JEE Advanced 2025 Revision Notes for Differential Calculus

Definition:

An equation that involves an independent variable $(x)$ say, dependent variable $(y)$ say and differential coefficients of dependent variable with respect to the independent variable that is $\dfrac{{dy}}{{dx}}$,$\dfrac{{{d^2}y}}{{{d^2}x}}$,…..etc called a differential equation.

For examples:

i) ${x^2}\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) + {x^3}{\left( {\dfrac{{dy}}{{dx}}} \right)^3}7{x^2}{y^2}$

ii) $({x^2} + {y^2})dx = ({x^2} - {y^2})dy$

Order:

The order of a differential equation is the order of the highest derivative occurring in the equation. The order of a differential equation is always a positive integer is called the order of the differential equation.

Degree:

The degree of a differential equation is the degree (exponent) of the derivative of the highest order derivative occurring in the equation when the differential coefficients are made free from radicals and fractions. It is written as a polynomial in differential coefficients after the equation is free from negative and fractional powers of the derivatives is called degree of differential equations.

For example:

${\left( {\dfrac{{{d^2}y}}{{{d^2}x}}} \right)^3} + \sin \left( {\dfrac{{dy}}{{dx}}} \right) = 0$

Here order is 2, but this differential equation can’t be written as a polynomial in differential coefficient; hence its degree is not defined.

Linear and Nonlinear Differential Equations:

A differential equation is said to be linear if the dependent variable and all of its derivatives occur in the first power and there are no products. A linear equation of nth order can be written in the form.

${P_0}\dfrac{{{d^n}y}}{{d{x^n}}} + {P_1}\dfrac{{{d^{n - 1}}y}}{{d{x^{n - 1}}}} + {P_2}\dfrac{{{d^{n - 2}}y}}{{d{x^{n - 2}}}} + ...... + {P_{n - 1}}\dfrac{{dy}}{{dx}} + {P_n}y = Q$

Where, ${P_0},{P_1},{P_2},........,{P_{n - 1}}$ and $Q$ must be either constant or functions of $x$ only.

A linear differential equation is always of the first degree, but every differential equation of the first degree need not be linear.

Example:

The equations are non-linear.

$\dfrac{{{d^2}y}}{{d{x^2}}} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} + xy = 0$ and

$x\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) + y\left( {\dfrac{{dy}}{{dx}}} \right) + y = {x^3},\left( {\dfrac{{dy}}{{dx}}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0$

Formation of Differential Equations:

• We are formulating a differential equation from a given $x$ and $y$ containing arbitrary constants.

• Suppose, we have a given equation with $n$ arbitrary constants $f(x,y,{c_1},{c_2},..........,{c_n}) = 0$

• Differentiate the equation successively $n$ times to get $n$ equations.

• Eliminating the arbitrary constants from these $n + 1$ equations lead to the required differential equations.

Solutions of Differential Equations of the First Order and First Degree:

The following method can solve a differential equation of first degree and first order.

Inspection Method:

If the differential equation can written as 

$f[{f_1}(x,y)d\{ {f_1}(x,y)\} ] + \varphi [{f_2}(x,y)d\{ {f_2}(x,y)\} ] + .....0]$ then each term can be integrated separately.

Initial Value Problem:

An initial value problem is when some initial conditions are given to solve a differential equation.

• Differentiate the given equation as often as the number of arbitrary constants involved.

• Eliminate the arbitrary constant from the equations of $y,y',y''$ etc.

General Solutions:

Suppose the solution of the differential equation contains as many independent arbitrary constants as the order of the differential equation. In that case, it is called the general solution or the complete integral of the differential equation.

For example the general solution of $\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0$ is $y = A\cos x + B\sin x$ because it contains two arbitrary constants A and B, which is equal to the order of the equation.

Particular Solution:

A solution obtained by giving particular values to the arbitrary constants in the general solution is called a particular solution. e.g., In the previous example, if $A = B = 1$, then $y = \cos x + \sin x$  is a particular solution of the differential equation $\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0$.

The solution of a differential equation is also called its primitive.

Solution of Differential Equations:

A differential equation solution is a relation between the variables, not involving the differential coefficients, such that this relation and the derivative obtained from it satisfy the given differential equation.

For example: Let $\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0$

Integrating above equation 2 times, we get $y = A\cos x + B\sin x$.

In these equations, we have three types of equations:

1. Variable separable form

2. Homogenous equations

3. Linear Differential Equations

Variable Separable Form:

 If in the equation, it is possible to get all terms containing $x$ and $dx$ to one side and all the terms containing $y$ and $dy$ to the other, the variables are said to be separable.

Procedure to Solve:

Consider the equation $\dfrac{{dy}}{{dx}} = X.Y$, where $X$ is a function of $x$ only and $Y$ is the function of $y$ only.

Put the equation in the form $\dfrac{1}{Y}dy = X.dx$

Integrating both sides and we get the answer.

Differential Equation Reducible To Variable Separable Method:

A differential equation of the form $\dfrac{{dy}}{{dx}} = f(ax + by + c)$ can be reduced to variable separable form by substituting.

