= Solution
For $x\ne0$, multiply by $x$ to obtain the first-order <linear ordinary differential equation> $y'+xy=5xe^{-x^2}$. Its <integrating factor> is $\mu=e^{x^2/2}$. The <product rule> then gives
$$
(ye^{x^2/2})'=5xe^{-x^2/2}.
$$
Integration yields $ye^{x^2/2}=-5e^{-x^2/2}+C$, so
$$
\boxed{y(x)=Ce^{-x^2/2}-5e^{-x^2},\qquad x\ne0.}
$$
The <constant of integration> is arbitrary on each connected interval of the original equation's domain. The formula extends smoothly through zero, but the printed equation itself contains $1/x$ and is not defined there.
Back to article page
