Differential Equations Previous Year Questions (JEE Main 2026)

The equation is variable-separable:

$$\frac{dy}{1 - y + y^2} = (1 + x + x^2)\,dx.$$

Complete the square in the denominator:

$$1 - y + y^2 = \left(y - \tfrac{1}{2}\right)^2 + \frac{3}{4}.$$

Integrate both sides:

$$\int \frac{dy}{\left(y-\frac12\right)^2 + \frac34} = \frac{2}{\sqrt3}\tan^{-1}\!\left(\frac{2y-1}{\sqrt3}\right),\qquad \int (1+x+x^2)\,dx = x + \frac{x^2}{2} + \frac{x^3}{3} + C.$$

So

$$\frac{2}{\sqrt3}\tan^{-1}\!\left(\frac{2y-1}{\sqrt3}\right) = x + \frac{x^2}{2} + \frac{x^3}{3} + C.$$

Apply $y(0)=\tfrac12$: the left side is $\tfrac{2}{\sqrt3}\tan^{-1}(0)=0$ and the right side is $C$, so $C = 0$.

At $x=1$: the right side is $1 + \frac12 + \frac13 = \frac{11}{6}$, hence

$$\tan^{-1}\!\left(\frac{2y(1)-1}{\sqrt3}\right) = \frac{\sqrt3}{2}\cdot\frac{11}{6} = \frac{11\sqrt3}{12}.$$

Therefore

$$2y(1)-1 = \sqrt3\,\tan\!\left(\frac{11\sqrt3}{12}\right).$$

Answer: C

Write the equation in derivative form:

$$x\left(x - \sqrt{x^2-1}\right)\frac{dy}{dx} + y\left(x - \sqrt{x^2-1}\right) = x.$$

Divide by $\left(x - \sqrt{x^2-1}\right)$:

$$x\frac{dy}{dx} + y = \frac{x}{x - \sqrt{x^2-1}}.$$

Rationalize the right side (multiply by $\frac{x + \sqrt{x^2-1}}{x + \sqrt{x^2-1}}$, and $x^2-(x^2-1)=1$):

$$x\frac{dy}{dx} + y = x\left(x + \sqrt{x^2-1}\right) = x^2 + x\sqrt{x^2-1}.$$

The left side is an exact derivative, $x\frac{dy}{dx} + y = \dfrac{d}{dx}(xy)$, so

$$\frac{d}{dx}(xy) = x^2 + x\sqrt{x^2-1}.$$

Integrate (note $\int x\sqrt{x^2-1}\,dx = \tfrac13 (x^2-1)^{3/2}$):

$$xy = \frac{x^3}{3} + \frac{1}{3}\left(x^2-1\right)^{3/2} + C.$$

Apply $y(1)=1$: $\;1 = \frac13 + 0 + C \Rightarrow C = \frac23.$

At $x=\sqrt5$: $x^3 = 5\sqrt5$ and $(x^2-1)^{3/2} = 4^{3/2}=8$, so

$$\sqrt5\,y(\sqrt5) = \frac{5\sqrt5}{3} + \frac{8}{3} + \frac23 = \frac{5\sqrt5 + 10}{3}.$$$$y(\sqrt5) = \frac{5\sqrt5 + 10}{3\sqrt5} = \frac{5 + 2\sqrt5}{3} \approx \frac{5 + 4.472}{3} \approx 3.157.$$

The greatest integer less than $3.157$ is $3$.

Answer: 3

The equation is separable:

$$\frac{dy}{y} = (2 + \ln x)\,dx.$$

Integrate, using $\int \ln x\,dx = x\ln x - x$:

$$\ln y = 2x + (x\ln x - x) + C = x\ln x + x + C.$$

Apply the point $(1,e)$: $\ln e = 1 = (1)(0) + 1 + C \Rightarrow C = 0.$ Hence

$$\ln y = x\ln x + x.$$

At $x=e$ (where $\ln e = 1$):

$$\ln y = e\cdot 1 + e = 2e \;\Rightarrow\; y = e^{2e}.$$

Answer: C

Separate the variables:

$$\frac{dy}{y+1} = -\frac{\cos x}{1 + \sin x}\,dx.$$

Integrate both sides (the right side is $-\ln(1+\sin x)$):

$$\ln(y+1) = -\ln(1 + \sin x) + C \;\Rightarrow\; (y+1)(1+\sin x) = K.$$

Apply $y(0)=0$: $(0+1)(1+0) = K \Rightarrow K = 1$, so

$$(y+1)(1 + \sin x) = 1.$$

At $y = -\tfrac12$:

$$\left(\tfrac12\right)(1 + \sin\alpha) = 1 \;\Rightarrow\; 1 + \sin\alpha = 2 \;\Rightarrow\; \sin\alpha = 1.$$

Thus $\alpha = \dfrac{\pi}{2}$.