$ax + by + c = z$

$\Rightarrow a + b\dfrac{{dy}}{{dx}} = \dfrac{{dz}}{{dx}}$

The given equation becomes

$\dfrac{1}{b}\lgroup\dfrac{dz}{dx}-a \rgroup f(z)$

$\Rightarrow \dfrac{{dz}}{{dx}} = a + bf(z)$

$\Rightarrow dz/a + bf(z) = dx$

Hence, the variables are separated by the term $z$ and $x$.

Homogeneous Differential Equations: 

A differential equation which can be expressed in the form $\dfrac{{dy}}{{dx}} = f(x,y)$ or $\dfrac{{dy}}{{dx}} = g(x,y)$ where, $f(x,y)$ and $g(x,y)$ are homogeneous functions of degree zero are called a homogeneous differential equation.

 In this equation we put $y = vx$

$\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

The given equation, reduces to

$v + x\dfrac{{dv}}{{dx}} = F(v)$

$x\dfrac{{dv}}{{dx}} = F(v) – v$

$\therefore \dfrac{{dv}}{{F(v) - v}} = \dfrac{{dx}}{x}$ 

Hence, the variables are separated in terms of $v$ and $x$.

Differential Equations Reducible To Homogeneous Equation:

The differential equation of the form

$\dfrac{{dy}}{{dx}} = \dfrac{{{a_1}x + {b_1}y + {c_1}}}{{{a_2}x + {b_2}y + {c_2}}}......(i)$

Put $x = X + h$ and $y = Y + k$

$\therefore \dfrac{{dY}}{{dX}} = \dfrac{{{a_1}X + {b_1}Y + ({a_1}h + {b_1}k + {c_1})}}{{{a_2}X + {b_2}Y + ({a_2}h + {b_2}k + {c_2})}}........(ii)$

We choose $h$ and $k$ , so as to satisfy ${a_1}h + {b_1}k + {c_1} = 0$ and ${a_2}h + {b_2}k + {c_2} = 0$

On solving we get 

$\dfrac{h}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{k}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$

$h = \dfrac{{{b_1}{c_2} - {b_2}{c_1}}}{{{c_1}{a_2} - {c_2}{a_1}}}$ and $k = \dfrac{{{c_1}{a_2} - {c_2}{a_1}}}{{{a_1}{b_2} - {a_2}{b_1}}}$

Provided ${a_1}{b_2} - {a_2}{b_1} \ne 0,$

$\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_1}}}$

Then equation $(ii)$ reduces to $\dfrac{{dY}}{{dX}} = \dfrac{{\left( {{a_1}X + {b_1}Y} \right)}}{{\left( {{a_2}X + {b_2}Y} \right)}}$ , which is a homogeneous form and will be solved easily.

Exact Differential Equation:

Differential Equation $M(x,y)dy + N(x,y)dy = 0$ is called an exact differential equation.

In a function $u(x,y)$ exist such that.

$du = Mdx + Ndy$.

Linear Differential Equation:

A first order linear differential equation can be written in the form

$\dfrac{{dy}}{{dx}} + Py = Q$

Where $P$ and $Q$ are constant or function of $x$ only or 

$\dfrac{{dy}}{{dx}} + Px = Q$

Where $P$ and $Q$ are constant or function of $y$ only.

Differential Equation Reducible to Linear Form:

Bernoulli’s Equation: An equation of the form $\dfrac{{dy}}{{dx}} + Py = Q{y^n}$, where $P$ and $Q$ are functions of $x$ along or constant is called Bernoulli’s equation.

Divided both the side by ${y^n}$, we get

${y^{ - n}}\dfrac{{dy}}{{dx}} + P{y^{ - n + 1}} = Q$

Put $P{y^{ - n + 1}} = z$

$\Rightarrow ( - n + 1){y^{ - n}}\dfrac{{dy}}{{dx}} = \dfrac{{dz}}{{dx}}$

The equation reduces to 

$\dfrac{1}{1-n}\dfrac{{dy}}{{dx}} + Pz = Q$

$\Rightarrow \dfrac{{dz}}{{dx}} + (1 - n)Pz = Q(1 - n)$

Which is linear in $z$ and can be solved in the usual manner.

Clairaut Form for Differential Equation:

Differential equation $y = Px + f(p),$ where $P = \dfrac{{dy}}{{dx}}......(i)$ is called clairaut form of differential equation. In which, get its general solution by replacing $P$ from $C$.

Now, differential on both sides of equation,(i) with respect to $x$ and put $\dfrac{{dy}}{{dx}} = P$

$P = P + x\dfrac{{dp}}{{dx}} + f’(P)\dfrac{{dp}}{{dx}} = 0$

$\Rightarrow \left[ {x + f`(p)} \right]\dfrac{{dp}}{{dx}} = 0$

$\Rightarrow \dfrac{{dp}}{{dx}} = 0$

$\Rightarrow p = c$

Orthogonal Trajectory:

Any curve, which cuts every member of a given family of curves at right angles, is called an orthogonal trajectory of the family.

