I'm reading a paper and I need to reproduce the following reduction of a second order ODE to a system of two first order ODEs.
Second order ODE:
$$ (a(f))'' + (b(f))' + c(f) + \sigma f' = 0 $$
where f(x) and ' refers to differentiation with respect to x. $\sigma$ is a constant.
According to the paper, if I set
$$ g(x) := -(a(f))'(x)$$
I should get:
$$ f' = -g/a'(f) $$ $$ g' = -g( \sigma + b'(f) )/a'(f) + c(f) $$
This seems like it should be simple enough, but I'm getting confused somewhere in the chain rule. I started by trying to derive the equation for $f'$ by doing this:
$$g=-\frac{da}{dx} = -\frac{da}{df} \frac{df}{dx} = -\frac{da}{df} f' $$
But then I get
$$ f' = -g / \left(\frac{da}{df}\right) $$
instead of
$$ f' = -g/a'(f) $$
What am I doing wrong?
Here, $a(f(x))$ is just a composite function. The chain rule for composite functions is
$$ \frac{d}{dx} \Big(a \big( f(x) \big) \Big) = \frac{df}{dx}\big(f(x)\big)\cdot \frac{da}{dx}(x)= a'(f)\cdot f'(x). $$
Necessarily,
$$g(x) = - (a(f))' = -a'(f)\cdot f'(x).$$