What is the relation between two functions? Suppose that $f:\mathbb{R}^{+}\rightarrow\mathbb{R}$ and $g:\mathbb{R}^{+}\rightarrow\mathbb{R}$
Moreover we know that
$$
f''(x)g'(x) + g''(x)f(x) = 0.
$$
It is a second order differential equation with two unknown functions.
Is there any relation between functions $f(x)$ and $g(x)$?
Any hints?
 A: Let $h=g'$. Then $\frac {h '} h=-\frac {f''} f$. So $log (h(x))=C-\int_0^{x} \frac {f''(t)} {f(t)}\, dt$. Take exponential and integrate again to write $g$ in terms of $f$. 
A: First, let's discuss what this equation means.
$$f''(x)g'(x) + g''(x)f(x) = 0$$
It is by itself a relation between the two functions.
If we fix $f(x)$ and two initial (or boundary) conditions, we can use the ODE to find $g(x)$.
On the other hand, if we fix $g(x)$ and two initial (or boundary) conditions, we can use the ODE to find $f(x)$.
I doubt we can derive an explicit function $g(f)$ or $f(g)$, even if we fix the necessary constants.
Kavi Rama Murthy provided one way to describe $g(x)$ in terms of $f(x)$.

There ways to transform the equation, for example the Fourier transform:
$$f(x)= \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^\infty e^{i u x} F(u) du \\ g(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{i v x} G(v) dv$$
The inverse transform is:
$$F(u)= \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^\infty e^{-i u x} f(x) dx \\ G(v)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty e^{-i v x} g(x) dx$$
Then our equation becomes:
$$i \int_{-\infty}^\infty \int_{-\infty}^\infty  e^{i (u+v) x} u^2 v F(u) G(v) du dv+\int_{-\infty}^\infty \int_{-\infty}^\infty  e^{i (u+v) x} v^2 F(u) G(v) du dv=0$$
Let's make a substitution:
$$u=w-v$$
$$i \int_{-\infty}^\infty \int_{-\infty}^\infty  e^{i w x} (w-v)^2 v F(w-v) G(v) dw dv+\int_{-\infty}^\infty \int_{-\infty}^\infty  e^{i w x} v^2 F(w-v) G(v) dw dv=0$$
Since this should be valid for any $x$, we can get rid of one of the integrals (see convolution theorem) and write:
$$i  \int_{-\infty}^\infty  (w-v)^2 v F(w-v) G(v) dv+\int_{-\infty}^\infty   v^2 F(w-v) G(v) dv=0$$

$$ \int_{-\infty}^\infty \left(v+i (w-v)^2 \right) v F(w-v) G(v) dv=0$$

This is a complex integral equation, which is equivalent to two real integral equations for four functions:
$$F(v)=F_r(v)+i F_i(v) \\ G(v)=G_r(v)+i G_i(v)$$
I know this is way too complicated, but sometimes integral equations are better than ODEs, for example they could be more stable numerically.