Procedure for finding the Orthogonal Trajectory:

• Let $f(x,y,c) = 0$ be the equation of the given family of curves, where $c$ is an arbitrary parameter.

• Differentiate $f = 0$,with respect to $x$ and eliminate 0 from a differential equation.

• Substitute $\left( { - \dfrac{{dy}}{{dx}}} \right)$ for $\left( {\dfrac{{dy}}{{dx}}} \right)$ in the above differential equation.

• This will give the differential equation of the orthogonal trajectories.

• By solving this differential equation, we get the required orthogonal trajectories.

Example: 1

Find the general solution of the differential equation

$({x^2} + 1)\dfrac{{dy}}{{dx}} + 2xy = \sqrt {{x^2} + 4}$

Solution:

$\dfrac{{dy}}{{dx}} + \left( {\dfrac{{2x}}{{{x^2} + 1}}} \right)y = \dfrac{{\sqrt {{x^2} + 4} }}{{{x^2} + 1}}$( compare it with $\dfrac{{dy}}{{dx}} + Py = Q$)

$IF = {e^{\int {pdx} }} = {e^{\int {\dfrac{{2x}}{{{x^2} + 1}}dx} }} = {e^{\log |{x^2} + 1|}} = {x^2} + 1$

Required solution is  $y({x^2} + 1) = \int {\dfrac{{\sqrt {{x^2} + 4} }}{{{x^2} + 1}}} ({x^2} + 1)dx$

$\Rightarrow y({x^2} + 1) = \int {\sqrt {{x^2} + 4} } dx$

$y({x^2} + 1) = \dfrac{x}{2}\sqrt {{x^2} + 4}  + 2\log |x + \sqrt {{x^2} + 4} | + c$

Example: 2

If $y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2^{2}}\right)$ then $\dfrac{d y}{d x} \text { at } x=0$ is

Answer: 

$y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2}\right)$

Multiplying numerator and denominator by $(1-x)$

$\Rightarrow y=\dfrac{\left.(1-x)(1+x)\left(1+x^{2}\right) \ldots .1+x^{2^{2}}\right)}{(1-x)}$

$\Rightarrow y=\left(\dfrac{1-x^{2^{-1}}}{(1-x)}\right)$

$\therefore \dfrac{d y}{d x}=\dfrac{(1-x) \cdot\left\{-2^{n+1} x^{z^{-1}-1}\right)-\left(1-x^{2^{n-1}}\right)(-1)}{(1-x)^{2}}$

So $\left.\dfrac{d y}{d x}\right|_{x-0}=\dfrac{-2^{-i+1} \cdot 01+1-0}{1^{2}}=1$

Example: 3

If $e^{y}+x y=e$ then $\dfrac{d^{2} y}{d x^{2}}$ at $x=0$ is $e^{-\lambda}$, then numerical quantity $\lambda$ will become.

Answer: 

$\mathrm{e}^{y}+x y=\mathrm{e}$

On putting $x=0$, we get $e^{y}=e$ $y=1$ when $x=0$

On differentiating the relation (i) we get

$\dfrac{d y}{d x}+1 \cdot y+x \cdot \dfrac{d y}{d x}=0$

On putting $x=0, y=1$ we get

$e^{y}\left(\dfrac{d y}{d x}\right)+1=0 \Rightarrow \dfrac{d y}{d x}=\dfrac{-1}{e}$

Now on differentiating solution (ii) we get 

$e^{y}\left(\dfrac{d y}{d x}\right)^{2}+e^{y} \dfrac{d^{2} y}{d x^{2}}+\dfrac{d y}{d x}+\dfrac{d y}{d x}+x \dfrac{d^{2} y}{d x^{2}}=0$

On putting $x=0, y=1$, $\dfrac{d y}{d x}=\dfrac{-1}{e}$ 

We get $\dfrac{d^{2} y}{d x^{2}}=\dfrac{1}{e^{2}}=e^{-2} \Rightarrow \lambda=2$

Importance of Differential Calculus - JEE Advanced

The introduction of calculus literally transforms the concepts of various mathematical operations a student has gained over the years. Previously, students have learnt how variables can affect the outcomes of a function. In this chapter, students will learn how a variable changes and influences the change of a function.

The formulae of this chapter will be explained using various mathematical techniques from scratch. Students who are preparing for JEE Advanced will have to master using these formulae and increase their answering efficiency.

This branch of applied Mathematics will teach students how to formulate derivatives of functions based on a particular variable. The formulas will focus on how these derivatives are calculated using the appropriate differential calculus formulae.

This chapter will explain what functions, relations, sets, etc. are. It will also explain what continuity, limits, and differentiability are through examples and exercises. This chapter is very important for completing the Class 12 Calculus syllabus and preparing for the IIT JEE exams. This is where the Differential Calculus JEE Advanced revision notes will become the best companion for your preparation.

Benefits of Differential Calculus JEE Advanced Revision Notes

• The revision notes are composed of simplifying all the basic and advanced concepts of Differential Calculus for your preparation. You will find a brilliant use of these notes to grab the concepts faster and better.

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