac{d y}{d x} - 3 y = x^{2}$

Ans:

$x \frac{d y}{d x} - 3 y = x^{2}$

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

$\frac{d y}{d x} + \frac{3}{x} \cdot y = x$

$y . I . F = \int I . F \cdot q (x) \cdot d x$

$p (x) = \frac{3}{x} q (x) = x$

$I . F_{.} = e^{\int \frac{3}{x} \cdot d x}$

$= e^{3 \ln x}$

$= {(e)}^{\ln x^{3}}$

$= x^{3}$

$y \cdot x^{3} = \int x^{3} \cdot x \cdot d x$

$y \cdot x^{3} = \frac{x^{5}}{5} + c$

$y = \frac{x^{2}}{5} + \frac{c}{x^{3}}$

(V) $\frac{d y}{d x} + y \tan x = \sec x$

Ans:

We have, $\frac{d y}{d x} + y \tan x = \sec x$

which is a linear differential equation Here, $P = \tan x, Q = \sec x$ ,

$∴$ I.F. $= e^{\int \tan x d x} = e^{\log \sec x} = \sec x$

$∴$ The general solution is

$y \sec x = \int \sec x \cdot \sec x + C$

$\Rightarrow y \sec x = \int \sec^{2} x d x + C$

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

4. Write order of the differential equation of the family of following curves

(I) $y = A e^{x} + B e^{x + c}$

Ans:

$y = A e^{x} + B e^{x + c}$

$y = e^{x} (A + B e^{c}) = {Re}^{x}$

$\frac{d y}{d x} = A e^{x} + B e^{x + c} \frac{d y}{d x} = y$

Therefore, order is 1.

(II) $A y = B x^{2}$

Ans:

$A y = B x^{2}$

Differentiating wrt x

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

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

Therefore, order is 1.

(III) ${(x - a)}^{2} + {(y - b)}^{2} = 9$

Ans:

${(x - a)}^{2} + {(y - b)}^{2} = 9$

Differentiate w.r.t x

$\frac{d}{d x} {(x - a)}^{2} + \frac{d}{d x} {(y - b)}^{2} = 9$

$2 (x - a) \frac{d}{d x} (x - a) + 2 (y - b) \frac{d}{d x} (y - b) = 0$

$2 (x - a) (\frac{d y}{d x} - 0) + 2 (y - b) (\frac{d y}{d x} - 0) = 0$

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

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

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

$\frac{d}{d x} (x - a) = \frac{d}{d x} [(b - y) \frac{d y}{d x}]$

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

$[\frac{d}{d x} (b - y)] 1 - 0 = (b - y) \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} (- \frac{d y}{d x})$

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

Hence, the order is 2.

(IV) $A x + B y^{2} = B x^{2} - A y$

Ans:

$A x + B y^{2} = B x^{2} - A y$

$A x + A y = B x^{2} - B y^{2}$

$A (x + y) = B (x^{2} - y^{2})$

$A (x + y) = B (x + y) (x - y)$

$A = B (x - y)$

$\frac{A}{B} = x - y$

$\frac{A}{B} + y = x$

Differentiate w.r.t x

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

Therefore, the order is 1.

(V) $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 0$

Ans:

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 0$

$\frac{x^{2}}{a^{2}} = \frac{y^{2}}{b^{2}}$

$x^{2} = \frac{a^{2}}{b^{2}} y^{2}$

$x = \pm \frac{a}{b} y$

$x = \frac{a}{b} y$

$1 = \frac{a}{b} \frac{d y}{d x}$

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

$- \frac{a}{b} y = x$

$- \frac{a}{b} \frac{d y}{d x} = 1$

$\frac{d y}{d x} = \frac{- b}{a}$

Therefore, the order is 1.

(VI) $y = a c o s (x + b)$

Ans:

Given $y = a c o s (x + b)$

Differentiating it w.r.t x

$\frac{d x}{d y} = - a s i n (x + b)$

Therefore, the order is 1.

(VII) $y = a + b e^{x + c}$

Ans:

$y = a + b e^{x + c}$

$b e^{x + c} = y - a$

$\frac{d y}{d x} = b e^{x + c}$

$\frac{d y}{d x} = b e^{x + c} = y - a$

$diff - wrt x$

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

Hence, the order is 2.

5. (i) Show that $y = e^{m \sin^{- 1} x}$ is a solution of

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

Ans:

Using the Chain Rule, we get,

$\Rightarrow y_{1} = \frac{d y}{d x} = (e^{m \sin^{- 1} x}) \cdot \frac{d}{d x} (m \sin^{- 1} x)$

$= y \cdot {\frac{m}{\sqrt{1 - x^{2}}}}$

$\Rightarrow y_{1} \cdot \sqrt{1 - x^{2}} = m y .$

$y_{1}^{2} (1 - x^{2}) = m^{2} y^{2}$

W.r.t x, using the product and chain rule

$y_{1}^{2} \cdot \frac{d}{d x} (1 - x^{2}) + (1 - x^{2}) \cdot \frac{d}{d x} (y_{1}^{2}) = m^{2} \frac{d}{d x} (y^{2}),$

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

$\Rightarrow y_{1}^{2} (- 2 x) + (1 - x^{2}) (2 y_{1}) (y_{2}) = m^{2} (2 y) (y_{1})$

Dividing by $2 y_{1} \neq 0$ , we get,

$(1 - x^{2}) y_{2} - x y_{1} = m^{2} y$

Hence, the Proof.

(ii) Show that $y = \sin (\sin x)$ is a solution of differential equation

$\frac{d^{2} y}{d x^{2}} + (\tan x) \frac{d y}{d x} + y \cos^{2} x = 0$

Ans:

$y = sin (sin x)$

$dy/dx = cos (sinx) . cosx$

$and d^2y/d x^{2} = -sin (sinx) . cos2x - sinx cos (sinx)$

$LHS = - sin (sinx) cos2x - sinx cos (sinx) + sinx/ cosx cos (sinx) cosx + sin (sinx) cos2 x$

$= 0 = RHS$

$y = sin(sinx)$

$\Rightarrow dy/dx = cosxcos(sinx)$

$\Rightarrow d^{2} y/d x^{2} = -cosx \cdot cosx \cdot sin(sinx) - sinx \cdot cos(sinx)$

$\Rightarrow d^{2} y/d x^{2} = -yco s^{2} x - (sinx/cosx)dy/dx$

$= d^{2} y/d x^{2} + tanx(dy/dx) + yco s^{2} x$

$= 0$

(III) Show that $y = A x + \frac{B}{x}$ is a solution $\frac{x^{2} d^{2} y}{d x^{2}} + x \frac{d y}{d x} - y = 0$

Ans:

Here ‘a’ and ‘b’ are arbitrary

Given solution

$y = a x + (b / x), x \neq 0$

$y = a x + (b / x) . . . (1)$

$x y = a x^{2} + b$

Differentiate with respect to ‘x’

$x y + y . 1 = a (2 x) = 2 a x . . . (2)$

Differentiate with respect to ‘x’

Differentiate again w.r.t ‘x’

