General Solution of $y'(x)+p(x)e^{r(x) y(x)}=q(x)$

Question

I solved the case for the non-homogenous constant coefficients case and I wondered if there is a way to find a general solution to a non-constant coefficient case. I don't know how to approach this at all, the substitution $y(x)=\frac{\log (v(x))}{r(x)}$ gets problematic immediately.

Answer 1

Here is what I have tried thus far..

$$ y' + p(x)\mathrm{e}^{r(x)y} = q(x) $$ lets set $$ y' - q(x) = v'\implies y =v + \int q(x')dx' $$ therefore we obtain

$$ v' +p(x)\mathrm{e}^{r\left(v + \int q(x')dx' \right)} = 0 $$ since $p(x)\mathrm{exp}(\int q(x')dx') = f(x)$ I can re-write the equation to yeild $$ v' + \phi(x)\mathrm{e}^{rv} = 0 $$ where $$ \phi(x) = p(x)\mathrm{exp}(\int q(x')dx') $$

this is where I become stuck, as setting $w = \mathrm{e}^{-rv}$

we find $$ w' -\frac{r'}{r}w\ln(w) - r\phi = 0 $$

i will keep at it.