Differential equation with "backwards product rule".

Question

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?

$$f(x)y(x) + f'(x)y'(x) = h(x)$$

Answer 1

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).

Answer 2

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.