$x y + y . 1 + y = 2 a \Rightarrow x y + 2 y = 2 a \dots (3)$

Substitute (3) in (2)

$x y + y = (x y + 2 y) x$

$x y + y = x^{2} y + 2 x y$

$= x^{2} y + x y y$

$= 0$

Hence, Proved

(IV) Show that $y = a \cos (\log x) + b \sin (\log x)$ is a solution of

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

Ans:

$y = a \cos (\log x) + b \sin (\log x)$

Differentiating w.r.t. x,

$\frac{d y}{d x} = - a \sin (\log x) \times \frac{1}{x} + b \cos (\log x) \times \frac{1}{x} \frac{d^{2} y}{d x^{2}}$

$= - a [\frac{\cos (\log x) \times \frac{1}{x} \times x - 1 \cdot \sin (\log x)}{x^{2}}] + b [\frac{- \sin (\log x) \times \frac{1}{x} \times x - 1 \cdot \cos (\log x)}{x^{2}}]$

$\Rightarrow x^{2} \frac{d^{2} y}{d x^{2}} = - a \cos (\log x) + a \sin (\log x) - b \sin (\log x) - b \cos (\log x)$

$\Rightarrow x^{2} \frac{d^{2} y}{d x^{2}} = - [a \cos (\log x) + b \sin (\log x)] - [- a \sin (\log x) + b \cos (\log x)]$

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

$[(1) and (2)]$

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

(V) Verify that $y = \log (x + \sqrt{x^{2} + a^{2}})$ satisfies the differential equation:

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

Ans: Given

$y = l o g (x + \sqrt{x^{2} + a^{2}})$

On differentiating with x, we get

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

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

On differentiating again with $x$ , we get

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

Now let's see what is the value of $\frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x}$

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

Conclusion: Therefore,

$y = \log (x + \sqrt{x^{2} + a^{2}})$

Is not the solution of $\frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = 0$

(VI) Find the differential equation of the family of curves $y = e^{x} (A c o s x + B s i n x)$ , where A and B are arbitrary constants.

Ans:

$y = e^{x} (A c o s x + B s i n x) . . . . . (1)$

$∴ (d y / d x) = e^{x} [A s i n x + B c o s x] + [A c o s x + B s i n x] e x$

$∴ (d y / d x) = e^{x} [(B A) s i n x + (B + A) c o s x]$

$∴ (d y / d x) = e^{x} [(A + B) c o s x (A B) s i n x] . . . . . (2)$

$∴ (d^{2} y / d x^{2}) = e^{x} [(A + B) c o s x (A B) s i n x] + e^{x} [(A + B) s i n x (A B) c o s x]$

$∴ (d^{2} y / d x^{2}) = e^{x} [2 B c o s x 2 A s i n x]$

$∴ (d^{2} y / d x^{2}) = 2 e^{x} [B c o s x A s i n x]$

$∴ (d^{2} y / d x^{2}) \times (1 / 2) = e^{x} [B c o s x A s i n x] . . . . . (3)$

$(1) + (3) y + (1 / 2) (d^{2} y / d x^{2}) = e^{x} [(A + B) c o s x (A B) s i n x]$

$∴ y + (1 / 2) (d^{2} y / d x^{2}) = (d y / d x) - - - - - - - (2)$

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

(vii) Find the differential equation of an ellipse with major and minor axes 2a and 2b respectively.

Ans:

Equation of ellipse

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Differentiate w.r.t x

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

$y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = \frac{- p b^{2}}{k a^{2}} = \frac{- b^{2}}{a^{2}} = \frac{y}{x} \frac{d y}{d x}$

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

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

(viii) Form the differential equation representing the family of curves ${(y - b)}^{2} = 4 (x - a) .$

Ans:

The general equation of the given family of curves

$(y - b)^{2} = 4 (x - a) \dots (i)$

Differentiating (i) w.r.t.x, we

get

$2 (y - b) \frac{d y}{d x} = 4 \Rightarrow (y - b) y^{'} = 2 . . . (i i)$

Where $\frac{d y}{d x} = y^{'} .$

Differentiating (ii) w.r.t.y, we

get

$(y-b){y}''+{{\left( {{y}'} \right)}^{2}}=0\quad\text{ }\!\!~\!\!\text{ }\ldots

(iii)$,

where $\frac{d^{2} y}{d x^{2}} = y^{″}$

Putting $(y - b) = \frac{2}{y^{'}}$ from (ii) in (iii),

we get $\frac{2 y^{″}}{y^{'}} + {(y^{'})}^{2} = 0 \Rightarrow 2 y^{″} + {(y^{'})}^{3} = 0$

Hence, $2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{3} = 0$ is the required differential equation.

6. Solve the following differential equations.

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

Ans:

$\frac{d y}{d x} + y \cot x = \sin 2 x$

Comparing equation (1) by $\frac{d y}{d x} + P y = Q$

$P = \cot x, Q = \sin x$

$∴$ I.F. $= e^{\int \cot x d x}$

$\Rightarrow$ I.F $= e^{\log \sin x} = \sin x$

Multiplying equation (1) by $\sin x$ $\sin x \frac{d y}{d x} + \sin x \cdot \cot x y = \sin^{2} x$

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

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

$\int [\frac{d}{d x} (\sin x \cdot y)] d x = \int 1 d x - \int \cos^{2} x d x$

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

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

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

$\Rightarrow y \sin x = \frac{1}{2} x - \frac{1}{4} \sin 2 x + C$

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

Ans:

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

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

Comparing equation (1) by $\frac{d y}{d x} + P y = Q$

$P = \frac{2}{x}, Q = x \log x$

. I.F. $= e^{\int \frac{2}{x} d x} = e^{2 \log x} = e^{\log x^{2}} = x^{2}$

Multiplying equation (1) by $x^{2}$

$x^{2} \frac{d y}{d x} + x^{2} \times \frac{2}{x} y = x^{3} \log x$

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

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

Integrating both sides w.r.t. $x$

$x^{2} y = \int x^{3} \log x d x + C$

$= \log x \int x^{3} d x - \int {\frac{d}{d x} (\log x) \int x^{3} d x} d x + C$

(Taking $\log x$ as first function) $= \frac{x^{4}}{4} \log x - \int \frac{1}{x} \times \frac{x^{4}}{4} d x + C$

$= \frac{x^{4}}{4} \log x - \frac{1}{4} \int x^{3} d x + C$

$= \frac{x^{4}}{4} \log x - \frac{1}{4} \times \frac{x^{4}}{4} + C$

$\Rightarrow x^{2} y = \frac{1}{16} x^{4} [4 \log x - 1] + C$

$16 x^{2} y = 4 x^{4} \log x - x^{4} + C$

(III) $\frac{d x}{d y} + \frac{1}{x} . y = \cos x + \frac{\sin x}{x}, x > 0$

Ans: $\frac{d x}{d y} + \frac{1}{x} . y = \cos x + \frac{\sin x}{x}, x > 0$

$P = \frac{1}{x}, Q = \cos x + \frac{\sin x}{x}$

$I .F = e^{\int P d x} = e^{\int \frac{1}{x} d x} = e^{\log x} = x$

$y (I .F) = \int Q \cdot (I .F) d x$

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

$= \int \cos x x d x + \int \sin x d x$

$= x \sin x - \int \sin x d x + \int \sin x d x + c$

$= x \sin x + c$

(Iv) $\cos^{3} x \frac{d y}{d x} + \cos x = \sin x$

Ans:

