JEE Advanced 2025 · Paper 2 · Mathematics Q9

MathematicsNumerical+4Question 9 of 48

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

$x^2\frac{dy}{dx} + xy = x^2 + y^2,\qquad x > \frac1e,$

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

Your answer

Answer and worked solution

Official answer: 0.7

Substitute. The equation is homogeneous: $y'=1+\frac{y^2}{x^2}-\frac yx$ . Put $y=vx$ :
$xv'=(v-1)^2\implies-\frac{1}{v-1}=\ln x+C.$

Constant. $y(1)=0$ means $v(1)=0$ , so $C=1$ . Then $v=\frac{\ln x}{1+\ln x}$ and
$y=\frac{x\ln x}{1+\ln x}.$

Values. $y(e)=\frac e2$ and $y(e^2)=\frac{2e^2}3$ , so
$2\cdot\frac{e^2/4}{2e^2/3}=\frac34=0.75.$

JEE Advanced 2025 · Paper 2, Mathematics question 9