A calculus problem with functions such that $f''(x) = g(x)$ and $g''(x) = f(x)$ Let:


*

*$f(x)$ and $g(x)$ be twice differentiable, non-decreasing functions. 

*$f''(x) = g(x)$ and $g''(x) = f(x)$.

*$f(x) \cdot g(x)$ is a linear function. 


Then we have to show that $f(x) = g(x) = 0$.
I am really not able to do anything useful, any help would be appreciated :)

We have
$$\left\{\begin{array}{ll}
(i)&f(x)g(x)=Ax+B\\
(ii)&f''(x)=g(x)\\
(iii)&f(x)=g''(x).
\end{array}\right.$$
Let us omit the first condition (i) and first consider the system of ODE represented in conditions (ii) and (iii).
(ii) implies $$f(x)=\left(\int\int g\right)(x)+\alpha x+\beta$$ for constants $\alpha,\beta$, and this with (iii) implies 
$$g''(x)=\left(\int\int g\right)(x)+\alpha x+\beta.$$
Differentiating twice then yields
$$g^{(4)}(x)=g(x)$$
This linear (homogeneous) ODE has four independent solutions: $e^{x},e^{-x},\cos x,\sin x$ (upto multiplicative constants) in addition to the trivial solution $0$.  Applying (iii) repeatedly then gives the following non-trivial solution pairs $(f,g)$:
$$\left\{\begin{array}{l}(e^{x},e^{x})\\ (e^{-x},e^{-x})\\ (\cos x, -\cos x)\\ (\sin x, -\sin x)\end{array}\right.$$
It is clear that none of these combinations satisfy (i), and therefore only the trivial solution pair $(0,0)$ holds, which proves the claim.