$\cos^{3} x \frac{d y}{d x} + \cos x = \sin x$

Differentiate w.r.t y

$\frac{d y}{d x} + \frac{y}{\cos^{2} n} = \frac{\sin x}{\cos^{3} x}$

$\frac{d y}{d x} + y \cdot \sec^{2} x = \frac{\sin x}{\cos^{3} x}$

$I \cdot F \cdot = e^{\int \sec^{2} x d x} = e^{\tan x}$

$y \cdot e^{\tan x} = \int e^{t \tan x} \tan x \cdot \sec^{2} x d x$

$= \int e^{t} \cdot t d t$

$= e^{t} (t - 1) + c$

$y \cdot \tan x = e^{\tan x} (\tan x - 1) + c$

(V) $y d x + (x - y^{3}) d y = 0$

Ans:

The given differential equation is

$y d x + (x - y^{3}) d y = 0$

$\Rightarrow y \frac{d x}{d y} + (x - y^{3}) = 0$

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

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

which is a linear differential equation

Thus, $IF = e^{\int \frac{d y}{y}} = e^{\log y} = y$

Multiplying both sides of E (i) by IF and integrating,

we get

$x \cdot (IF) = \int Q \cdot (IF) d y + c$

$\Rightarrow x \cdot y = \int y^{3} d y + c$

$= \frac{y^{4}}{4} + c$

which is the required solution.

(VI) $y e^{y} d x = (y^{3} + 2 x e^{y}) d y$

Ans:

$y e^{y} d x = (y^{3} + 2 x e^{y}) d y$

$\frac{y e^{y}}{(y^{3} + 2 x e^{y})} = \frac{d y}{d x}$ or, $\frac{d x}{d y} + (- \frac{2}{y}) x = \frac{y^{2}}{e^{y}}$ ,

which is linear in $x$ .

I.F. $= e^{\int \frac{- 2}{y} d y} = e^{- 2 \log y} = \frac{1}{y^{2}}$

Multiplying both sides by the I.F. and integrating, we get,

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

$x \frac{1}{y^{2}} = - e^{- y} + C$ or $x = - y^{2} e^{- y} + c y^{2}$

When $x = 0, y = 1.0 = - e^{- 1} + cor c = \frac{1}{e}$ .

Hence, the required particular solution is

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

7. Solve each of the following differential equations:

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

Ans:

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

Given differential equation can be written as

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

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

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

$∴ \log | y | - \log | 1 - a y |$

$= \log | x + 2 | + \log C$

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

$y = C (x + 2) (1 - 2 y)$

(II) $c o s y d x + (1 + 2 e^{- x}) s i n y d y = 0.$

Ans:

$c o s y d x + (1 + 2 e^{- x}) s i n y d y = 0.$

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

$\Rightarrow \int \frac{e^{x}}{2 + e^{x}} d x = \int \frac{- \sin y}{\cos y} d y$

$\Rightarrow \ln (e^{x} + 2) = \ln | \cos y | + \ln C$

$\Rightarrow \ln (e^{x} + 2) = \ln | \cos y | C$

$\Rightarrow e^{x} + 2 = C \cos y \dots (1)$

$\Rightarrow e^{x} + 2 = \pm C \cos y \Rightarrow e^{x} + 2 = k \cos y$

$x = 0, y = \frac{π}{4} in (1)$

We get

$1 + 2 = kcos \frac{π}{4}$

$\Rightarrow k = 3 \sqrt{2}$

$∴ e^{x} + 2 = 3 \sqrt{2} \cos y$

Is the particular solution.

(III) $x \sqrt{1 + y^{2}} d x + y \sqrt{1 + x^{2}} d y = 0$

Ans:

$x \sqrt{1 + y^{2}} d x + y \sqrt{1 + x^{2}} d y = 0$

It is given that

$x \sqrt{1 + y^{2}} d x + y \sqrt{1 + x^{2}} d y = 0$

We can write it as

$\frac{x}{\sqrt{1 + x^{2}}} d x + \frac{y}{\sqrt{1 + y^{2}}} d y = 0$

By integrating the given terms, we get

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

Using the formula

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

We get

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

(IV) $\sqrt{(1 - x^{2}) (1 - y^{2})} d y + x y d x = 0$

Ans:

$\sqrt{(1 - x^{2}) (1 - y^{2})} d y + x y d x = 0$

By simplifying the equation, we get

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

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

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

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

Integrating both sides, we get

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

$1 - y^{2} = t \Rightarrow 2 y d y = d t$

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

$∴ (i) \Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{t}} d t = - \int \frac{m}{m^{2} - 1} \cdot m d m$

$\Rightarrow \frac{1}{2} \frac{t^{1 / 2}}{1 / 2} + \int \frac{m^{2}}{m^{2} - 1} d m = 0 \Rightarrow \sqrt{t} + \int \frac{m^{2} + 1 - 1}{m^{2} - 1} d m = 0$

$\Rightarrow \sqrt{t} + \int (1 + \frac{1}{m^{2} - 1}) d m = 0 \Rightarrow \sqrt{t} + m + \frac{1}{2} \log | \frac{m - 1}{m + 1} | = 0$

Now substituting this value of $t$ and $m$ , we get

$\sqrt{1 - y^{2}} - \sqrt{1 - x^{2}} + \frac{1}{2} \log | \frac{\sqrt{1 - y^{2}} - 1}{\sqrt{1 - y^{2}} + 1} | + C = 0$

(V) $(x y^{2} + x) d x + (y x^{2} + y) d y = 0; y (0) = 1.$

Ans:

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

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

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

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

Integrating both sides

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

This is the integrals of the type

$\int \frac{f^{'} (x)}{f (x)} d x = \log | f (x) | + c \log | x^{2} + 1 |$

$= - \log | y^{2} + 1 | + \log c \log | x^{2} + 1 | + \log | y^{2} + 1 |$

$= \log c \log | x^{2} + 1 | | y^{2} + 1 | = \log c$

$\Rightarrow c = | x^{2} + 1 | | y^{2} + 1 |$

(VI) $\frac{d y}{d x} = y \sin^{3} x c o x^{2} x + x e^{x}$

Ans:

$\frac{d y}{d x} = y \sin^{3} x c o x^{2} x + x e^{x}$

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

$d y = [(\sin^{3} x \cdot \cos^{2} x) + x \cdot e^{x}] \cdot d x \to 2$

Taking the integral sign in both side of equation (2)

$\int d y = \int \sin^{3} x \cdot \cos^{2} x d x + \int x \cdot e^{x} \cdot d x + c$

