Differential Equations

26 JEE Main questions · Maths

Linear differential equations

MathsJEE Main 21 Jan 2026, Shift 1·Q8·MCQ·Moderate
Open
Let $y = y(x)$ be the solution curve of the differential equation $(1 + x^2)\,dy + (y - \tan^{-1}x)\,dx = 0$ , $y(0) = 1$ . Then the value of $y(1)$ is:
1. (1) $\dfrac{2}{e^{\frac{\pi}{4}}} + \dfrac{\pi}{4} - 1$
2. (2) $\dfrac{2}{e^{\frac{\pi}{4}}} - \dfrac{\pi}{4} - 1$
3. (3) $\dfrac{4}{e^{\frac{\pi}{4}}} + \dfrac{\pi}{2} - 1$
4. (4) $\dfrac{4}{e^{\frac{\pi}{4}}} - \dfrac{\pi}{2} - 1$
MathsJEE Main 21 Jan 2026, Shift 2·Q16·MCQ·Moderate
Open
Let $y = y(x)$ be the solution of the differential equation $\sec x\dfrac{dy}{dx} - 2y = 2 + 3\sin x$ , $x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ , $y(0) = -\dfrac{7}{4}$ . Then $y\left(\dfrac{\pi}{6}\right)$ is equal to:
1. (1) $-\dfrac{5}{2}$
2. (2) $-\dfrac{5}{4}$
3. (3) $-3\sqrt3 - 7$
4. (4) $-3\sqrt2 - 7$
MathsJEE Main 23 Jan 2026, Shift 2·Q21·Numerical·Moderate
Open
If the solution curve $y = f(x)$ of the differential equation $(x^2 - 4)\,y' - 2xy + 2x(4 - x^2)^2 = 0$ $x > 2$ , passes through the point $(3, 15)$ , then the local maximum value of $f$ is .
MathsJEE Main 24 Jan 2026, Shift 2·Q16·MCQ·Moderate
Open
Let $y = y(x)$ be a differentiable function in the interval $(0, \infty)$ such that $y(1) = 2$ . and $\lim_{t\to x}\left(\frac{t^2y(x) - x^2y(t)}{x - t}\right) = 3 \quad\text{for each } x > 0.$ Then $2y(2)$ is equal to
1. (1)18
2. (2)23
3. (3)27
4. (4)12
MathsJEE Main 28 Jan 2026, Shift 1·Q20·MCQ·Moderate
Open
Let $y = y(x)$ be the solution of the differential equation $x\frac{dy}{dx} - \sin 2y = x^3(2 - x^3)\cos^2 y,\quad x \ne 0.$ If $y(2) = 0$ , then $\tan(y(1))$ is equal to
1. (1) $\dfrac{3}{4}$
2. (2) $\dfrac{7}{4}$
3. (3) $-\dfrac{7}{4}$
4. (4) $-\dfrac{3}{4}$
We corrected a printing error in the original paper
Source printed “If $y(2) = x$ ”; corrected to the intended “ $y(2) = 0$ ” (the source's own solution uses $y(2) = 0$ , and the key (2) follows from it).
MathsJEE Main 4 Apr 2026, Shift 2·Q19·MCQ·Tough
Open
