JEE Main Study Notes for Differential Equations: Check Important Topics, Preparation Tips, and Solved Questions

JEE Main Study Notes for Differential Equation: Differential Equation is an equation that contains one or more functions with their derivatives. Any Differential Equation can be expressed as dy/dx = f(x). Here ‘y’ is a dependent variable and ‘x’ is an independent variable. Differential equation contains two types of derivatives i.e., partial derivatives and ordinary derivatives.

• The topic of Differential Equations has less weightage in JEE Main. Quadratic Equation has around 1%-2% of weightage in JEE main. Check JEE Main 2025 Question Paper
• The study notes on Differential equations cover some important parts of JEE Main Mathematics Syllabus such as the order and degree of differential equations, formation of differential equations, the solution of differential equations by the method of separation of variables, and the solution of homogeneous and linear differential equations.

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Order and degree of Differential Equation

• Order: Highest differential coefficient
• Degree: Differential equation when expressed as [Polynomial in the derivatives] is the power of highest order derivative.

f(xy)dmy/dxm +g(xm)dm-1y/dxm-1 +... = 0

Note: Degree of differential equation is only defined in polynomial form.

Must Read:

• JEE Main Study Notes for Matrices & Determinants
• JEE Main Study Notes for Mathematical Reasoning

Formation of Differential Equations

An equation that has n arbitrary constants, a differential equation of nth order can be obtained by eliminating these arbitrary constants.

Steps:

• Suppose we have a given equation with n arbitrary constants f (x,y,c1,c2,....cn)=0
• Differentiate the equation successively n times to get n equations.
• The arbitrary constants should be eliminated from these n+1 equations leading to the required differential equation.

Example:

Form a differential equation of the family of curves represented by

 , where a and b are arbitrary constants.

Solution:

The given equation is  ...(i)

This equation has two arbitrary constants, therefore we will get a second-order differential equation. Differentiating the equation (i) twice with respect to x, we get

...(ii)

...(iii)

We have to eliminate a and b from these three equations.

From (i) , put in (iii)

we get which is the required differential equation.

Solution of Differential Equations

Followings are the differential equation form of the first order and first degree.

Variables Separable Form

If the equation can be reduced to the form of f(x)dx + g(y) dy=0, we may say that the variables have been separated. Integrate the reduced form, and you will get , where c is an arbitrary constant.

Differential Equation Reducible to variables Separable Form

A differential equation of the form ...(i)

can be reduced to variables separable form by substituting

The equation (i) then becomes

Hence the variables are separated in terms of z and x.

Differential Equation reducible to variable to Variables Separable Form

A differential equation of the form ...(i)

can be reduced to variables separable form by substituting

The equation (i) then becomes

Hence the variables are separated in terms of z and x.

Also Read:

• JEE Main Study Notes for Trigonometry
• JEE Main Study Notes for Integral Calculus

Homogeneous Differential Equations

If , then the function f(x,y) is said to be homogeneous of degree n.

Alternatively, a homogeneous function f(x,y) of degree n can be expressed as

Suppose  is a differential equation expression form.

Where f (x, y) and g (x, y) are homogeneous functions of the same degree. To solve such types of equations we put . The given equation then reduces to 

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

Linear Differential Equation

A linear differential equation can be found in the following forms

(i), where P and Q are functions of x or constants.

(ii) , where R and S are functions of y or constants.

Consider the differential equation (i) i.e

Multiply both the sides by to get

Integrating, we get

which is the general solution of the given equation. The expression is called the Integrating Factor of the equation.

Therefore, The solution can also be written as

where, I.F. = Integrating Factor =

[For the second differential equation , the integrating factor, I.F. = and the general solution is

x (I.F.) = ]

Read: JEE Main Probability Study Notes

Solution of Differential Equation by Inspection

Previous Years’ Solved Question on Differential Equations

Question 1: Equation of curve through the point (1, 0) which satisfies the differential equation (1 + y2) dx − xy dy = 0, is ________.

Solution:

We have [dx / x] = [y dy] / [1 + y2]

Integrating, we get log |x| = ([1 / 2] * log [1 + y2]) + log c

|x| = c √(1 + y2)−−−−−−−

But it passes through (1, 0), so we get c = 1

Therefore, the solution is x2= y2+ 1 or x2− y2 = 1.

Question 2: A function y = f (x) has the second-order derivatives f′′(x) = 6 (x − 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x − 5, then the function is __________.

Solution:

Given f′′(x) = 6 (x − 1)

f′(x) = 3 (x-1)2 + c1 ..(i)

But at point (2, 1) the line y = 3x − 5 is tangent to the graph y = f(x).

Hence, dy / dx|x = 2 = 3 or

f′ (2) = 3.

Then from (i) f′ (2) = 3 (2-1)2 + c1

3 = 3 + c1

c1 = 0 i.e.,

f′ (x) = 3 (x-1)2

Given f (2) = 1

f (x) = (x-1)3 + c2

f (2) = 1 + c2

1 = 1 + c2

c2 = 0

Hence, f (x) = (x-1)3.

Question 3: An integrating factor for the differential equation (1 + y2) dx − (tan-1 y − x) dy = 0

Solution:

(1 + y2) dx − (tan-1 y − x) dy = 0

dy / dx = (1 + y2) / (tan-1 y − x)

dx / dy = [tan-1 y / 1 + y2] − [x / 1 + y2] [dx / dy] + [x / 1 + y2] = [tan-1 y] / [1 + y2]

This is equation of the form dx / dy + Px = Q

So, I.F. = e∫Pdy

= e∫1 / 1 + y2.dy

= etan−1y

Question 4: Solution of differential equation 2xy dy / dx = x2 + 3y2 is __________.

Solution:

It is homogeneous equation dy / dx = [x2 + 3y2] / 2xy

Put y = vx and dy / dx = v + x * [dv / dx]

So, we get x * [dv / dx] = [1 + v2] / 2v

2v dv / 1 + v2 = dx / x

On integrating, we get

x2 + y2 = px3. (where p is a constant)

Question 5: The solution of the differential equation dy / dx = 1 + x + y + xy is __________.

Solution:

dy / dx = 1 + x + y + xy

dy / dx = (1 + x) (1 + y)

dy / dx + sin ([x + y] / 2)

= sin ([x − y] / 2)

On integrating, we get

log (1 + y) = x2 / 2 + x + c

Preparation Tips for Differential Equations

• Conventional Differential Equations, Formation and Solution of Differential Equations are the most important units of differential equations. Candidates should prepare themselves in these topics.
• Have limited resources because too many resources will only alleviate the stress. Candidates are advised to research well and gather some solid resources and stick to it.
• Revise the topics you have studied on every alternative day. Make some notes for the last minute and just don’t panic.
• Candidates should attempt mock tests to identify his/her strengths and weaknesses.
• Maintain a healthy lifestyle during the preparation otherwise you will lose your productivity. Eat healthy and sleep for at least 6-7 hours.Check JEE Main Preparation Tips

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