Differential equation with "backwards product rule". If we have the following differential equation, ($h,f$ known, $y$ unknown):
$$f'(x)y(x) + f(x)y'(x) = h(x)$$
it would be easy, since we could spot the derivative for a product:
$$(f(x)y(x))' = h(x)$$
and conclude
$$y(x) = \frac{1}{f(x)}\left(C + \int h(x) dx\right)$$
But what if we have it the "other way around", like this?
$$f(x)y(x) + f'(x)y'(x) = h(x)$$
 A: Let $h$ be an anti-derivative of $\frac f {f'}$ Then $(e^{h}y)'=e^{h}(y'+\frac f {f'} y)=e^{h}\frac 1 {f'} (y'f'+fy)=e^{h}\frac h {f'} $. Now integrate this. 
A: If the equation
$f'(x)y(x) + f(x)y'(x) = h(x) \tag 1$
is written "the other way around",
$f(x)y(x) + f'(x)y'(x) = h(x), \tag 2$
then there is no obvious way to apply the product rule; if we observe, however, that on an interval $J$ with
$f'(x) \ne 0, \; x \in J, \tag 3$
then we may write (2) in the form
$y'(x) + \dfrac{f(x)}{f'(x)}y(x) = \dfrac{h(x)}{f'(x)}, \tag 4$
which is a first-order system with varying coefficients, which has a well-known solution
$y(x)$
$= \exp \left ( -\displaystyle \int_{x_0}^x \dfrac{f(s)}{f'(s)} \; ds \right ) \left (y(x_0) +  \exp \left (\displaystyle \int_{x_0}^x \dfrac{f(s)}{f'(s)} \; ds \right)  \displaystyle \int_{x_0}^x  \exp \left ( -\displaystyle \int_{x_0}^s \dfrac{f(u)}{f'(u)} \; du \right ) \dfrac{h(s)}{f'(s)} \; ds\right ), \tag 5$
with $x_0 \in J$.  
The formula (5) may in fact also be applied to (1) if we assume
$f(x) \ne 0, x \in J, \tag 6$
and divide (1) by $f(x)$:
$y'(x) + \dfrac{f'(x)}{f(x)} y(x) = \dfrac{h(x)}{f(x)}; \tag 7$
we obtain a formula which effectively interchanges $f(x)$ and $f'(x)$ in (5).  It  appears that the formula
$\ln f(x) - \ln f(x_0) = \displaystyle \int_{x_0}^x \dfrac{f'(s)}{f(s)} \; ds \tag 8$
may help further simplify (5) when applied to the case of (7).
