If $f\left(x\right)\text{ and }g\left(x\right)$ are two solutions of the differential eouation $a\frac{d^{2}y}{d{x}^2}+x^2\frac{dy}{dx}+y=e^x,\text{ then }f\left(x\right)-g\left(x\right)$ is the solution of 

It is given that $f\left(x\right)\text{ and }g\left(x\right)$are solutions of the

differential equation

$\begin{array}{l}a\frac{d^{2}y}{d{x}^2}+x^2\frac{dy}{dx}+y=e^x\\ \therefore a\frac{d^2}{d{x}^2}\left\{f\right(x\left)\right\}+x^2\frac{d}{dx}\left\{f\right(x\left)\right\}+f\left(x\right)=e^x\end{array}$

and $\begin{array}{l}\frac{d^2}{d{x}^2}\left\{g\right(x\left)\right\}+x^2\frac{d}{dx}\left\{g\right(x\left)\right\}+g\left(x\right)e^x\\ \frac{d^2}{d{x}^2}\left\{f\right(x)-g(x\left)\right\}+x^2\frac{d}{dx}\left\{f\right(x)-g(x\left)\right\}+\left\{f\right(x)-g(x\left)\right\}=0\end{array}$

$\Rightarrow f\left(x\right)-g\left(x\right)$ is a solution of the differential equation

$a\frac{d^{2}y}{d{x}^2}+x^2\frac{dy}{dx}+y=0$