Differential Equations JEE Main previous year questions with solutions

5 solved JEE Main questions on Differential Equations, free to read — no sign-in needed. The full chapter has 241 questions; sign in to attempt the remaining 236 in the exam simulator.

1. Q1JEE Main 2026 (05 Apr, Shift 1)Homogeneous DE
   Let $y = y(x)$ be the solution of the differential equation $x\sin\left(\dfrac{y}{x}\right)dy = \left(y\sin\left(\dfrac{y}{x}\right) - x\right)dx$, $y(1) = \dfrac{\pi}{2}$ and let $\alpha = \cos\left(\dfrac{y(e^{12})}{e^{12}}\right)$. Then the number of integral values of $p$, for which the equation $x^2 + y^2 - 2px + 2py + \alpha + 2 = 0$ represents a circle of radius $r \leq 6$, is __________.
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   Answer: 6

   Given the differential equation: $x\sin\left(\dfrac{y}{x}\right)dy = \left(y\sin\left(\dfrac{y}{x}\right) - x\right)dx$ Rearranging the terms, we get: $x\sin\left(\dfrac{y}{x}\right)dy - y\sin\left(\dfrac{y}{x}\right)dx = -xdx$ Dividing both sides by $x^2$: $\sin\left(\dfrac{y}{x}\right)\left(\dfrac{xdy - ydx}{x^2}\right) = -\dfrac{dx}{x}$ $\sin\left(\dfrac{y}{x}\right)d\left(\dfrac{y}{x}\right) = -\dfrac{dx}{x}$ Integrating both sides: $-\cos\left(\dfrac{y}{x}\right) = -\ln|x| - C$ $\cos\left(\dfrac{y}{x}\right) = \ln|x| + C$ Using the initial condition $y(1) = \dfrac{\pi}{2}$: $\cos\left(\dfrac{\pi}{2}\right) = \ln(1) + C \Rightarrow C = 0$ Thus, the solution to the differential equation is: $\cos\left(\dfrac{y}{x}\right) = \ln|x|$ We need to find $\alpha = \cos\left(\dfrac{y(e^{12})}{e^{12}}\right)$. Substituting $x = e^{12}$: $\alpha = \ln(e^{12}) = 12$ The given equation of the circle is: $x^2 + y^2 - 2px + 2py + \alpha + 2 = 0$ Substituting $\alpha = 12$: $x^2 + y^2 - 2px + 2py + 14 = 0$ The radius $r$ of the circle is given by: $r = \sqrt{(-p)^2 + p^2 - 14} = \sqrt{2p^2 - 14}$ For the equation to represent a real circle, $r \gt 0$: $2p^2 - 14 \gt 0 \Rightarrow p^2 \gt 7$ We are given that $r \leq 6$: $\sqrt{2p^2 - 14} \leq 6$ $2p^2 - 14 \leq 36$ $2p^2 \leq 50 \Rightarrow p^2 \leq 25$ Combining the inequalities: $7 \lt p^2 \leq 25$ Since $p$ is an integer, the possible values for $p^2$ are $9, 16, 25$. This gives $p \in \{-5, -4, -3, 3, 4, 5\}$. Therefore, there are $6$ integral values of $p$. Answer: $6$
2. Q2JEE Main 2025 (08 Apr, Shift 2)Linear Differential Equation
   Let $f(x)=x-1$ and $g(x)=e^x$ for $x \in \mathbb{R}$. If $\frac{d y}{d x}=\left(e^{-2 \sqrt{x}} g(f(f(x)))-\frac{y}{\sqrt{x}}\right), y(0)=0$, then $y(1)$ is :-
   1. A.$\frac{1-e^2}{e^4}$
   2. B.$\frac{2 e-1}{e^3}$
   3. C.$\frac{e-1}{e^4}$
   4. D.$\frac{1-e^3}{e^4}$
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   Answer: (C)