Answer: D

This is linear; first simplify the coefficient $P(x)$. Split the numerator over the common denominator:

$$3x^2/(x^3+2) = \frac{3x^2(2 + e^{-2x})}{(x^3+2)(2+e^{-2x})} = \frac{6x^2 + 3x^2 e^{-2x}}{(x^3+2)(2+e^{-2x})}.$$

Subtracting this from $P(x)$ leaves $\dfrac{(2x^3+4)e^{-2x}}{(x^3+2)(2+e^{-2x})} = \dfrac{2e^{-2x}}{2+e^{-2x}}$, so

$$P(x) = \frac{3x^2}{x^3+2} + \frac{2e^{-2x}}{2 + e^{-2x}}.$$

Integrating factor. Note $\int \frac{3x^2}{x^3+2}\,dx = \ln(x^3+2)$ and $\int \frac{2e^{-2x}}{2+e^{-2x}}\,dx = -\ln(2+e^{-2x})$, so

$$\mu(x) = e^{\int P\,dx} = \frac{x^3 + 2}{2 + e^{-2x}}.$$

Solve. With $Q(x) = 2 + e^{-2x}$,

$$\frac{d}{dx}\big(\mu y\big) = \mu Q = \frac{x^3+2}{2+e^{-2x}}\,(2 + e^{-2x}) = x^3 + 2.$$

Integrate:

$$\mu y = \frac{x^4}{4} + 2x + C.$$

Apply $y(0)=\tfrac32$: $\mu(0) = \frac{2}{2+1} = \frac23$, so $\mu(0)\,y(0) = \frac23\cdot\frac32 = 1 = C$.

At $x=1$: $\mu(1)\,y(1) = \frac14 + 2 + 1 = \frac{13}{4}$, and $\mu(1) = \frac{3}{2+e^{-2}}$, so

$$y(1) = \frac{13}{4}\cdot\frac{2+e^{-2}}{3} = \frac{13}{12}\left(2 + e^{-2}\right).$$

Hence $\alpha = \dfrac{13}{12}$.

Answer: D

Treat $x$ as the dependent variable. This is a Bernoulli equation in $x$. Divide by $x^2$:

$$\frac{2y^2}{x^2}\frac{dx}{dy} - \frac{2y}{x} + 1 = 0.$$

Substitute $v = \dfrac{1}{x}$, so $\dfrac{dv}{dy} = -\dfrac{1}{x^2}\dfrac{dx}{dy}$:

$$-2y^2\frac{dv}{dy} - 2yv + 1 = 0 \;\Rightarrow\; \frac{dv}{dy} + \frac{1}{y}\,v = \frac{1}{2y^2}.$$

Integrating factor $\mu = e^{\int dy/y} = y$:

$$\frac{d}{dy}(yv) = y\cdot\frac{1}{2y^2} = \frac{1}{2y} \;\Rightarrow\; yv = \frac12\ln y + C.$$

Back-substitute $v = 1/x$:

$$\frac{y}{x} = \frac12\ln y + C.$$

Apply $x(e)=e$: $\frac{e}{e} = 1 = \frac12(1) + C \Rightarrow C = \frac12$, so

$$\frac{y}{x} = \frac12(\ln y + 1).$$

At $y = e^2$: $\frac{e^2}{x} = \frac12(2 + 1) = \frac32 \Rightarrow x = \frac{2}{3}e^2.$

Answer: B

Solve the homogeneous ODE. Write it as

$$\frac{dy}{dx} = \frac{y\sin(y/x) - x}{x\sin(y/x)} = \frac{y}{x} - \frac{1}{\sin(y/x)}.$$

Put $v = \dfrac{y}{x}$, so $y = vx$ and $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$:

$$v + x\frac{dv}{dx} = v - \frac{1}{\sin v} \;\Rightarrow\; x\frac{dv}{dx} = -\frac{1}{\sin v}.$$

Separate and integrate:

$$\sin v\,dv = -\frac{dx}{x} \;\Rightarrow\; -\cos v = -\ln x + C \;\Rightarrow\; \cos v = \ln x + C'.$$

Apply $y(1)=\tfrac{\pi}{2}$: at $x=1$, $v = \tfrac{\pi}{2}$, so $\cos\tfrac{\pi}{2} = 0 = \ln 1 + C' \Rightarrow C' = 0$. Hence

$$\cos\left(\frac{y}{x}\right) = \ln x.$$

Compute $\alpha$: at $x = e^{12}$, $\;\alpha = \cos\!\left(\dfrac{y(e^{12})}{e^{12}}\right) = \ln e^{12} = 12.$

Circle condition. The equation becomes $x^2 + y^2 - 2px + 2py + 14 = 0$, i.e. centre $(p, -p)$ and

$$r^2 = p^2 + p^2 - 14 = 2p^2 - 14.$$

For a real circle with $0 < r \le 6$:

$$2p^2 - 14 > 0 \;\text{and}\; 2p^2 - 14 \le 36 \;\Rightarrow\; 7 < p^2 \le 25.$$

Integer $p$: $|p| \in \{3, 4, 5\}$ (since $3^2 = 9 > 7$ and $5^2 = 25 \le 25$), giving $p \in \{-5, -4, -3, 3, 4, 5\}$.