$\int \sin^{3} x \cdot \cos^{2} x d x wt \cos x = t$

$\int \sin x \cdot \sin^{2} x \cdot \cos^{2} x d x$

$\int \sin x (1 - \cos^{2} x) \cdot \cos^{2} x d x$

$- \int (1 - t^{2}) \cdot t^{2} d t \Rightarrow - \int t^{2} - t^{4} d t$

$= - [\frac{t^{3}}{3} - \frac{t^{5}}{5}]$

$\Rightarrow - [\frac{\cos^{3} x}{3} - \frac{\cos^{5} t}{s}]$

$\int x \cdot e^{x} d x = x \int e^{n} d x - \int \frac{d}{d x} x \cdot \int e^{x} d x] d x$

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

$= x e^{x} - e^{x}$

$= e^{x} (x - 1)$

From Equation (3)

$y = - [\frac{\cos^{3} x}{3} - \frac{\cos^{5} x}{5}] + e^{n} (x - 1) + C$

(VII) $\sec^{2} x \tan y d x + \sec^{2} y \tan x d y = 0$

Ans:

Given equation:

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

Dividing both sides by $\tan y \tan x$

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

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

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

Integrating both sides

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

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

$u = \tan x, v = \tan y$

Diff u w.r.t. x and v w.r.t y

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

$\frac{d u}{\sec^{2} x} = dx$

$\frac{d v}{d y} = \sec^{2} y$

$\frac{d v}{\sec^{2} y} = dy$

Therefore, our equation becomes

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

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

$\int \frac{d u}{u} + \int \frac{d v}{v} = 0$

$\log | u | + \log | v | = \log c$

$u = \tan x and v = \tan y$

$\log | \tan x | + \log | \tan y | = \log c$

$\log | \tan x + \tan y | = \log c$

$\tan x \tan y = C$

8. Solve the following differential equations:

(I) $x^{2} y d x - (x^{3} + y^{3}) d y = 0$

Ans:

$(x^{3} + y^{3}) d y - x^{2} y d x = 0$

Is rearranged as

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

$\frac{y}{x} = v \Rightarrow y = v x$

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

$∴ v + x \frac{d v}{d x} = \frac{v}{1 + v^{3}}$

$\Rightarrow x \frac{d v}{d x} = \frac{v}{1 + v^{3}} - v = \frac{- v^{4}}{1 + v^{3}}$

$\Rightarrow \frac{1 + v^{3}}{v^{4}} d v = - \frac{d x}{x}$

Integrating both sides, we get $\int (\frac{1}{v^{4}} + \frac{1}{v}) d v = - \int \frac{d x}{x}$

$\Rightarrow - \frac{1}{3 v^{3}} + \log | v | = - \log | x | + C$

$\Rightarrow - \frac{x^{3}}{3 y^{3}} + \log | (\frac{y}{x}) | = - \log | x | + C$

$\Rightarrow \frac{- x^{3}}{3 y^{3}} + \log | y | = C$

Is the solution of the given differential equation.

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

Ans:

Solution of the given differential equation Re-writing the given equation as

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

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

It is clearly a homogenous differential equation

Assuming $y = v x$

Differentiating both sides

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

Substituting $\frac{dy}{dx}$ from the given equation

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

$1 + v^{2} + v = v + x \frac{dv}{dx}$

$1 + v^{2} = x \frac{dv}{dx}$

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

Now integrating both sides

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

$\ln x = \arctan v + c$

Formula

$\int \frac{dx}{x} = \ln x \int \frac{dv}{1 + v^{2}} = \tan^{- 1} v$

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

$\ln x = \tan^{- 1} \frac{y}{x} + c$

Is the solution of given differential equation.

(III) $(x^{2} - y^{2}) d x + 2 x y d y = 0, y (1) = 1$

Ans:

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

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

It is homogeneous differential equation.

Putting $y = u x \Rightarrow u + \frac{x d u}{d x} = \frac{d y}{d x}$

From (I)

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

$\Rightarrow \frac{x d u}{d x} = - [\frac{1 - u^{2}}{2 u} + u]$

$\Rightarrow \frac{x d u}{d x} = - [\frac{1 + u^{2}}{2 u}]$

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

Integrating both sides, we get

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

$\Rightarrow \log | 1 + u^{2} | = - \log | x | + \log C$

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

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

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

Given that $y = 1$ when $x = 1$ $\Rightarrow 1 + 1 = C \Rightarrow C = 2$

$∴$ Solution is $x^{2} + y^{2} = 2 x$ .

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

Ans:

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

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

$y = v x$

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

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

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

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

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

$\log \sin v = \log x + \log c$

$\log \sin (\frac{y}{x}) = \log \frac{c}{x}$

$\Rightarrow \sin (\frac{y}{x}) = \frac{c}{x}$

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

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

Ans:

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

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

$\frac{d y}{d x} = (x^{2} + y^{2}) / 2 x y \dots . (i)$

Let $x = v y$

Here, differentiating w.r.t. $y$ ,

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

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

Here, from e (i),

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

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

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

$y (\frac{d v}{d y}) = ((v^{2} + 1) / 2 v) - (v / 1)$

$y (\frac{d v}{d y}) = (v^{2} + 1 - 2 v^{2}) / 2 v$

$y (\frac{d v}{d y}) = (- v^{2} + 1) / 2 v$

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

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

Integrating both sides

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

$\log | v^{2} - 1 | = - \log y + \log c$

$\log | v^{2} - 1 | y = \log c$

$v^{2} y - y = c$

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

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

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

$x^{2} - y^{2} = c y$

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

Ans:

From the integral

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

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

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

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

$\Rightarrow \log e^{y} = e^{x} + \frac{x^{3}}{3} + C$

$∴ y = e^{x} + \frac{x^{3}}{3} + C$

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

Ans:

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

It is given that

$\frac{d y}{d x} - \sqrt{\frac{1 - y^{2}}{1 - x^{2}}} = 0$

We can write it as

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

By cross multiplication

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

By integrating both sides w.r.t $x$

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

We get

$\sin^{- 1} y = \sin^{- 1} x + c$

(VIII) $(3 x y + y^{2}) d x + (x^{2} + x y) d y$

Ans:

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

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

$v + x \frac{d v}{d x} = - (\frac{3 x \cdot v x + v^{2} x^{2}}{x^{2} + x \cdot v x})$

$x \frac{d v}{d x} = \frac{- 3 v - v^{2}}{1 + v} - v$

$x \frac{d v}{d x} = \frac{- 3 v - v^{2} - v - v^{2}}{1 + v}$

$x \frac{d v}{d x} = \frac{- 2 v^{2} - 4 v}{1 + v}$

$\frac{1 + v}{2 v^{2} + 4 v} d v = - \frac{1}{x} d x$

$\int \frac{1 + v}{2 v^{2} + 4 v} d v = \int - \frac{1}{x} d x$

$\frac{1}{4} \int \frac{2 + 2 v}{2 v + v^{2}} d v = - \int \frac{1}{x} d x$

$\frac{1}{4} \log | v^{2} + 2 v | = - \log | x | + c$

$\frac{1}{4} \log (\frac{y^{2}}{x^{2}} + 2 \frac{y}{x}) \cdot x = c$

$\log (\frac{y^{2}}{x} + 2 y) = 4 c$

9. (I) Form the differential equation of the family of circles touching y-axis at (0, 0).

