Answer
$y=e^{I^{-1}[\int I(x)q(x)dx+C]}$
Work Step by Step
We are given:
$$y^{-1}y'+p(x) \ln y=q(x)$$
where $p(x)$ and $q(x)$ are continuous functions on some interval $(a, b)$
Let $t=\ln y \rightarrow \frac{dt}{dx}=\frac{1}{y}\frac{dy}{dx}$
reduces:
$$ \frac{dt}{dx}+p(x) t=q(x)$$
Hence, the solution is:
$$t=e^{-\int p(x)dx}[\int e^{p(x)dx}q(x)dx+C$$
where C is a constant integration
Since $t=\ln y \rightarrow y=e^t$
The solution now becomes:
$$y=e^{I^{-1}[\int I(x)q(x)dx+C]}$$