Let $y = y(x)$ be the solution of the differential equation: $\frac{dy}{dx} + \left(\frac{6x^2 + \left(3x^2 + 2x^3 + 4\right)e^{-2x}}{\left(x^3 + 2\right)\left(2 + e^{-2x}\right)}\right)y = 2 + e^{-2x},$ $x \in (-1, 2)$ , satisfying $y(0) = \dfrac32$ . If $y(1) = \alpha\left(2 + e^{-2}\right)$ , then $\alpha$ is equal to:
1. (1) $\dfrac{13}{8}$
2. (2) $\dfrac{6}{13}$
3. (3) $\dfrac{12}{13}$
4. (4) $\dfrac{13}{12}$
We corrected a printing error in the original paper
Source printed “than $\alpha$ is equal to”.
MathsJEE Main 5 Apr 2026, Shift 2·Q19·MCQ·Moderate
Open
Let $f : [1, \infty) \to \mathbb{R}$ be a differentiable function defined as $f(x) = \int_1^x f(t)\,dt + (1 - x)(\log_e x - 1) + e.$ Then the value of $f(f(1))$ is:
1. (1) $(1 + e^{e})$
2. (2) $(1 + e)$
3. (3) $(1 + e + e^{e})$
4. (4) $1 + 2e$
We corrected a printing error in the original paper
Source printed “Then the value fo $f(f(1))$ ”.
MathsJEE Main 5 Apr 2026, Shift 2·Q25·Numerical·Moderate
Open
Let $y = y(x)$ be the solution of the differential equation $(\tan x)^{1/2}\,dy = \left(\sec^3x - (\tan x)^{3/2}y\right)dx,$ $0 < x < \dfrac{\pi}{2}$ , $y\left(\dfrac{\pi}{4}\right) = \dfrac{6\sqrt2}{5}$ . If $y\left(\dfrac{\pi}{3}\right) = \dfrac45\alpha$ , then $\alpha^4$ equals .
MathsJEE Main 6 Apr 2026, Shift 2·Q18·MCQ·Tough
Open
Let $f : \mathbb{R} \to \mathbb{R}$ be such that $f(xy) = f(x)f(y)$ , for all $x, y \in \mathbb{R}$ and $f(0) \neq 0$ . Let $g : [1, \infty) \to \mathbb{R}$ be a differentiable function such that $x^2g(x) = \int_1^x \left(t^2f(t) - tg(t)\right)dt$ Then $g(2)$ is equal to:
1. (1) $\dfrac{13}{8}$
2. (2) $\dfrac{11}{16}$
3. (3) $\dfrac{15}{32}$
4. (4) $\dfrac{17}{64}$
MathsJEE Main 8 Apr 2026, Shift 2·Q17·MCQ·Moderate
Open
Let $y = y(x)$ be the solution of the differential equation $x\sqrt{1-x^2}\,dy + \left(y\sqrt{1-x^2} - x\cos^{-1}x\right)dx = 0,$ $x \in (0, 1)$ , $\displaystyle\lim_{x\to1^-} y(x) = 1$ . Then $y\left(\dfrac12\right)$ equals:
1. (1) $3 - \dfrac{\pi}{\sqrt3}$
2. (2) $4 - \sqrt3\,\pi$
3. (3) $4 - \dfrac{2\pi}{\sqrt3}$
4. (4) $3 - \dfrac{\pi}{2\sqrt3}$

