Antiderivative problem: find f and g I can't solve the following problem:
$f'(x)g'(x) = f(x)g(x) = e^{2x}, \forall x \in \mathbb{R}$
$f(0) = g(0) = 1$
I have to prove that $f(x) = g(x) = e^x$.
Does anybody have an idea?
 A: Let $f$ and $g$ be defined such that $f:=f(x)$ and $g:=g(x)$. So far we have $f^\prime g^\prime = fg = e^{2x}$, $\forall x\in \mathbb R$, and the initial values of $f(0)=1$, and $g(0)=1$.
Since we have $fg=e^{2x}$ lets find the derivative of $h=fg=e^{2x}$.
Thus
$$ h^\prime=(fg)^\prime = fg^\prime+f^\prime g= 2e^{2x}$$
Since $fg = e^{2x}$, we have
$$ fg^\prime+f^\prime g= 2fg= fg+fg$$
$\Rightarrow$
$$\begin{array}{ccc}
0 &=& (f^\prime g-fg)+(fg^\prime-fg)\\
&=&g(f^\prime-f)+f(g^\prime-g)
\end{array}$$
for all $x\in \mathbb R$. Likewise $f^\prime g^\prime = e^{2x}$. So,
$$ fg^\prime+f^\prime g= 2f^\prime g^\prime= f^\prime g^\prime+f^\prime g^\prime$$
$\Rightarrow$
$$\begin{array}{ccc}
0 &=& (f^\prime g^\prime-fg^\prime)+(f^\prime g^\prime-f^\prime g)\\
&=&g^\prime(f^\prime-f)+f^\prime(g^\prime-g)
\end{array}$$
for all $x\in \mathbb R$. This implies that 
$$g^\prime(f^\prime-f)+f^\prime(g^\prime-g)=g(f^\prime-f)+f(g^\prime-g)$$
for all $x\in \mathbb R$. Thus $g^\prime =g=C_g e^x$ and $f^\prime=f=C_fe^x$. Since $f(0)=1$, and $g(0)=1$, this implies that $1=C_g e^0=C_fe^0=C_g=C_f$. So $f(x)=e^x=g(x)$ as desired.
