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

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

It is given that $f(x)$ and $g(x)$ are solutions of the differential equation

$a \frac{d^2 y}{d x^2}+x^2 \frac{d y}{d x}+y=e^x$

∴  $a \frac{d^2}{d x^2}\{f(x)\}+x^2 \frac{d}{d x}\{f(x)\}+f(x)=e^x$

and, $\frac{d^2}{d x^2}\{g(x)\}+x^2 \frac{d}{d x}\{g(x)\}+g(x)= e^x$

$\Rightarrow a \frac{d^2}{d x^2}\{f(x)-g(x)\}+x^2 \frac{d}{d x}\{f(x)-g(x)\}+\{f(x)-g(x)\}=0$

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

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