Variable separable differential equations

MathsJEE Main 21 Jan 2026, Shift 1·Q21·Numerical·Tough
Open
Let $f : \mathbb{R} \to \mathbb{R}$ be a twice differentiable function such that the quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$ in $m$ , has two equal roots for every $x \in \mathbb{R}$ . If $f(0) = 1$ , $f'(0) = 2$ and $(\alpha, \beta)$ is the largest interval in which the function $f(\log_e x - x)$ is increasing, then $\alpha + \beta$ is equal to .
MathsJEE Main 22 Jan 2026, Shift 2·Q6·MCQ·Moderate
Open
If $y = y(x)$ satisfies the differential equation $16\left(\sqrt{x}\,\sqrt{9+\sqrt{x}}\right)\left(4 + \sqrt{9 + \sqrt{x}}\right)\cos y\,dy = (1 + 2\sin y)\,dx,\quad x > 0$ and $y(256) = \dfrac{\pi}{2}$ , $y(49) = \alpha$ , then $2\sin\alpha$ is equal to:
1. (1) $2\sqrt2 - 1$
2. (2) $2\left(\sqrt2 - 1\right)$
3. (3) $3\left(\sqrt2 - 1\right)$
4. (4) $\sqrt2 - 1$
We corrected a printing error in the original paper
Source printed the first factor as “ $\sqrt{x + 9\sqrt{x}}$ ”, with which none of the options is correct ( $2\sin\alpha \approx 1.466$ ); corrected to the intended “ $\sqrt{x}\,\sqrt{9+\sqrt{x}}$ ” (the source's own solution integrates with this factor), which gives the printed key (1).
MathsJEE Main 24 Jan 2026, Shift 1·Q21·Numerical·Tough
Open
Let a differentiable function $f$ satisfy the equation $\int_0^{36} f\left(\dfrac{tx}{36}\right)dt = 4\alpha f(x).$ If $y = f(x)$ is a standard parabola passing through the points $(2, 1)$ and $(-4, \beta)$ , Then $\beta^{\alpha}$ is equal to .
MathsJEE Main 2 Apr 2026, Shift 1·Q17·MCQ·Easy–moderate
Open
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:
1. (1) $e^{e}$
2. (2) $e^{e^2}$
3. (3) $e^{2e}$
4. (4) $e^{2^e}$
MathsJEE Main 4 Apr 2026, Shift 1·Q20·MCQ·Moderate
Open
Let $y = y(x)$ be the solution of the differential equation $\dfrac{dy}{dx} = (1 + x + x^2)(1 - y + y^2)$ , $y(0) = \dfrac12$ . Then $(2y(1) - 1)$ is equal to
1. (1) $\sqrt3\tan\left(\dfrac{11\sqrt3}{6}\right)$
2. (2) $\dfrac{\sqrt3}{2}\tan\left(\dfrac{11\sqrt3}{12}\right)$
3. (3) $\sqrt3\tan\left(\dfrac{11\sqrt3}{12}\right)$
4. (4) $\dfrac{\sqrt3}{2}\tan\left(\dfrac{11\sqrt3}{6}\right)$
Flawed question. Ill-posed: the solution with $y(0) = 1/2$ blows up near $x = 0.99$ (its arctangent would have to reach $11\sqrt3/12 > \pi/2$ ), so $y(1)$ does not exist; the keyed (3) is only a formal value (about $-102$ , impossible for an increasing $y$ starting at $1/2$ ), and no minimal change gives a valid question with these options. It is shown for completeness and is left out of practice tests.

Integral equation to differential equation

MathsJEE Main 22 Jan 2026, Shift 1·Q9·MCQ·Moderate
Open
Let $f : [1, \infty) \to \mathbb{R}$ be a differentiable function, If $6\int_1^x f(t)\,dt = 3xf(x) + x^3 - 4$ for all $x \ge 1$ , then the value of $f(2) - f(3)$ is
1. (1) $-4$
2. (2) $-3$
3. (3)4
4. (4)3
MathsJEE Main 23 Jan 2026, Shift 1·Q23·Numerical·Moderate
Open
Let $f$ be a twice differentiable non-negative function such that $(f(x))^2 = 25 + \int_0^x\left((f(t))^2 + (f'(t))^2\right)dt.$ Then the mean of $f(\log_e(1)), f(\log_e(2)), \ldots, f(\log_e(625))$ is equal to .
MathsJEE Main 4 Apr 2026, Shift 2·Q25·Numerical·Tough
Open
Let $f$ be a twice differentiable function such that $f(x) = \int_0^x\tan(t - x)\,dt - \int_0^x f(t)\tan t\,dt,\quad x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ Then $f''\left(\dfrac{\pi}{6}\right) + 12f'\left(-\dfrac{\pi}{6}\right) + f\left(\dfrac{\pi}{6}\right)$ is equal to .

Exact differential equations

MathsJEE Main 23 Jan 2026, Shift 1·Q7·MCQ·Easy–moderate
Open
Let $y = y(x)$ be the solution of the differential equation $x^4\,dy + (4x^3y + 2\sin x)\,dx = 0,\quad x > 0,\quad y\left(\frac{\pi}{2}\right) = 0.$ Then $\pi^4 y\left(\dfrac{\pi}{3}\right)$ is equal to:
1. (1)81
2. (2)92
3. (3)64
4. (4)72
MathsJEE Main 2 Apr 2026, Shift 1·Q20·MCQ·Easy–moderate
Open
Let $y = y(x)$ be the solution curve of the differential equation $(1 + \sin x)\dfrac{dy}{dx} + (y + 1)\cos x = 0$ , $y(0) = 0$ . If the curve $y = y(x)$ passes through the point $\left(\alpha, \dfrac{-1}{2}\right)$ , then a value of $\alpha$ is:
1. (1) $\dfrac{\pi}{6}$
2. (2) $\dfrac{\pi}{4}$
3. (3) $\dfrac{\pi}{3}$
4. (4) $\dfrac{\pi}{2}$