   $\begin{aligned} & f(x)=x-1 \\ & f(f(x))=f(x)-1=x-1-1=x-2 \\ & g(f(f(x)))=e^{x-2} \\ & \therefore \frac{d y}{d x}=e^{-2 \sqrt{x}} \times e^{x-2}-\frac{1}{\sqrt{x}} y \\ & \frac{d y}{d x}+\frac{1}{\sqrt{x}} y=e^{x-2 \sqrt{x}-2} \text { which is L.D.E } \\ & \text { I.F. }=e^{\int \frac{d y}{\sqrt{x}}}=e^{2 \sqrt{x}} \end{aligned}$ Its solution is $\begin{aligned} & y \times e^{2 \sqrt{x}}=\int e^{2 \sqrt{x}} \times e^{x-2 \sqrt{x}-2} d x+c \\ & y \times e^{2 \sqrt{x}}=\int e^{x-2} d x+c \\ & y \times e^{2 \sqrt{x}}=e^{x-2}+c \end{aligned}$ $\text { Given } \mathrm{x}=0, \mathrm{y}=0 \Rightarrow 0=\mathrm{e}^{-2}+\mathrm{c} \quad ; \mathrm{c}=-\mathrm{e}^{-2}$ $\therefore \mathrm{y} \times \mathrm{e}^{2 \sqrt{\mathrm{x}}}=\mathrm{e}^{\mathrm{x}-2}-\mathrm{e}^{-2}$ $\text { when } \mathrm{x}=1, \mathrm{y} \times \mathrm{e}^2=\mathrm{e}^{-1}-\mathrm{e}^{-2}$ $y=\frac{e^{-1}-e^{-2}}{e^2}=\frac{\frac{1}{e}-\frac{1}{e^2}}{e^2}=\frac{e^2-e}{e^5}=\frac{e-1}{e^4}$ Option (1) is correct
3. Q3JEE Main 2024 (31 Jan, Shift 2)Applications of DE
   The temperature $T\left(t\right)$ of a body at time $t=0$ is ${160}^{^{\circ}}F$ and it decreases continuously as per the differential equation $\dfrac{dT}{dt}=-K\left(T-80\right)$, where $K$ is positive constant. If $T\left(15\right)={120}^{^{\circ}}F$, then $T\left(45\right)$ is equal to
   1. A.${85}^{^{\circ}}F$
   2. B.${95}^{^{\circ}}F$
   3. C.${90}^{^{\circ}}F$
   4. D.${80}^{^{\circ}}F$
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   Answer: (C)

   Given: $\dfrac{dT}{dt}=-K\left(T-80\right)$ $\Rightarrow \dfrac{dT}{\left(T-80\right)}=-Kdt$ $\Rightarrow \int _{160}^{T}\dfrac{dT}{\left(T-80\right)}=\int _{0}^{t}-Kdt$ $\Rightarrow {\left[\log \left|T-80\right|\right]}_{160}^{T}=-Kt$ $\Rightarrow \log \left|T-80\right|-\log 80=-Kt$ $\Rightarrow \log \left|\dfrac{T-80}{80}\right|=-Kt$ $\Rightarrow T=80+80{e}^{-Kt}$ Now, using the value $T\left(15\right)={120}^{^{\circ}}$ we get, $\Rightarrow 120=80+80{e}^{-K\cdot 15}$ $\Rightarrow \dfrac{40}{80}={e}^{-15k}$ $\Rightarrow {e}^{-15k}=\dfrac{1}{2}$ $∴T\left(45\right)=80+80{e}^{-45k}$ $\Rightarrow T\left(45\right)=80+80{\left({e}^{-15k}\right)}^{3}$ $\Rightarrow T\left(45\right)=80+80\times \dfrac{1}{8}$ $\Rightarrow T\left(45\right)={90}^{^{\circ}}F$
4. Q4JEE Main 2023 (11 Apr, Shift 1)Exact Forms
   Let $y=y\left(x\right)$ be a solution curve of the differential equation, $\left(1-{x}^{2}{y}^{2}\right)dx=ydx+xdy$, If the line $x=1$ intersects the curve $y=y\left(x\right)$ at $y=2$ and the line $x=2$ intersects the curve $y=y\left(x\right)$ at $y=\alpha$, then a value of $\alpha$ is
   1. A.$\dfrac{1-3{e}^{2}}{2\left(3{e}^{2}+1\right)}$
   2. B.$\dfrac{1+3{e}^{2}}{2\left(3{e}^{2}-1\right)}$
   3. C.$\dfrac{3{e}^{2}}{2\left(3{e}^{2}-1\right)}$
   4. D.$\dfrac{3{e}^{2}}{2\left(3{e}^{2}+1\right)}$
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   Answer: (B)

