= Solution
For $b\ne-1$ and initially $y\ne0$, use the <Bernoulli power substitution> $v=y^{b+1}$. The equation becomes
$$
v'-a(b+1)v=(b+1)e^{-2x}.
$$
Write $\kappa=a(b+1)$. A <first-order linear differential equation> then gives
$$
\boxed{y^{b+1}=\begin{cases}Ce^{\kappa x}-\dfrac{b+1}{\kappa+2}e^{-2x},&\kappa\ne-2,\\e^{-2x}[C+(b+1)x],&\kappa=-2.\end{cases}}
$$
The second line is the resonant case of a <Bernoulli equation with exponential forcing>. Recover $y$ using real branches for which the original power $y^{-b}$ is defined; for example, unrestricted noninteger powers are unambiguous on $y>0$.
The exceptional exponent $b=-1$ makes the original equation linear: $y'=(a+e^{-2x})y$. Its solutions are
$$
\boxed{y=C\exp\left(ax-\frac12e^{-2x}\right).}
$$
Domain restrictions matter when the solution reaches zero. If $b<0$ and the real power at zero is defined, $y\equiv0$ is also a solution. For $-1<b<0$, <waiting-time solutions of a sublinear Bernoulli equation> may stay zero until an arbitrary $x_0$ and then become positive, with
$$
y^{b+1}=(b+1)e^{\kappa x}\int_{x_0}^{x}e^{-(\kappa+2)s}\,ds\quad(x>x_0),\qquad y=0\quad(x\le x_0).
$$
The joining derivative is zero because $1/(b+1)>1$. For $b=0$, the linear formula can pass through zero normally; for $b>0$, the original negative power excludes zero. These qualifications prevent dividing out a valid zero solution or inventing one where the equation is undefined.
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