Ans:

The centre of the circle touching the y-axis at origin lies on the x-axis.

Let (a, 0) be the centre of the circle.

Since it touches the y-axis at origin, its radius is a.

Now, the equation of the circle with centre (a, 0) and radius (a) is

${(x - a)}^{2} + y^{2} = a^{2} .$

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

differential equation of the family of circles touching y-axis at (0, 0

Differentiating equation (1) with respect to $x$ , we get:

$2 x + 2 y y^{'} = 2 a$

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

Now, on substituting the value of a in equation (1), we get:

$x^{2} + y^{2} = 2 (x + y y^{'}) x$

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

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

This is the required differential equation.

(ii) Form the differential equation of family of parabolas having vertex at $(0, 0)$ and axis along the

(i) positive y-axis

Ans:

The equation of the parabola having the vertex at origin and the axis along the positive

$y$ -axis is:

$x^{2} = 4 a y$ …….1

Differentiating equation (1) with respect to $x$ , we get:

$2 x = 4 a y^{'}$ ……(2)

Dividing equation (2) by equation (1), we get:

$\frac{2 x}{x^{2}} = \frac{4 a y^{'}}{4 a y}$

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

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

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

This is the required differential equation.

(ii) Positive x-axis

Ans:

Since parabola has axis along positive $x$ -axis,

its Equation: $y^{2} = 4 a x$ ...(1)

Diff. w.r.t. $x$

$\frac{d}{d x} (y^{2}) = \frac{d}{d x} (4 a x)$

$2 y \frac{d y}{d x} = 4 a$

Putting Value of 4 a in $(1)$

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

$y^{2} - 2 x y \frac{d y}{d x} = 0$

(iii) Form differential equation of family of circles passing through origin and whose Centre lie on x-axis.

Ans:

Equation of circle is $(x - a) + y = a . . . (1)$ ( ' a' arbitrary constant)

Differentiate with respect to 'x'

$2 (x - a) + 2 y \frac{d y}{d x} = 0$

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

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

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

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

(iv) Form the differential equation of the family of circles in the first quadrant and touching the coordinate axes.

Ans:

Let the equation of the circle be

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

$\Rightarrow x^{2} + y^{2} - 2 a (x + y) + a^{2} = 0$

$\Rightarrow 2 x + 2 y - 2 a (1 + y_{1}) = 0$

$\Rightarrow x + y - a (1 + y_{1}) = 0$

$\Rightarrow a = \frac{x + y}{1 + y_{1}}$

Hence, the required differential equation is

$x^{2} + y^{2} - 2 (\frac{x + y}{1 + y_{1}}) (x + y) + {(\frac{x + y}{1 + y_{1}})}^{2} = 0$

$\Rightarrow {(1 + y_{1})}^{2} (x^{2} + y^{2}) - 2 (x + y)^{2} (1 + y_{1})$ $+ (x + y)^{2} = 0$

10. Show that the differential equation $\frac{d y}{d x} = \frac{x + 2 y}{x - 2 y}$ is homogeneous and solve it.

Ans:

Step 1: Find $\frac{d y}{d x}$

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

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

Step 2: Put $F (x, y) = \frac{d y}{d x}$ Find $F (λ x, λ y)$

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

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

Finding

$F (λ x, λ y)$

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

$= \frac{λ (x + 2 y)}{λ (x - 2 y)}$

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

$= F (x, y)$

Thus, $F (λ x, λ y) = F (x, y) = λ^{\circ} F (x, y)$

Thus, $F (x, y)$ is Homogeneous function of degree zero

Therefore, the given Differential Equation is Homogeneous

differential Equation

Step 3: Solving $\frac{d y}{d x}$ by Putting $y = v x$

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

Let $y = v x$

So, $\frac{d y}{d x} = \frac{d (v x)}{d x}$

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

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

Putting in eqn, (1)

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

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

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

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

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

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

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

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

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

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

Integrating Both Sides

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

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

$\int \frac{(2 v - 1) d v}{2 v^{2} + v + 1} d v = - \log | x | + c$

By equating

$\log | x^{2} + x y + y^{2} | = 2 \sqrt{3} \tan^{- 1} (\frac{x + 2 y}{\sqrt{3 x}}) + c$

12. Solve the following differential equations:

(i) $\frac{d y}{d x} - 2 y = \cos 3 x$

Ans:

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

This is a linear differential equation of the form

$\frac{d y}{d x} + P y = Q$ , where $P = - 2$ and $Q = \cos 3 x$

Multiplying both sides by (1), we get

$e^{- 2 x} \frac{d y}{d x} - 2 y e^{- 2 x} = \cos 3 x e^{- 2 x}$

Integrating both sides w.r.t. $x$ , we get

$y e^{- 2 x} = \int e^{- 2 x} \cos 3 x d x + C$

$\Rightarrow y e^{- 2 x} = I + C,$

$I = e^{- t x} \cos 3 x \dots .$

$I = \int e^{- 2 x} \cos 3 x d x$

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

$= \frac{1}{3} e^{- 2 x} \sin 3 x + \frac{2}{3} [\frac{- 1}{3} e^{- 2 x} \cos 3 x - \int (- 2) e^{- 2 t} (\frac{- \cos 3 x}{3}) d x$

$= \frac{1}{3} e^{- 2 π} \sin 3 x + \frac{2}{3} [\frac{- 1}{3} e^{- 2 x} \cos 3 x - \frac{2}{3} \int e^{- 2 x} \cos 3 x a x]$

$= \frac{1}{3} e^{- 2 x} \sin 3 x - \frac{2}{9} e^{- 2 x} \cos 3 x - \frac{4}{9} I$

$\Rightarrow (I + \frac{4}{9} I) = \frac{e^{- 2 x}}{9} (3 \sin 3 x - 2 \cos 3 x)$

$\Rightarrow I = \frac{e^{- 2 x}}{13} (3 \sin 3 x - 2 \cos 3 x)$

Substituting the value of 1 in (2)

$y e^{- 2 x} = \frac{e^{- 2 x}}{13} (3 \sin 3 x - 2 \cos 3 x) + C$

$y = \frac{3 \sin 3 x}{13} - \frac{2 \cos 3 x}{13} + c e^{2 x}$

(II) $\sin x \frac{d y}{d x} + y \cos x = 2 \sin^{2} x \cos x if y (\frac{π}{2}) = 1$

Ans:

Given

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

$\Rightarrow \frac{1}{\sin x} [(\sin x) \frac{d y}{d x} + y \cos x] = 2 \sin^{2} x \cos x \times \frac{1}{\sin x}$

