JEE Advanced 2025 Paper 2 · Q09 · Differential Equations

Question

Let $y(x)$ be the solution of the differential equation $$x^2 \frac{dy}{dx} + xy = x^2 + y^2,\quad x > \frac{1}{e},$$ satisfying $y(1) = 0$. Then the value of $\dfrac{2 (y(e))^2}{y(e^2)}$ is __________.

SolveKar AI · Accepted Answer

Substitute $y = vx$. Then $\frac{dy}{dx} = v + x v'$. The ODE becomes $x^2(v + xv') + x \cdot (vx) = x^2 + v^2 x^2$ $	o$ $x^3 v' + 2x^2 v = x^2(1 + v^2)$ $	o$ $x v' = 1 + v^2 - 2v = (1 - v)^2$. Separable: $\frac{dv}{(1-v)^2} = \frac{dx}{x}$. Integrate: $\frac{1}{1 - v} = \ln x + C$. $y(1) = 0$ gives $v(1) = 0$, so $C = 1$. Hence $v = 1 - \frac{1}{1 + \ln x} = \frac{\ln x}{1 + \ln x}$, and $y(x) = \frac{x \ln x}{1 + \ln x}$. $y(e) = e \cdot \frac{1}{2} = \frac{e}{2}$. $y(e^2) = e^2 \cdot \frac{2}{3} = \frac{2e^2}{3}$. $\frac{2 (y(e))^2}{y(e^2)} = \frac{2 (e^2/4)}{2e^2/3} = \frac{e^2/2 \cdot 3}{2e^2} = \frac{3}{4} = 0.75$.

JEE Advanced 2025 Paper 2 · Q09 · Differential Equations

Solution