Show that $ f^2 − g^2 = C $ for some real constant C Suppose $f$ and $g$ are continuously differentiable functions such that $f(x) = g'(x)$ and $g(x) = f'(x)$ and that any product of $f, f', g$ and $g'$ is commutative for all $x ∈ R$. Show that $ f^2 − g^2 = C $ for some real constant C
I have actually no clue how to solve this, and would be really greatfull for all the help i can get
 A: Define a new function $\gamma(x) = f^2(x) - g^2(x)$, then the derivative is:
$$ \frac{\mathrm{d}\gamma}{\mathrm{d}x} = 2 f(x) \frac{\mathrm{d}f}{\mathrm{d}x} - 2 g(x) \frac{\mathrm{d}g}{\mathrm{d}x}  =  2 \Big(\underbrace{\frac{\mathrm{d}g}{\mathrm{d}x}}_{=f(x)}\frac{\mathrm{d}f}{\mathrm{d}x} - \underbrace{\frac{\mathrm{d}f}{\mathrm{d}x}}_{=g(x)} \frac{\mathrm{d}g}{\mathrm{d}x}  \Big) $$
And since these commute, so that:
$$\frac{\mathrm{d}g}{\mathrm{d}x}\frac{\mathrm{d}f}{\mathrm{d}x}  =  \frac{\mathrm{d}f}{\mathrm{d}x} \frac{\mathrm{d}g}{\mathrm{d}x} $$ 
The term in the brackets is zero, thus:
$$\frac{\mathrm{d}\gamma}{\mathrm{d}x} =0  $$
Which implies that $\gamma(x) = f^2(x) - g^2(x)$ is a constant, say, $C$.
$$ f^2(x) - g^2(x) = C$$ 
A: The expected answer was already given but, after studying ordinary differential equations, you will get the following solution:
From the assumptions we conclude that
$$f''=g'=f$$
and thus
\begin{aligned}f&=c_1e^x+c_2e^{-x},\\ g&=f'=c_1e^x-c_2e^{-x}.
\end{aligned}
As a result,
\begin{aligned}f^2-g^2&=(f+g)(f-g)\\ &=(2c_1e^x)(2c_2e^{-x})\\ &=4c_1c_2\\ &=C.
\end{aligned}