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

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

This is a first order linear differential equation of the form

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

Here, $P = \cot x$ and $Q = 2 \sin x \cos x$

The integrating factor (I.F) of this differential equation is,

I. $F = e^{\int P d x}$

$\Rightarrow$ I. $F = e^{\int \cot x d x}$

We have

$\int \cot x d x = \log (\sin x) + c$

$\Rightarrow I . F = e^{\log (\sin x)}$

$∴ I . F = \sin x [∵ e^{\log x} = x]$

Hence, the solution of the differential equation is,

$y ($ I. $F) = \int (Q \times I . F) d x + c$

$\Rightarrow y (\sin x) = \int (2 \sin x \cos x \times \sin x) dx + c$

$\Rightarrow y \sin x = 2 \int \sin^{2} x \cos xdx + c$

Let $\sin x = t$

$\Rightarrow \cos x d x =$ dt (Differentiating both sides)

By substituting this in the above integral, we get

$y t = 2 \int t^{2} d t + c$

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c$

$\Rightarrow y t = 2 (\frac{t^{2 + 1}}{2 + 1}) + c$

$\Rightarrow y t = 2 (\frac{t^{3}}{3}) + c$

$\Rightarrow y t = \frac{2 t^{3}}{3} + c$

$\Rightarrow y = \frac{2 t^{2}}{3} + \frac{c}{t}$

$\Rightarrow y = \frac{2 {(\sin x)}^{2}}{3} + \frac{c}{\sin x} [∵ t = \sin x]$

$∴ y = \frac{2}{3} \sin^{2} x + c cosec x$

Thus, the solution of the given differential equation is

$y = \frac{2}{3} \sin^{2} x + c cosec x$

(III) $3 e^{x} \tan y d x + (1 - e^{x}) \sec^{2} y d y = 0$

Ans:

The given differential equation is

$3 e^{x} \tan y d x + (1 - e^{x}) \sec^{2} y d y = 0$

$\frac{3 e^{x}}{(1 - e^{x})} d x = \frac{\sec^{2} y}{\tan y} d y$

On Integrating, we get

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

$- 3 \log | 1 - e^{x} | = \log | \tan y | + c$

$\tan y = k {(1 - e^{x})}^{3}$ which is the required solution of the given differential equation.

13. Solve the following differential equations:

(i) $(x^{3} + y^{3}) d y - x^{2} y d x = 0$

Ans:

$(x^{3} + y^{3}) d y - x^{2} y d x = 0$

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

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

$Let \frac{y}{x} = v \Rightarrow y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

So,our differential equations becomes

$v + x \frac{d v}{d x} = \frac{1}{\frac{1}{v} + v^{2}} = \frac{v}{1 + v^{3}}$

$\Rightarrow x \frac{d v}{d x} = \frac{v}{1 + v^{3}} - v = - \frac{v^{4}}{1 + v^{3}}$

$\Rightarrow - (\frac{1 + v^{3}}{v^{4}}) d v = \frac{d x}{x}$

Integrating both sides

$\Rightarrow - (\frac{1 + v^{3}}{v^{4}}) d v = \frac{d x}{x}$

$\Rightarrow \int - (\frac{1 + v^{3}}{v^{4}}) d v = \int \frac{d x}{x}$

$\Rightarrow \int - (v^{- 4} + \frac{1}{v}) d v = \int \frac{d x}{x}$

$\Rightarrow \frac{v^{- 3}}{3} - \log v = \log x + c$

$\Rightarrow \frac{1}{3 v^{3}} - c = \log x + \log v$

$\Rightarrow \frac{1}{3 v_{0}^{3}} - c = \log (v x)$

$\Rightarrow \frac{x^{3}}{3 y^{3}} = \log y + c,$

Which is required solution.

(ii) $x d y - y d x = \sqrt{x^{2} + y^{2}} d x$

Ans:

$Given$

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

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

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

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

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

$= λ^{\circ} . F (x, y)$

$F (x, y)$ is a homogeneous function of degree zero.

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

Let $y = v x$

$\Rightarrow \frac{d y}{d x} = v + x \cdot \frac{d v}{d x}$

Putting above value, we have

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

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

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

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

Integrating both sides, we get

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

$\log x + \log c = \log | v + \sqrt{1 + v^{2}} | [∵ \int \frac{d x}{\sqrt{x^{2} + a^{2}}} = \log | x + \sqrt{x^{2} + a^{2}} | + c]$

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

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

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

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

11. Show that the differential equation $(x^{2} + 2 x y - y^{2}) d x + (y^{2} + 2 x y - x^{2}) d y = 0$

is homogeneous and solve it.

Ans:

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

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

$y = V \cdot X$

$\frac{d y}{d x} = v .1 + x \cdot \frac{d v}{d x}$

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

$= (v^{2} - 2 v - 1) / (v^{2} + 2 v - 1)$

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

$x \cdot \frac{d v}{d x} = \frac{v^{2} - 2 v - 1 - v^{3} - 2 v^{2} + v}{v^{2} + 2 v - 1}$

$or . [\frac{v^{\land} 2 + 2 v - 1}{- v^{3} - v^{2} - v - 1}] \cdot d v = \frac{1}{x} \cdot d x$

$\frac{v^{2} + 2 v - 1}{v + 1} \cdot (v^{2} + 1) = \frac{A}{v + 1} + \frac{B v + C}{(v^{2} + 1)}$

Equating the coeff. of $v^{2}, v$ and constant terms.

$A + B = 1 \dots \dots \dots \dots \dots . . (1)$

$B + C = 2 \dots \dots \dots \dots \dots . . (2)$

$A + C = - 1 \dots \dots \dots \dots \dots . . (3)$

Subtracting eqn. (2) from (1)

$A - C = - 1 \dots \dots \dots \dots . . (4)$

by eqn. (3) and (4) $A = - 1, C = 0$

But

$A + B = 1.$

$- 1 + B = 1 => B = 2$

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

or. Integ.of $[1 / (v + 1) - 2 v / (v^{\land} 2 + 1)] \cdot d v =$ integ. of $1 / x . d x$

or. $\log (v + 1) - \log (v^{} + 1) = \log x + \log C$

or. $\log (v + 1) / (v^{2} + 1) = \log x . C$

(IV) $x^{2} d y + y (x + y) d x = 0$ given that $y = 1$ when $x = 1.$

Ans:

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

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

Put $y = v x$

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

The differential equation becomes

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

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

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

or. $\frac{1}{2} \log \frac{v}{v + 2} = - \log x + \log C$

or $\frac{C}{x} = \sqrt{\frac{y}{y + 2 x}}$

$If x = 1, y = 1, C = \frac{1}{\sqrt{3}}$

$or$

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

(V) $x e^{\frac{y}{x}} + y + x \frac{d y}{d x} = 0$ if $y (e) = 0$

Ans: Given differential equation is, $(x e^{\frac{y}{x}} + y) d x = x d y$

$\Rightarrow \frac{d y}{d x} = \frac{x \cdot e^{\frac{y}{x}} + y}{x} \dots . . (i)$

Let $F (x, y) = \frac{x \cdot e^{\frac{y}{x}} + y}{x}$

$∴ F (λ x, λ y) = \frac{λ x \cdot e^{\frac{λ y}{λ x} + λ y}}{λ x} = λ^{0} \frac{x \cdot e^{\frac{y}{x}} + y}{x} = λ^{0} F (x, y)$

Hence, given differential equation (i) is homogenous.