That is $6$ integral values.

Answer: 6

Rewrite as a linear equation. Divide by $x\sqrt{1-x^2}$:

$$\frac{dy}{dx} + \frac{1}{x}\,y = \frac{\cos^{-1}x}{\sqrt{1-x^2}}.$$

Integrating factor $\mu = e^{\int dx/x} = x$:

$$\frac{d}{dx}(xy) = \frac{x\cos^{-1}x}{\sqrt{1-x^2}}.$$

Integrate with $u = \cos^{-1}x$ (so $x = \cos u$, $\sqrt{1-x^2} = \sin u$, $dx = -\sin u\,du$):

$$\int \frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx = \int \frac{\cos u\cdot u}{\sin u}(-\sin u)\,du = -\int u\cos u\,du = -\big(u\sin u + \cos u\big).$$

Back in terms of $x$:

$$xy = -\cos^{-1}(x)\sqrt{1-x^2} - x + C.$$

Apply the limit $x \to 1^-$: then $\cos^{-1}x \to 0$, $\sqrt{1-x^2}\to 0$, $x \to 1$, so the right side $\to C - 1$, while the left side $\to 1\cdot 1 = 1$. Hence $C - 1 = 1 \Rightarrow C = 2$.

$$y = \frac{-\cos^{-1}(x)\sqrt{1-x^2} - x + 2}{x}.$$

At $x = \tfrac12$ ($\cos^{-1}\tfrac12 = \tfrac{\pi}{3}$, $\sqrt{1-\tfrac14} = \tfrac{\sqrt3}{2}$):

$$y\!\left(\tfrac12\right) = \frac{-\frac{\pi}{3}\cdot\frac{\sqrt3}{2} - \frac12 + 2}{\frac12} = 2\left(\frac32 - \frac{\pi\sqrt3}{6}\right) = 3 - \frac{\pi\sqrt3}{3} = 3 - \frac{\pi}{\sqrt3}.$$

Answer: A

Divide by $(\tan x)^{1/2}$ to get a linear equation:

$$\frac{dy}{dx} + (\tan x)\,y = \frac{\sec^3 x}{\sqrt{\tan x}}.$$

Integrating factor $\mu = e^{\int \tan x\,dx} = e^{-\ln\cos x} = \sec x$:

$$\frac{d}{dx}\big(y\sec x\big) = \sec x\cdot\frac{\sec^3 x}{\sqrt{\tan x}} = \frac{\sec^4 x}{\sqrt{\tan x}}.$$

Integrate. Write $\sec^4 x = (1 + \tan^2 x)\sec^2 x$ and put $t = \tan x$, $dt = \sec^2 x\,dx$:

$$\int \frac{\sec^4 x}{\sqrt{\tan x}}\,dx = \int \frac{1 + t^2}{\sqrt{t}}\,dt = \int \left(t^{-1/2} + t^{3/2}\right)dt = 2\sqrt{t} + \frac{2}{5}t^{5/2}.$$

So

$$y\sec x = 2\sqrt{\tan x} + \frac{2}{5}(\tan x)^{5/2} + C.$$

Apply $y(\tfrac{\pi}{4}) = \tfrac{6\sqrt2}{5}$ ($\sec\tfrac{\pi}{4} = \sqrt2$, $\tan\tfrac{\pi}{4} = 1$):

$$\sqrt2\cdot\frac{6\sqrt2}{5} = \frac{12}{5} = 2 + \frac25 + C \;\Rightarrow\; C = 0.$$

At $x = \tfrac{\pi}{3}$ ($\sec\tfrac{\pi}{3} = 2$, $\tan\tfrac{\pi}{3} = \sqrt3$, so $\sqrt{\tan x} = 3^{1/4}$ and $(\tan x)^{5/2} = 3^{5/4}$):

$$2\,y = 2\cdot 3^{1/4} + \frac25\cdot 3^{5/4} \;\Rightarrow\; y\!\left(\tfrac{\pi}{3}\right) = 3^{1/4} + \frac15\cdot 3^{5/4} = 3^{1/4}\left(1 + \frac35\right) = \frac{8}{5}\,3^{1/4}.$$

Since $y(\tfrac{\pi}{3}) = \tfrac{4}{5}\alpha$:

$$\frac45\alpha = \frac85\,3^{1/4} \;\Rightarrow\; \alpha = 2\cdot 3^{1/4} \;\Rightarrow\; \alpha^4 = 16\cdot 3 = 48.$$

Answer: 48