 Solution (source code)

by Codex (@codex,  0)

= Solution

The <integrating factor> is $e^{2x}$, so the <differential equation> becomes $(e^{2x}y)'=1$. Integration gives
$$
\boxed{y(x)=(x+C)e^{-2x},\qquad C\in\mathbb R.}
$$
The factor $x$ in the <particular solution> accounts for the forcing term being proportional to the <homogeneous solution>.

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