Let $y = vx \Rightarrow \frac{d y}{d x} = v + x \cdot \frac{d v}{d x}$

Now, given differential equation (i) would become

$v + x \cdot \frac{d v}{d x} = \frac{x \cdot e^{\frac{v x}{x} + v x}}{x}$

$\Rightarrow v + x \cdot \frac{d v}{d x} = e^{v} + v$

$\Rightarrow x \cdot \frac{d v}{d x} = e^{v}$

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

$\Rightarrow \int e^{- v} d v = \int \frac{d v}{v}$

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

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

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

$\Rightarrow e^{\frac{y}{x}} \cdot \log x + C e^{\frac{y}{x}} + 1 = 0$

Putting $y (e) = 0$ , we get

$∴ 1^{\frac{y}{x}} \log x + C 1^{\frac{y}{x}} + 1 = 0$

$\Rightarrow C = - \frac{1}{e}$

The required particular solution is

$e^{\frac{y}{x}} \cdot \log x - \frac{1}{e} e^{\frac{y}{x}} + 1 = 0$

or $e^{\frac{y}{x}} \log x - e^{\frac{y}{x} - 1} + 1 = 0$

(VI) $(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$

Ans:

Given differential equation is

$(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$

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

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

$\Rightarrow v + x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v}, Let v = \frac{y}{x})$

$\Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v} - v$

$= \frac{1 - 3 v^{2} - v^{4} + 3 v^{2}}{v^{3} - 3 v}$

$= \frac{1 - v^{4}}{v^{3} - 3 v}$

$\Rightarrow (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = \frac{d x}{x}$

$\Rightarrow (\frac{v^{3}}{1 - v^{4}}) d v - (\frac{3 v}{1 - v^{4}}) d v = \frac{d x}{x}$

Integrating, we get

$\Rightarrow (\frac{v^{3}}{1 - v^{4}}) d v - (\frac{3 v}{1 - v^{4}}) d v = \frac{d x}{x}$

Integrating, we get

$- \frac{1}{4} \log | 1 - v^{4} | + \frac{3}{4} \log | \frac{v^{2} - 1}{v^{2} + 1} | = \log | x | + \log c$

$- \frac{1}{4} \log | 1 - {(\frac{y}{x})}^{4} | + \frac{3}{4} \log | \frac{{(\frac{y}{x})}^{2} - 1}{{(\frac{y}{x})}^{2} + 1} | = \log | x c |$

which is the required solution

(VII) $\frac{d y}{d x} - \frac{y}{x} + cosec (\frac{y}{x}) = 0$ given that $y = 0$ when $x = 1$

Ans:

Differential equation is

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

Let $F (x, y) = \frac{d y}{d x} = \frac{y}{x} - cosec (\frac{y}{x})$

Finding $F (λ x, λ y)$

$F (λ x, λ y) = \frac{λ y}{λ x} - cosec (\frac{λ y}{λ x}) = \frac{y}{x} - cosec (\frac{y}{x}) = λ^{\circ} F (x, y)$

$∴ F (x, y)$ is $a$ homogenous function of degree zero

$F (λ x, λ y) = λ^{\circ} F (x, y)$

Putting $y = v x$

Diff w.r.t. $x$

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

Putting value of $\frac{d y}{d x}$ and $y = v x$ in (1)

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

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

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

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

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

Integrating both sides

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

$\int - \sin v d v = \log | x | + c$

Put value of $v = \frac{y}{x}$

$\cos \frac{y}{x} = \log | x | + C$

Putting $x = 1 y = 0$

$\cos \frac{0}{1} = \log 1 + C$

$1 = 0 + C [\log 1 = 0]$

Putting value in (2)

$c = 1$

$\cos \frac{y}{2} = \log | x | + 1$

$\cos \frac{y}{2} = \log | x | + \log e$

$\cos \frac{y}{2} = \log | x |$

16. Solve the following differential equations:

(I) $\cos^{2} x \frac{d y}{d x} = \tan x - y$

Ans:

Given differential equation is

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

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

Given differential equation is a linear differential equation of the type $\frac{d y}{d x} + p y = Q$

I.F. $= e^{\int P d x} = e^{\int \sec^{2} x d x} = e^{\tan x}$

$∴$ Solution is given by $e^{\tan x} y = \int \tan x \cdot \sec^{2} x \cdot e^{\tan x} d x$

Let $I = \int \tan x \cdot \sec^{2} x \cdot e^{\tan x} d x$

Let $\tan x = t, \sec^{2} x d x = d t$

$\Rightarrow I = \int t e^{t} d t$

Integrating by parts $∴ I = t e^{t} - \int e^{t} d t = t e^{t} - e^{t} + C$

$\Rightarrow I = \tan x e^{\tan x} - e^{\tan x} + C$ ,

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

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

(II) $x \cos x \frac{d y}{d x} + y (x \sin x + \cos x) = 1$ .

Ans:

Given $x \cos x (d y / d x) + y (x \sin x + \cos x) = 1$ .

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

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

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

It is linear differential equation in the form $\frac{d y}{d x} + P y = Q$

where $P = \tan x + (\frac{1}{x})$ and $\frac{\sec x}{x}$

$∴ (I . F .) = e^{\int (\tan x + (1 / x)) dx}$

$= e^{\int \tan x + \int d x / x} = e^{\log \sec x + \log x} = e^{\log (x \sec x)}$

$= x \sec x$

Now, multiplying (1) by I.F. and integration, we get

$y \times I . F = \int Q \times I . F + C$

$y x \sec x = \int (\sec x / x) x (x \sec x) d x + c$

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

$x y \sec x = \tan x + C$

Which is the required solution

(III). $(1 + e^{\frac{x}{y}}) d x + e^{\frac{x}{y}} (1 - \frac{x}{y}) d y = 0$

Ans:

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

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

Hence, homogeneous

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

Equating Homogeneous,

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

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

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

$\log_{e} | e^{v} + v | = - \log_{e} | y | + \log_{e} C$

$\log_{e} | (e^{v} + v) y | = \log_{c} C$

$(e^{σ} + v) y = C = A$

$(\frac{x}{e^{y}} + \frac{x}{y}) y = A,$

The General solution

(IV) $(y - \sin x) d x + \tan x d y = 0, y (0) = 0.$

Ans:

The given diff. equation can be written as

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

This is linear differential equation.

I.F. $= e^{\int \cot x d y} = e^{\log \sin x} = \sin x$

The solution is:

$y \sin x = \int \sin x \cos x d y + C$

$= \frac{1}{2} \int \sin 2 x d y + C y \sin x$

$= \frac{- 1}{4} \cos 2 x + C$

It is given that , when

$c - \frac{1}{4} = 0 or c = \frac{1}{4}$

$y \sin x = \frac{1}{4} (1 - \cos 2 x) = \frac{1}{2} \sin^{2} x$

$2 y = \sin x$

which is the required solution.

