Answer to Question #264693 in Differential Equations for Hassan

Question

the solution of the given linear differential equation x^2 y'+x(x+2)=e^x for x>0

Answer 1

$\begin{gathered} x^{2} y^{\prime}+x(x+2)=e^{x} \\ x^{2} y^{\prime}=e^{x}-x(x+2) \\ y^{\prime}=\frac{e^{x}-x^{2}-2 x}{x^{2}} \\ \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{e^{x}-x^{2}-2 x}{x^{2}} \\ \mathrm{~d} y=\frac{\left(e^{x}-x^{2}-2 x\right) \mathrm{d} x}{x^{2}} \\ \int 1 \mathrm{~d} y=\int \frac{e^{x}}{x^{2}}-\frac{2}{x}-1 \mathrm{~d} x \\ y=-2 \ln (x)+\mathbf{E i}(x)-\frac{e^{x}}{x}-x+C \end{gathered}$