Answer
$y(x)=f^{-1}[I^{-1}[\int I(x)q(x)dx + C]]$
Work Step by Step
We are given:
$$f'(y)\frac{dy}{dx}+p(x)f(y)=q(x)$$
with $p(x),q(x) \in C(a,b)$
Let $t= f( y) \rightarrow t'=f'(y)y'$
Substituting:
$$t'+p(x)t=q(x)$$
The solution can be found by obtaning:
$$t=e^{-\int p(x)dx}[\int e^{\int p(x)dx}q(x)dx+C$$
$C$ is a constant of integration.
Since $t= f( y) \rightarrow y=f^{-1}y$
and f is an invertible function
The solution is:
$$y(x)=f^{-1}[I^{-1}[\int I(x)q(x)dx + C]]$$
