JEE Main 2025 — 7 April, Morning Shift — Question 26

$x\left(x^{2}+e^{x}\right) d y+\left(e^{x}(x-2) y-x^{3}\right) d x=0$

$$\begin{aligned} & \frac{d y}{d x}+\frac{e^{x}(x-2)}{x\left(x^{2}+e^{x}\right)} y=\frac{x^{3}}{x\left(x^{2}+e^{x}\right)} \\& \text { I.F. }=e^{\int \frac{e^{x}(x-2)}{x\left(x^{2}+e^{x}\right)} d x} \\& =e^{\int \frac{e^{x} x+2 x}{e^{x}+x^{2}} d x-\int \frac{2}{x} d x} \\& =e^{\ln e^{2}+x^{2} \mid-2 \ln x} \\& =\frac{e^{x}+x^{2}}{x^{2}} \\& \therefore \quad y\left(\frac{e^{2}+x^{2}}{x^{2}}\right)=\int d x+c \\& \Rightarrow \quad y\left(\frac{e^{x}+x^{2}}{x^{2}}\right)=x+c \end{aligned}$$

Also, $y(1)=0$

$\Rightarrow c=-1$

$\therefore \quad y\left(\frac{e^{2}+x^{2}}{x^{2}}\right)=x-1$

Hence, $y(2)=\frac{4}{e^{2}+4}$