Differential Equations JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Differential Equations, free to read — no sign-in needed. The full chapter has 29 questions; sign in to attempt the remaining 24 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Let $y = f(x)$ be the real valued function defined on the interval $(0, \infty)$, satisfying $y(1) = 0$ and the differential equation $x\dfrac{dy}{dx} = y - x^3$. Then which of the following statements is (are) TRUE ?
   1. A.The function $f$ has a local minimum at $x = \dfrac{1}{\sqrt{3}}$
   2. B.The function $f$ has a local maximum at $x = \dfrac{1}{\sqrt{3}}$
   3. C.The function $f$ is increasing in the interval $(1, 2)$
   4. D.If $g(x) = 4x^3 - 5x^2 + \dfrac{3}{2}x$ for $x \gt 0$, then the number of elements in the set $\{x \in (0, \infty) : f(x) = g(x)\}$ is $2$
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   Answer: B,D

   The given differential equation is $x\dfrac{dy}{dx} = y - x^3$. Dividing by $x$, we get: $\dfrac{dy}{dx} - \dfrac{1}{x}y = -x^2$ This is a linear differential equation of the form $\dfrac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -\dfrac{1}{x}$ and $Q(x) = -x^2$. Integrating Factor (IF) = $e^{\int -\dfrac{1}{x} dx} = e^{-\ln x} = \dfrac{1}{x}$. Multiplying the differential equation by the IF: $\dfrac{d}{dx} \left( y \cdot \dfrac{1}{x} \right) = -x$ Integrating both sides with respect to $x$: $\dfrac{y}{x} = -\dfrac{x^2}{2} + C$ $y = -\dfrac{x^3}{2} + Cx$ Given $y(1) = 0$: $0 = -\dfrac{1}{2} + C \Rightarrow C = \dfrac{1}{2}$ Thus, $f(x) = \dfrac{x}{2} - \dfrac{x^3}{2}$. To find local extrema, we find $f'(x)$: $f'(x) = \dfrac{1}{2} - \dfrac{3x^2}{2}$ Setting $f'(x) = 0$ gives $3x^2 = 1 \Rightarrow x = \dfrac{1}{\sqrt{3}}$ (since $x \gt 0$). Now, $f''(x) = -3x$. At $x = \dfrac{1}{\sqrt{3}}$, $f''\left(\dfrac{1}{\sqrt{3}}\right) = -\sqrt{3} \lt 0$. So, $f(x)$ has a local maximum at $x = \dfrac{1}{\sqrt{3}}$. For $x \in (1, 2)$, $x^2 \gt 1$, so $f'(x) = \dfrac{1 - 3x^2}{2} \lt 0$. Thus, $f(x)$ is decreasing in $(1, 2)$. For the intersection of $f(x)$ and $g(x)$: $\dfrac{x}{2} - \dfrac{x^3}{2} = 4x^3 - 5x^2 + \dfrac{3}{2}x$ $x - x^3 = 8x^3 - 10x^2 + 3x$ $9x^3 - 10x^2 + 2x = 0$ $x(9x^2 - 10x + 2) = 0$ Since $x \in (0, \infty)$, $x \neq 0$. We solve $9x^2 - 10x + 2 = 0$. Discriminant $\Delta = (-10)^2 - 4(9)(2) = 100 - 72 = 28 \gt 0$. Sum of roots $= \dfrac{10}{9} \gt 0$ and Product of roots $= \dfrac{2}{9} \gt 0$. Both roots are real and positive. Therefore, there are exactly $2$ elements in the set. Answer: The function $f$ has a local maximum at $x = \dfrac{1}{\sqrt{3}}$; If $g(x) = 4x^3 - 5x^2 + \dfrac{3}{2}x$ for $x \gt 0$, then the number of elements in the set $\{x \in (0, \infty) : f(x) = g(x)\}$ is $2$
2. Q2JEE Advanced Adv 2026 (Paper 2)
   Let $y : (-\infty, \infty) \rightarrow (0, \infty)$ be the solution of the differential equation $\dfrac{dy}{dx} = \dfrac{e^{5x} y^3 + y^3}{e^x + e^x y^4}$, satisfying $y(0) = \dfrac{1}{\sqrt{2}}$. Then the value of $y(\log_e 2)$ is
   1. A.$\sqrt{\dfrac{5 + \sqrt{35}}{2}}$
   2. B.$\sqrt{\dfrac{7 + \sqrt{53}}{2}}$
   3. C.$\dfrac{7 + \sqrt{53}}{2}$
   4. D.$\dfrac{5 + \sqrt{35}}{2}$
   Show answer & solution

   Answer: (B)

