Application of Taylor's Formula

Question

If we are given that $f''(x) = f(x)$, how do we show that there exist constants $a$ and $b$ such that $f(x) = ae^x + be^{-x}$ for all $x$?

A hint is given: We can define another function $g$ by $g(x) = f(x) - ae^x - be^{-x}$, and choose constants $a$ and $b$ such that $g(0) = g'(0) = 0$. I don't really understand this hint, though... could anyone explain?

Answer 1

HINT : $g''(x)=g(x)-f(x)$

Second derivertive same $g-f$

Answer 2

If $f''=f'$, then $f(x)=a\mathrm{e}^x+b$, for some $a,b$ constant.

If $f''=f$, then $f(x)=a\mathrm{e}^x+b\mathrm{e}^{-x}$, for some $a,b$ constant.

The way to show it in the second case is:

$f''=f$ implies that $f''+f'=f'+f$ and hence $g=f'+f$ satisfies $g'=g$, which means that $g(x)=a\mathrm{e}^x$, and $f'+f=a\mathrm{e}^{x}$, and thus $$\mathrm{e}^x(f'+f)=a\mathrm{e}^{2x}$$ or $$ \big(\mathrm{e}^xf(x)\big)'=a\mathrm{e}^{2x}, $$ etc.