JEE Advanced 2025 — Paper 2 — MATH-III solution

JEE Advanced2025Paper 2MATH-III

Let $y (x)$ be the solution of the differential equation

$x^{2} \frac{d y}{d x} + x y = x^{2} + y^{2}, x > \frac{1}{e},$

satisfying $y (1) = 0$ . Then the value of $\frac{2 ( y ( e ) ) ^{2}}{y ( e ^{2} )}$ is ________.

Your answer

Solution

Step 1: Homogenise.

Divide by $x^{2}$ : $\frac{d y}{d x} + \frac{y}{x} = 1 + \frac{y ^{2}}{x ^{2}}$ . Substitute $y = v x$ so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .

Step 2: Separable form.

The equation becomes $x \frac{d v}{d x} = 1 + v^{2} - 2 v = (v - 1)^{2}$ . Hence $\frac{d v}{( v - 1 ) ^{2}} = \frac{d x}{x}$ .

Step 3: Integrate.

$- \frac{1}{v - 1} = ln ∣ x ∣ + C$ , i.e. $\frac{x}{x - y} = ln x + C$ (taking $x > 1/ e > 0$ ).

Step 4: Apply $y (1) = 0$ .

$\frac{1}{1} = 0 + C \Rightarrow C = 1$ . So $\frac{x}{x - y} = ln (e x)$ .

Step 5: Evaluate at $x = e$ and $x = e^{2}$ .

At $x = e$ : $\frac{e}{e - y} = 2 \Rightarrow y (e) = \frac{e}{2}$ .

At $x = e^{2}$ : $\frac{e ^{2}}{e ^{2} - y} = 3 \Rightarrow y (e^{2}) = \frac{2 e ^{2}}{3}$ .

Step 6: Final ratio.

$\frac{2 ( y ( e ) ) ^{2}}{y ( e ^{2} )} = \frac{2 \cdot e ^{2} /4}{2 e ^{2} /3} = \frac{3}{4} = 0.75$ .

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