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For a nonzero real solution use the Bernoulli power substitution $u=\sqrt[3]{y}$. Dividing the differential equation by $3u^2$ gives $u^{\prime }+u/3=x/3$. The integrating factor $e^{x/3}$ yieldsDirect substitution shows this formula also solves the equation at its zeros. Division temporarily discarded the additional solution $\boxed{y\equiv 0}$. Since the real power $y^{2/3}$ is not locally Lipschitz at zero, the complete real solution family also permits waiting-time solutions of a sublinear Bernoulli equation: join a zero interval to a displayed branch at a root of $x-3+C{e}^{-x/3}$. At a joining point the branch and its derivative are both zero, so the joined function satisfies the original equation. On each open nonzero interval every solution has the displayed form.

For $b\ne -1$ and initially $y\ne 0$, use the Bernoulli power substitution $v=y^{b+1}$. The equation becomesWrite $\kappa =a(b+1)$. A first-order linear differential equation then givesThe 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: $y^{\prime }=(a+e^{-2x})y$. Its solutions areDomain restrictions matter when the solution reaches zero. If $b<0$ and the real power at zero is defined, $y\equiv 0$ 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, withThe 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.