Answer
The solution of the differential equation
$$
x^{2} y^{\prime}-y=2 x^{3} e^{-1 / x}
$$
is
$$
y=e^{-1 / x}\left(\frac{1}{2} x^{2}+C\right)
$$
where $ C $ is an arbitrary constant
Work Step by Step
$$
x^{2} y^{\prime}-y=2 x^{3} e^{-1 / x}
$$
$$
\Rightarrow y^{\prime}-\frac{1}{x^{2}} y=2 x e^{-1 / x} \quad\quad (1)
$$
This is a linear equation and the integrating factor is
$$
I(x)=e^{ \int \frac{-1}{x^{2}} d x}=e^{\frac{1}{x}}
$$
Multiplying Eq.(1) by $e^{\frac{1}{x}}$ gives
$$
e^{1 / x} y^{\prime}-e^{1 / x} \cdot \frac{1}{x^{2}} y=2 x \Rightarrow\left(e^{1 / x} y\right)^{\prime}=2 x \Rightarrow
$$
Then we integrate both sides of the last equation we get:
$$
e^{1 / x} y=x^{2}+C \Rightarrow y=e^{-1 / x}\left(x^{2}+C\right) $$
So the solution of the differential equation
$$
x^{2} y^{\prime}-y=2 x^{3} e^{-1 / x}
$$
is
$$
y=e^{-1 / x}\left(x^{2}+C\right)
$$
where $ C $ is an arbitrary constant