Given $f^{\prime} = g , \, g^{\prime} = f$, Prove $f = \sinh, \, g = \cosh$ I'm having some trouble proving these rigorously
Given $f, g\,\colon \mathbb{R} \to \mathbb{R}\,$ differentiable with $$f^{\prime} = g \quad \text{and}\quad g^{\prime} = f$$
First, if $f(0) = g(0) = 0$, Prove that $f = g = 0$ 
Second, if $f(0) = 0$, $\,g(0) = 1$, Prove that $f = \sinh$, $\,g = \cosh$
Any suggestions?
 A: Consider $F=f+g$ and $G=f-g$; then
$$
F'=f'+g'=g+f=F
\qquad
G'=f'-g'=g-f=-G
$$
What can you say about $F$ and $G$?
A: Hint: Each of these functions satisfies $h''=h$. Do you know the general form for functions that satisfy this?
A: "egreg" directed your attention to a certain function called (capital) $F$ and the fact that $F'=F,$ and then asked what you can say about $F.$
From one less-than-rigorous point of view, one might say that if $\dfrac{dF}{dx} = F\vphantom{\frac{}{\displaystyle\int}}$ then $\dfrac{dF} F = dx$ and then integrate both sides, getting $\log |F| = x +\text{constant},$ so that $|F| = (e^x\times\text{positive constant}),$ so that $F= (e^x\times\text{nonzero constant}).$ Of course, writing $\dfrac{dF} F \vphantom{\frac{\displaystyle\int}{\displaystyle\int}}$ assumes $F$ is not $0$, so you have to examine whether $F=0$ is a solution (and of course it is), etc., etc.
However, if one wishes to be logically rigorous, one may ask how we know that $F(x) = ce^x$ are the only solutions, and here's one way:
Suppose $F' = F.$
Let $H(x) = \dfrac{F(x)}{e^x}.$
Then by the quotient rule,
\begin{align}
H'(x) & = \frac{e^x F'(x) - e^x F(x)}{e^{2x}} \\[10pt]
& = \frac{e^x F(x) - e^x F(x)}{e^{2x}} \text{ since } F'=F \\[10pt]
& = 0.
\end{align}
Thus $H$ is constant and so $F(x) = (\text{constant}\times e^x).$
Hence there are no other solutions.
The above tacitly uses the mean value theorem, so let us make that not so tacit:
$$
\frac{H(a)-H(b)}{a-b} = H'(c) = 0,
$$
therefore $H(a) = H(b),$ for all $a,b;$ hence $H$ is constant.
(The reason for working with $\dfrac{F(x)}{e^x} \vphantom{\dfrac{}{\displaystyle\int}}$ rather than with $\dfrac{e^x}{F(x)}$ is that the former assures us that the denominator will nowhere be $0.$)
A: You can rewrite this problem as a differential equation. For example, we know that $f''=f$ which is equivalent to saying $f''-f=0$ which constitutes a differential equation. If you write out the general solution to this equation, which happens to be $f(x) = A\sinh(x) + B\cosh(x)$ , and then apply the initial conditions given to you, you will find the answer.
