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 243 questions; sign in to attempt the remaining 238 in the exam simulator.

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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 (26 Jul, Shift 1)Area bounded by Miscellaneous Curves
   Let a curve $y=y\left(x\right)$ pass through the point $\left(3,3\right)$ and the area of the region under this curve, above the $x$-axis and between the abscissae $3$ and $x\left(\gt 3\right)$ be ${\left(\dfrac{y}{x}\right)}^{3}$. If this curve also passes through the point $\left(\alpha ,6\sqrt{10}\right)$ in the first quadrant, then $\alpha$ is equal to _______.
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   Answer: 6

   Given, the area of the region under this curve, above the $x$-axis and between the abscissae $3$ and $x\left(\gt 3\right)$ be ${\left(\dfrac{y}{x}\right)}^{3}$ So, ${\int }_{3}^{x}ydx={\left(\dfrac{y}{x}\right)}^{3}$ $\Rightarrow {x}^{3}{\int }_{3}^{x}ydx={y}^{3}$ Now differentiating both side w.r.t $x$, we get $\Rightarrow {x}^{3}y+3{x}^{2}\dfrac{{y}^{3}}{{x}^{3}}=3{y}^{2}\dfrac{dy}{dx}$ $\left\{\text{as given}{\int }_{3}^{x}ydx={\left(\dfrac{y}{x}\right)}^{3}\right\}$ $\Rightarrow {x}^{4}+3{y}^{2}=3yx\dfrac{dy}{dx}$ $\Rightarrow 3xy\dfrac{dy}{dx}=3{y}^{2}+{x}^{4}$ Now put ${y}^{2}=t,y\dfrac{dy}{dx}=\dfrac{1}{2}\dfrac{dt}{dx}$ $\Rightarrow \dfrac{dt}{dx}-\dfrac{2}{x}t=\dfrac{2}{3}{x}^{3}$ Now $IF={e}^{\int -\dfrac{2}{x}dx}=\dfrac{1}{{x}^{2}}$ So, solution is given by $t\times IF=\int \dfrac{2}{3}{x}^{3}\times IF$ $\Rightarrow \dfrac{t}{{x}^{2}}=\dfrac{{x}^{2}}{3}+C$ Now given curve passes through $\left(3,3\right)$ so $C=-2$ $\Rightarrow \dfrac{{y}^{2}}{{x}^{2}}=\dfrac{{x}^{2}}{3}-2$ So the equation of curve is $3{y}^{2}={x}^{4}-6{x}^{2}$ Now curve also passes through $\left(\alpha ,6\sqrt{10}\right)$ So, ${\alpha }^{4}-6{\alpha }^{2}=1080$ $\Rightarrow \alpha =6$