LONG ANSWER TYPE QUESTIONS (6 MARKS EACH)

17. Solve the following differential equations:

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

Ans:

Given Differential equation can be written as

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

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

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

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

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

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

Put $(\frac{y}{x}) = v$ to get $y = v x$ and $\frac{d y}{d x} = v + x \frac{d y}{d x}$

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

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

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

or $\log | v \cos v | + \log x^{2}$

$= \log C$

or $x^{2} v \cos v = C$ or $x y \cos (\frac{y}{x}) = C$

(II)

$3 e^{x} \tan y d x + (1 - e^{x}) \sec^{2} y d y = 0$

Given that $y = \frac{π}{4}$ , when $x = 1$

Ans:

The given differential equation is

$3 e^{x} \tan y d x + (1 - e^{x}) \sec^{2} y d y = 0$

$\frac{3 e^{x}}{(1 - e^{x})} d x = \frac{\sec^{2} y}{\tan y} d y$

On Integrating, we get

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

$- 3 \log | 1 - e^{x} | = \log ltan y ∣ + c$

By putting

$y = \frac{π}{4}, a n d x = 1$

$(1 - e)^{3} \tan y = {(1 - e^{x})}^{3}$

which is the required solution of the given differential equation.

(III) $\frac{d y}{d x} + y \cot x = 2 x + x^{2} \cot x$ Given that $y (0) = 0$

Ans:

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

Let $P = \cot x, Q = 2 x + x^{2} \cot x$

I.F $= e^{\int pdx} = e^{\int \cot x dx} = e^{\int \log (\sin x)} = \sin x$

Sol is

$y (I . F) = \int (Q x ∣ . F) d x + C$

$y \sin x = \int (2 x + x^{2} \cot x) \sin d x + c$

$= \int (2 x + x^{2} \cot x) d x + c$

$= 2 \int x \sin x d x + \int x^{2} \cos x d x + c$

$= 2 [x (- \cos x) - (1) (- \sin x)] + [x^{2} \sin x - 2 x (- \cos x) + 2 c (- \sin x)]$

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

$= x^{2} \sin x + c$

Then Y (0) =0

$Y = x^{2}$

This is the Required General solution for the given differential Equation.

Significance of Important Questions for Class 12 Maths Chapter 9 Differential Equations

Apart from the exercises given in the NCERT textbook, these important questions will act as the perfect guide to learning how to use the concepts of differential equations and calculus to solve.

Students will study the syllabus of this chapter and will learn the concepts, fundamental principles and formulas of differential equations and will proceed to complete the syllabus accordingly. To test their knowledge and mathematical skills, they will need a vivid platform. This platform is provided with these important questions. They can solve and compare their answers to the solutions provided and find out where they need to study more.

Benefits of Important Questions for Class 12 Maths Chapter 9 Differential Equations

Use these important questions as an evaluation tool for your preparation.
Learn to efficiently use time and solve these questions.
Resolve doubts faster by using the solutions provided.
Focus on the stepwise answering formats and score more.

Conclusion

Practicing important questions from Chapter 9, Differential Equations, is essential for scoring well in your Class 12 Maths exams. These questions cover all the key concepts and problem-solving techniques you need to master. With consistent practice and understanding, you’ll be well-prepared to answer any question in this chapter. Download the free PDF, stay focused, and excel in your exams!

Important Study Materials for Class 12 Maths Chapter 9 Differential Equations

S.No	CBSE Class 12 Maths Differential Equations Other Study Materials
1	CBSE Class 12 Maths Differential Equations Revision Notes
2	CBSE Class 12 Maths Differential Equations NCERT Solutions
3	CBSE Class 12 Maths Differential Equations Formulas
4.	CBSE Class 12 Maths Differential Equations RD Sharma Solutions
5.	CBSE Class 12 Maths Differential Equations RS Aggarwal Solutions
6.	CBSE Class 12 Maths Differential Equations NCERT Exemplar Solutions

Download CBSE Class 12 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 12 Maths Important Questions for other chapters:

S. No	CBSE Class 12 Maths Important Questions
1	Chapter 1 - Relations and Functions Important Questions
2	Chapter 2 - Inverse Trigonometric Functions Important Questions
3	Chapter 3 - Matrices Important Questions
4	Chapter 4 - Determinants Important Questions
5	Chapter 5 - Continuity and Differentiability Important Questions
6	Chapter 6 - Application of Derivatives Important Questions
7	Chapter 7 - Integrals Important Questions
8	Chapter 8 - Application of Integrals Important Questions
10	Chapter 10 - Vector Algebra Important Questions
11	Chapter 11 - Three-Dimensional Geometry Important Questions
12	Chapter 12 - Linear Programming Important Questions
13	Chapter 13 - Probability Important Questions

Important Related Links for CBSE Class 12 Maths

S.No	CBSE Class 12 Maths Study Materials
1	CBSE Class 12 Maths NCERT Solutions
2	CBSE Class 12 Maths Revision Notes
3	CBSE Class 12 Maths NCERT Book
4	CBSE Class 12 Maths Previous year Question Paper
5	CBSE Class 12 Maths Sample Papers
6	CBSE Class 12 Maths NCERT Exemplar
7	CBSE Class 12 Maths RD Sharma Solutions
8	CBSE Class 12 Maths RS Aggarwal Solutions

FAQs on CBSE Class 12 Maths Important Questions - Chapter 9 Differential Equations

1. What are differential equations, and why are they important in Class 12 Maths?

Differential equations are mathematical equations that involve derivatives, representing rates of change. They are crucial for understanding real-world phenomena in physics, engineering, and economics.

2. How can I prepare effectively for Chapter 9 in Class 12 Maths through these Important Questions?

Start by understanding the concepts from the NCERT textbook. Then, practise important questions, focus on solving different types of problems, and review previous year question papers.

3. Are these Class 12 Maths Chapter 9 Differential Equations useful for competitive exams?

Yes, practising these questions strengthens your basics, which are often tested in entrance exams like JEE and other competitive tests.

4. Do these Important Questions on Class 12 Maths Chapter 9 Differential Equations include solutions?

The free PDF includes step-by-step solutions for better understanding and self-assessment.

5. How much weightage does Chapter 9 hold in the CBSE Class 12 Maths exam?

Typically, questions from this chapter carry significant weightage, often included in the long-answer section. It’s important to prepare thoroughly.

6. What are the key methods to solve Important Questions on Class 12 Maths Chapter 9 Differential Equations?

The key methods include separation of variables, integrating factors, and homogeneous equations. Mastering these techniques is crucial for solving different types of problems.

7. How can I identify which method to use for a given differential equation?

Analyzing the form of the equation helps determine the method. For instance, if variables can be separated, use the separation of variables method. If the equation is linear, apply the integrating factor method.

8. Are real-life applications of differential equations included in the class 12 Maths Chapter 9 Differential Equations syllabus?

Yes, the chapter includes real-life applications such as population growth, radioactive decay, and cooling laws, which help in understanding the practical use of differential equations.

9. Can I expect direct questions from NCERT exercises in the board exam?

Yes, CBSE often includes direct or slightly modified questions from NCERT exercises, making it essential to practice them thoroughly.

10. How do I approach solving long-answer questions from class 12 Maths Chapter 9 Differential Equations?

Break the problem into smaller steps. Identify the type of equation, choose the appropriate method, and solve step-by-step. Always include explanations for better clarity.