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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

$$y(x) = frac1f(x)left(C + int h(x) dxright)$$

But what if we have it the "other way around", like this?

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

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

$y'(x) + dfracf(x)f'(x)y(x) = dfrach(x)f'(x), tag 4$

which is a first-order system with varying coefficients, which has a well-known solution

$y(x)$
$= exp left ( -displaystyle int_x_0^x dfracf(s)f'(s) ; ds right ) left (y(x_0) + exp left (displaystyle int_x_0^x dfracf(s)f'(s) ; ds right) displaystyle int_x_0^x exp left ( -displaystyle int_x_0^s dfracf(u)f'(u) ; du right ) dfrach(s)f'(s) ; dsright ), tag 5$

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 $f(x)$:

$y'(x) + dfracf'(x)f(x) y(x) = dfrach(x)f(x); tag 7$

we obtain a formula which effectively interchanges $f(x)$ and $f'(x)$ in (5). It appears that the formula

$ln f(x) - ln f(x_0) = displaystyle int_x_0^x dfracf'(s)f(s) ; ds tag 8$

may help further simplify (5) when applied to the case of (7).

Let $h$ be an anti-derivative of $frac f f'$ Then $(e^hy)'=e^h(y'+frac f f' y)=e^hfrac 1 f' (y'f'+fy)=e^hfrac h f' $. Now integrate this.