Solution
= Solution
Divide by $x^2$, which is allowed because $x>0$:
$$
y'+\frac1xy=4e^{x^2}.
$$
The <integrating factor> for this <linear ordinary differential equation> is $e^{\int dx/x}=x$. Multiplying through makes the left side a <product rule> derivative:
$$
(xy)'=4xe^{x^2}.
$$
Part (b) now gives $xy=2e^{x^2}+C$. Therefore
$$
\boxed{y(x)=\frac{2e^{x^2}+C}{x},\qquad x>0.}
$$