Mathematics - Differential Equation Question with Solution | TestHub

$\begin{array}{rl}& x\left(x^2+e^x\right)dy+\left(e^x(x-2)y-x^3\right)dx=0\\ & x\left(x^2+e^x\right)\frac{dy}{dx}+e^x(x-2)y=x^3\\ & \frac{dy}{dx}+\frac{e^x(x-2)}{x\left(x^2+e^x\right)}y=\frac{x^2}{x^2+e^x}\\ & \text{ I.F. }=e^{\int \frac{e^x(x-2)}{x\left(x^2+e^x\right)}dx}=e^{\int \frac{e^x\left(\frac{1}{x^2}-\frac{2}{x^2}\right)dx}{\left(1+\frac{e^x}{x^2}\right)}dx}\end{array}$
Let $1+\frac{e^x}{x^2}=t\Rightarrow \frac{x^2{e}^x-e^{x}2x}{x^4}dx=dt$
$\Rightarrow \text{ I.F. }{\mathrm{e}}^{\ln \left(1+\frac{{\mathrm{e}}^2}{{\mathrm{x}}^2}\right)}=1+\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{x}}^2}$
Now $y\left(1+\frac{e^x}{x^2}\right)=\int \frac{x^2}{x^2+e^x}\cdot \frac{x^2+e^x}{x^2}dx+C$
$y\left(1+\frac{e^x}{x^2}\right)=x+C$
Passing through $(1,0)$
$\begin{array}{rl}& \Rightarrow C=-1\\ & y=\frac{x-1}{1+\frac{e^x}{x^2}}\\ & y(2)=\frac{1}{1+\frac{e^2}{4}}=\frac{4}{4+e^2}\end{array}$