   The given differential equation is $\dfrac{dy}{dx} = \dfrac{e^{5x} y^3 + y^3}{e^x + e^x y^4}$ Factoring the numerator and denominator, we get $\dfrac{dy}{dx} = \dfrac{y^3 (e^{5x} + 1)}{e^x (1 + y^4)}$ Separating the variables $x$ and $y$, we obtain $\dfrac{1 + y^4}{y^3} dy = \dfrac{e^{5x} + 1}{e^x} dx$ $\left( \dfrac{1}{y^3} + y \right) dy = (e^{4x} + e^{-x}) dx$ Integrating both sides yields $\int \left( y^{-3} + y \right) dy = \int (e^{4x} + e^{-x}) dx$ $-\dfrac{1}{2y^2} + \dfrac{y^2}{2} = \dfrac{e^{4x}}{4} - e^{-x} + C$ Multiplying by $2$, we get $y^2 - \dfrac{1}{y^2} = \dfrac{e^{4x}}{2} - 2e^{-x} + 2C$ Using the initial condition $y(0) = \dfrac{1}{\sqrt{2}}$, we substitute $x = 0$ and $y = \dfrac{1}{\sqrt{2}}$: $\left(\dfrac{1}{\sqrt{2}}\right)^2 - \dfrac{1}{\left(\dfrac{1}{\sqrt{2}}\right)^2} = \dfrac{e^0}{2} - 2e^0 + 2C$ $\dfrac{1}{2} - 2 = \dfrac{1}{2} - 2 + 2C$ This gives $2C = 0$, so $C = 0$. The equation becomes $y^2 - \dfrac{1}{y^2} = \dfrac{e^{4x}}{2} - 2e^{-x}$ To find $y(\log_e 2)$, we substitute $x = \log_e 2$ into the equation. We have $e^{4x} = e^{4 \log_e 2} = 16$ and $e^{-x} = e^{-\log_e 2} = \dfrac{1}{2}$. $y^2 - \dfrac{1}{y^2} = \dfrac{16}{2} - 2\left(\dfrac{1}{2}\right)$ $y^2 - \dfrac{1}{y^2} = 8 - 1 = 7$ Let $y^2 = t$. Since $y \in (0, \infty)$, $t \gt 0$. $t - \dfrac{1}{t} = 7$ $t^2 - 7t - 1 = 0$ Solving for $t$ using the quadratic formula gives $t = \dfrac{7 \pm \sqrt{49 - 4(1)(-1)}}{2} = \dfrac{7 \pm \sqrt{53}}{2}$ Since $t = y^2 \gt 0$, we must choose the positive root: $y^2 = \dfrac{7 + \sqrt{53}}{2}$ Taking the positive square root since $y \gt 0$, we get $y = \sqrt{\dfrac{7 + \sqrt{53}}{2}}$
3. Q3JEE Advanced Adv 2025 (Paper 2)
   Let $y(x)$ be the solution of the differential equation $x^2 \frac{d y}{d x}+x y=x^2+y^2, x\gt \frac{1}{e}$ satisfying $y(1)=0$. Then the value of $2 \frac{(y(e))^2}{y\left(e^2\right)}$ is _____ .
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   Answer: 0.75

   Put $\mathrm{y}=\mathrm{vx} \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$ D.E. $\begin{aligned} & \mathrm{x}^2\left(v+\mathrm{x} \frac{\mathrm{~d} v}{\mathrm{dx}}\right)+\mathrm{x}^2 v=\mathrm{x}^2\left(1+v^2\right) \\ & \Rightarrow v+x \frac{d v}{d x}+v=1+v^2 \\ & \Rightarrow x \frac{d v}{d x}=1+v^2-2 v \\ & \Rightarrow \int \frac{d v}{(v-1)^2}=\int \frac{d x}{x} \\ & \Rightarrow-\frac{1}{v-1}=\ln |x|+\mathrm{C} \end{aligned}$ $\Rightarrow \frac{\mathrm{x}}{\mathrm{x}-\mathrm{y}}=\ln |x|+\mathrm{C}=\ln \mathrm{x}+\mathrm{C} \quad\left(\right.$ Since $\left.\mathrm{x}\gt \frac{1}{\mathrm{e}}\right)$ Given $y(1)=0$ $\Rightarrow C=1$ So $\frac{x}{x-y}=\ln (e x)$ Now $y(e)=\frac{e}{2}$ and $y\left(e^2\right)=\frac{2 e^2}{3}$ $\therefore \frac{2(y(e))^2}{y\left(e^2\right)}=\frac{2 \cdot \frac{e^2}{4}}{\frac{2 e^2}{3}}=\frac{3}{4}=0.75$
4. Q4JEE Advanced Adv 2025 (Paper 1)
   For all $x\gt 0$, let $y_1(x), y_2(x)$, and $y_3(x)$ be the functions satisfying $\begin{aligned} & \frac{d y_1}{d x}-(\sin x)^2 y_1=0, y_1(1)=5 \\ & \frac{d y_2}{d x}-(\cos x)^2 y_2=0, y_2(1)=\frac{1}{3} \\ & \frac{d y_3}{d x}-\left(\frac{2-x^3}{x^3}\right) y_3=0, y_3(1)=\frac{3}{5 e}\end{aligned}$ respectively. Then $\lim _{x \rightarrow 0^{+}} \frac{y_1(x) y_2(x) y_3(x)+2 x}{e^{3 x} \sin x}$ is equal to _____
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   Answer: 2