   Given, $\left(1-{x}^{2}{y}^{2}\right)dx=ydx+xdy$ $\Rightarrow dx=\dfrac{ydx+xdy}{1-{\left(xy\right)}^{2}}$ $\Rightarrow dx=\dfrac{d\left(yx\right)}{1-{\left(xy\right)}^{2}}$ $\Rightarrow dx=\dfrac{1}{2}\left(\dfrac{d\left(xy\right)}{1-xy}+\dfrac{d\left(xy\right)}{1+xy}\right)$ Now integrating both side we get, $\Rightarrow 2x+c=\ln \left|\dfrac{1+xy}{1-xy}\right|$ $\Rightarrow \left|\dfrac{xy+1}{xy-1}\right|={e}^{c}{e}^{2x}$ Now given, $y\left(1\right)=2$ so putting the value in above equation we get, $3={e}^{c}{e}^{2}$ $\Rightarrow {e}^{c}=\dfrac{3}{{e}^{2}}$ Hence, the equation becomes $\left|\dfrac{xy+1}{xy-1}\right|=3{e}^{2x-2}$ Now finding $y\left(2\right)$ so putting $x=2$ in above equation we get, $\left|\dfrac{2y+1}{2y-1}\right|=3{e}^{2}$ $\Rightarrow 2y+1=2\cdot 3{e}^{2}y-3{e}^{2}$ $\Rightarrow 1+3{e}^{2}=2y\left(3{e}^{2}-1\right)$ $\Rightarrow y\left(2\right)=\dfrac{1+3{e}^{2}}{2\left(3{e}^{2}-1\right)}$
5. Q5JEE Main 2022 (29 Jul, Shift 2)Variable Separable Form
   If the solution curve of the differential equation $\dfrac{dy}{dx}=\dfrac{x+y-2}{x-y}$ passes through the point $\left(2,1\right)$ and $\left(k+1,2\right),k\gt 0$, then
   1. A.$2{\tan }^{-1}\left(\dfrac{1}{k}\right)={\log }_{e}\left({k}^{2}+1\right)$
   2. B.${\tan }^{-1}\left(\dfrac{1}{k}\right)={\log }_{e}\left({k}^{2}+1\right)$
   3. C.$2{\tan }^{-1}\left(\dfrac{1}{k+1}\right)={\log }_{e}\left({k}^{2}+2k+2\right)$
   4. D.$2{\tan }^{-1}\left(\dfrac{1}{k}\right)={\log }_{e}\left(\dfrac{{k}^{2}+1}{{k}^{2}}\right)$
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   Answer: (A)

   Given, $\dfrac{dy}{dx}=\dfrac{x+y-2}{x-y}=\dfrac{\left(x-1\right)+\left(y-1\right)}{\left(x-1\right)-\left(y-1\right)}$ Now let $x-1=X,y-1=Y$ So, $\dfrac{dy}{dx}=\dfrac{X+Y}{X-Y}........\left(1\right)$ Now let $Y=VX\dfrac{dY}{dX}=V+X\dfrac{dV}{dX}$ Putting the value in equation $\left(1\right)$ we get, $V+X\dfrac{dV}{dX}=\dfrac{1+V}{1-V}$ $\Rightarrow X\dfrac{dV}{dX}=\dfrac{{V}^{2}+1}{1-V}$ $\Rightarrow \int \dfrac{1-V}{1+{V}^{2}}dV=\int \dfrac{dX}{X}$ $\Rightarrow \int \dfrac{dV}{1+{V}^{2}}-\dfrac{1}{2}\int \dfrac{2VdV}{1+{V}^{2}}=\int \dfrac{dX}{X}$ $\Rightarrow {\tan }^{-1}V-\dfrac{1}{2}\ln \left(1+{V}^{2}\right)=\ln X+c$ $\Rightarrow {\tan }^{-1}\left(\dfrac{Y}{X}\right)-\dfrac{1}{2}\ln \left(1+\dfrac{{Y}^{2}}{{X}^{2}}\right)=\ln \left(X\right)+c$ $\Rightarrow {\tan }^{-1}\left(\dfrac{y-1}{x-1}\right)-\dfrac{1}{2}\ln \left(1+\dfrac{{\left(y-1\right)}^{2}}{{\left(x-1\right)}^{2}}\right)=\ln \left(x-1\right)+c$ Now given curve passes through $\left(2,1\right)$ So, $0-\dfrac{1}{2}\ln 1=\ln 1+c\Rightarrow c=0$ Now given curve also passes through $\left(k+1,2\right)$ So, ${\tan }^{-1}\left(\dfrac{1}{k}\right)-\dfrac{1}{2}In\left(1+\dfrac{1}{{k}^{2}}\right)=\ln k$ $\Rightarrow 2{\tan }^{-1}\left(\dfrac{1}{k}\right)=In\left(\dfrac{1+{k}^{2}}{{k}^{2}}\right)+2\ln k$ $\Rightarrow 2{\tan }^{-1}\left(\dfrac{1}{k}\right)=In\left(1+{k}^{2}\right)$