Differential Equations JEE Main PYQs (2020–2026)

Solved Differential Equations PYQs

3 recent questions with answers. The full set of 199 is free to practise.

1. JEE Main 2026EasyHomogeneous Differential Equations & Reducible Forms
   Q1

   Let$x=x(y)$be the solution of the differential equation$2y^2\frac{dx}{dy}-2xy+x^2=0$,$y>1$,$x(e)=e$.
   
   Then$x(e^2)$is equal to:

   • A$\frac{3}{2}{e}^2$
   • B$\frac{2}{3}{e}^2$
   • C$e^2$
   • D$2e^2$

   Show answer & solution

   Correct answer: (B)

   $$2y(ydx-xdy)+x^{2}dy=0$$
   
   $$\Rightarrow -2y{x}^{2}d\left(\frac{y}{x}\right)+x^{2}dy=0$$
   
   $$\Rightarrow -2yd\left(\frac{y}{x}\right)+dy=0$$
   
   $$\Rightarrow -2d\left(\frac{y}{x}\right)+\frac{1}{y}dy=0$$
   
   $$\Rightarrow -\frac{2y}{x}+{\log }_{e}y=C$$
   
   Given $x(e)=e$
   
   $$\Rightarrow C=-1$$
   
   $$\therefore -\frac{2y}{x}+{\log }_{e}y=-1$$
   
   $$\Rightarrow \frac{2y}{x}-{\log }_{e}y=1$$
   
   Put $y=e^2$ , we get
   
   $$\Rightarrow \frac{2e^2}{x}-2=1$$
   
   $$\to \frac{2e^2}{x}=3\to x=\frac{2e^2}{3}$$
   
   $$\Rightarrow x(e^2)=\frac{2e^2}{3}$$
2. JEE Main 2026EasyVariable Separable Method & Reducible Forms
   Q2

   If the curve $y=f(x)$ passes through the point $(1,e)$ and satisfies the differential equation $dy=y(2+{\log }_{e}x)dx$, $x>0$, then $f(e)$ is equal to :

   • A$e^e$
   • B$e^{e^2}$
   • C$e^{2e}$
   • D$e^{2e}$

   Show answer & solution

   Correct answer: (C)

   $$\frac{dy}{dx}=y(2+lnx)$$
   
   To solve this, first separate the variables by dividing both sides by $y$ :
   
   $$\int \frac{dy}{y}=\int (2+lnx)dx$$
   
   Now integrate both sides. The left side gives $\ln y$ . For the right side, integrate term by term:
   
   $$\ln y=2x+x\ln x-x+C$$
   
   Combine like terms $2x-x=x$ :
   
   $$\ln y=x+x\ln x+C$$
   
   Since it passes through $(1,e)$ , put $x=1$ and $y=e$ in the equation. Then $\ln y=lne=1$ and $\ln 1=0$ :
   
   $$1=1+0+C\to C=0$$
   
   So the required relation becomes:
   
   $$\ln y=x+x\ln x$$
   
   Taking antilog (exponential) on both sides:
   
   $$f(x)=y=e^{x+x\ln x}$$
   
   Now substitute $x=e$ . Also, $\ln e=1$ :
   
   $$\Rightarrow f(e)=e^{e+e}=e^{2e}$$
3. JEE Main 2026EasyLinear Differential Equations of First Order (Integrating Factor)
   Q3

   Let $y=y(x)$ be the solution curve of the differential equation $\left(1+x^2\right)dy+\left(y-{\tan }^{-1}x\right)dx=0,y(0)=1$ . Then the value of $y(1)$ is :

   • A$\frac{4}{e^{\pi /4}}-\frac{\pi }{2}-1$
   • B$\frac{2}{e^{\pi /4}}+\frac{\pi }{4}-1$
   • C$\frac{4}{e^{\pi /4}}+\frac{\pi }{2}-1$
   • D$\frac{2}{e^{\pi /4}}-\frac{\pi }{4}-1$

   Show answer & solution

   Correct answer: (B)

   Given
   
   $$\left(1+x^2\right)dy+\left(y-{\tan }^{-1}x\right)dx=0,$$
   
   $$y(0)=1,$$
   
   divide by $dx$ :
   
   $$(1+x^2)\frac{dy}{dx}+y-{\tan }^{-1}x=0$$
   
   $$\frac{dy}{dx}+\frac{1}{1+x^2}y=\frac{{\tan }^{-1}x}{1+x^2}.$$
   
   This is linear. Integrating factor:
   
   $$\mu (x)=exp\left(\int \frac{1}{1+x^2}dx\right)=e^{{\tan }^{-1}x}.$$
   
   So
   
   $$\frac{d}{dx}\left(y{e}^{{\tan }^{-1}x}\right)=e^{{\tan }^{-1}x}\cdot \frac{{\tan }^{-1}x}{1+x^2}.$$
   
   Let $t={\tan }^{-1}x$ , then $dt=\frac{dx}{1+x^2}$ . Hence
   
   $$y{e}^t=\int t{e}^{t}dt+C=e^t(t-1)+C.$$
   
   Therefore
   
   $$y=t-1+C{e}^{-t}$$
   
   $$={\tan }^{-1}x-1+C{e}^{-{\tan }^{-1}x}.$$
   
   Use $y(0)=1$ and ${\tan }^{-1}0=0$ :
   
   $$1=0-1+C⟹C=2.$$
   
   So
   
   $$y(1)=\frac{\pi }{4}-1+2e^{-\frac{\pi }{4}}$$
   
   $$=\frac{2}{e^{\pi /4}}+\frac{\pi }{4}-1.$$
   
   Correct option: B.

Topic-wise question count

Where JEE Main has asked from Differential Equations, most frequent first.

• Linear Differential Equations of First Order (Integrating Factor)70
• Variable Separable Method & Reducible Forms58
• Homogeneous Differential Equations & Reducible Forms47
• Applications of Differential Equations in Growth, Decay & Geometry12
• Formation of Differential Equations6
• Bernoulli's Differential Equation5
• Order and Degree of a Differential Equation1

Year-wise Differential Equations questions in JEE Main

Number of questions from this chapter in each year's paper(s).