   $\begin{aligned} & \frac{d y_1}{y_1}+\frac{d y_2}{y_2}+\frac{d y_3}{y_3}=\left(\sin ^2 x+\cos ^2 x+\frac{2-x^3}{x^3}\right) d x \\ & \ln \left(y_1 y_2 y_3\right)=\frac{-1}{x^2}+C \\ & \ln \left(y_1(x) y_2(x) y_3(x)\right)=\frac{-1}{x^2}+C \\ & \ln \left(5 \cdot \frac{1}{3} \cdot \frac{3}{5 e}\right)=\frac{-1}{x^2}+C \\ & \therefore C=0\end{aligned}$ $\begin{aligned} & y_1(x) y_2(x) y_3(x)=e^{\frac{-1}{x^2}} \\ & \lim _{x \rightarrow 0} \frac{e^{\frac{-1}{x^2}}+2 x}{e^{3 x} \sin x} \\ & \lim _{x \rightarrow 0} \frac{1}{e^{3 x+\frac{1}{x^2}} \sin x}+\lim _{x \rightarrow 0} \frac{2 x}{e^{3 x} \sin x}\end{aligned}$ $\lim _{x \rightarrow 0} \frac{1}{e^{3 x} \cdot \frac{\sin x}{x} \cdot \frac{e^{\frac{1}{x^2}}}{\frac{1}{x}}}+2$ $=0+2 \quad\left\{\lim _{\mathrm{y} \rightarrow \infty} \frac{\mathrm{e}^{\mathrm{y}^2}}{\mathrm{y}}=\frac{\mathrm{e}^{\mathrm{y}^2} \cdot 2 \mathrm{y}}{1}=\infty\right\}$ $=2$
5. Q5JEE Advanced Adv 2023 (Paper 2)
   Let $f:[1.\infty )\rightarrow ℝ$ be a differentiable function such that $f\left(1\right)=\dfrac{1}{3}$ and $3{\int }_{1}^{x}f\left(t\right)dt=xf\left(x\right)-\dfrac{{x}^{3}}{3},x\in [1,\infty )$. Let $e$ denote the base of the natural logarithm. Then the value of $f\left(e\right)$ is
   1. A.$\dfrac{{e}^{2}+4}{3}$
   2. B.$\dfrac{{\log }_{e}4+e}{3}$
   3. C.$\dfrac{4{e}^{2}}{3}$
   4. D.$\dfrac{{e}^{2}-4}{3}$.
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   Answer: (C)

   Given, $3{\int }_{1}^{x}f\left(t\right)dt=xf\left(x\right)-\dfrac{{x}^{3}}{3}$ Now differentiating both side we get, $\Rightarrow 3f\left(x\right)=f\left(x\right)+x{f}^{'}\left(x\right)-{x}^{2}$ $\Rightarrow x{f}^{'}\left(x\right)-2f\left(x\right)={x}^{2}$ $\Rightarrow {f}^{'}\left(x\right)-\dfrac{2}{x}f\left(x\right)=x$ Which is a linear differential equation, $\Rightarrow I.F.={e}^{-\dfrac{2}{x}dx}=\dfrac{1}{{x}^{2}}$ Now solution is given by, $\Rightarrow y\left(\dfrac{1}{{x}^{2}}\right)=\int x\times \dfrac{1}{{x}^{2}}dx=\ln ⁡x+C$ $\Rightarrow y={x}^{2}(\ln ⁡x+C)$ $\Rightarrow f\left(x\right)={x}^{2}(\ln x+C)$ Now using given value, $f\left(1\right)=\dfrac{1}{3}$ we get, $\Rightarrow 0+C=\dfrac{1}{3}\Rightarrow C=\dfrac{1}{3}$ So, $f\left(x\right)={x}^{2}\left(\ln x+C\right)$ $\Rightarrow f\left(e\right)={e}^{2}\left(\ln e+\dfrac{1}{3}\right)$ $\Rightarrow f\left(e\right)=\dfrac{4{e}^{2}}{3}$