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

$x\left(x^2+e^x\right) d y+\left(\mathrm{e}^x(x-2) y-x^3\right) \mathrm{d} x=0, x>0$, passing through the point $(1,0)$. Then $y(2)$ is equal to :

<p>$$\begin{aligned} & x\left(x^2+e^x\right) d y+\left(e^x(x-2) y-x^3\right) d x=0 \\ & \frac{d y}{d x}+\frac{e^x(x-2)}{x\left(x^2+e^x\right)} y=\frac{x^3}{x\left(x^2+e^x\right)} \\ & \text { I.F. }=e^{\iint \frac{e^x(x-2)}{x\left(x^2+e^x\right)} d x} \\ & =e^{\int \frac{e^x+2 x}{e^x+x^2} d x-\int \frac{2}{x} d x} \\ & =e^{\ln e^{-x}+x^2-2 \ln x} \end{aligned}$$</p> <p>$$\begin{aligned} & =\frac{e^x+x^2}{x^2} \\ & \therefore \quad y\left(\frac{e^2+x^2}{x^2}\right)=\int d x+c \\ & \Rightarrow \quad y\left(\frac{e^x+x^2}{x^2}\right)=x+c \end{aligned}$$</p> <p>Also, $y(1)=0$</p> <p>$$\begin{aligned} & \Rightarrow \quad c=-1 \\ & \therefore \quad y\left(\frac{e^2+x^2}{x^2}\right)=x-1 \end{aligned}$$</p> <p>Hence, $y(2)=\frac{4}{e^2+4}$</p>