Homogeneous differential equations

MathsJEE Main 22 Jan 2026, Shift 1·Q19·MCQ·Moderate
Open
Let the solution curve of the differential equation $x\,dy - y\,dx = \sqrt{x^2 + y^2}\,dx$ , $x > 0$ , $y(1) = 0$ , be $y = y(x)$ . Then $y(3)$ is equal to
1. (1)4
2. (2)6
3. (3)1
4. (4)2
MathsJEE Main 5 Apr 2026, Shift 1·Q25·Numerical·Moderate
Open
Let $y = y(x)$ be the solution of the differential equation $x\sin\left(\frac{y}{x}\right)dy = \left(y\sin\left(\frac{y}{x}\right) - x\right)dx,\quad y(1) = \frac{\pi}{2}$ and let $\alpha = \cos\left(\dfrac{y\left(e^{12}\right)}{e^{12}}\right)$ . Then the number of integral value of $p$ , for which the equation $x^2 + y^2 - 2px + 2py + \alpha + 2 = 0$ represents a circle of radius $r \le 6$ , is .
Flawed question. Ill-posed: the solution is $\cos(y/x) = \log_e x$ , which exists only for $e^{-1} \le x \le e$ , so $y(e^{12})$ does not exist and $\alpha = \cos(\cdot) = 12$ is impossible; even with the formal $\alpha = 12$ ( $r^2 = 2p^2 - 14$ ) only $p = \pm3, \pm4, \pm5$ give a circle (6 values), while the key 11 also counts $p$ with $r^2 < 0$ ; no minimal, evident change repairs both. It is shown for completeness and is left out of practice tests.

Linear first order differential equation

MathsJEE Main 6 Apr 2026, Shift 1·Q25·Numerical·Moderate
Open
Let $y = y(x)$ be the solution of the differential equation $\left(x^2 - x\sqrt{x^2 - 1}\right)dy + \left(y\left(x - \sqrt{x^2 - 1}\right) - x\right)dx = 0,$ $x \ge 1$ . If $y(1) = 1$ , then the greatest integer less than $y\left(\sqrt5\right)$ is .
MathsJEE Main 28 Jan 2026, Shift 2·Q14·MCQ·Easy–moderate
Open
Let $y = y(x)$ be the solution of the differential equation $x\dfrac{dy}{dx} - y = x^2\cot x$ , $x \in (0, \pi)$ . If $y\left(\dfrac{\pi}{2}\right) = \dfrac{\pi}{2}$ , then $6y\left(\dfrac{\pi}{6}\right) - 8y\left(\dfrac{\pi}{4}\right)$ is equal to :
1. (1) $3\pi$
2. (2) $-3\pi$
3. (3) $-\pi$
4. (4) $\pi$

Bernoulli differential equation

MathsJEE Main 2 Apr 2026, Shift 2·Q20·MCQ·Moderate
Open
Let $x = x(y)$ be the solution of the differential equation $2y^2\dfrac{dx}{dy} - 2xy + x^2 = 0$ , $y > 1$ , $x(e) = e$ . Then $x(e^2)$ is equal to:
1. (1) $\dfrac32e^2$
2. (2) $\dfrac23e^2$
3. (3) $e^2$
4. (4) $2e^2$

Solution by direct integration

MathsJEE Main 5 Apr 2026, Shift 2·Q20·MCQ·Easy–moderate
Open
Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f''(x) = g''(x)$ for all $x \in \mathbb{R}$ , $f'(1) = 2g'(1) = 4$ and $g(2) = 3f(2) = 9$ . Then $f(25) - g(25)$ is equal to:
1. (1)20
2. (2)40
3. (3) $-20$
4. (4) $-40$