When I have some ODE, for example:
$$ u''(t) + 5u(t) = 0 $$
I put together a characteristic equation:
$$ \lambda ^2 + 5 = 0 $$
Then I compute its roots $r_1$ and $r_2$.
And now there are some general solutions:
1) If $r_1, r_2 \in \mathbb R$ and $r_1 = r_2$, then the general solution is: $$u(t) = C_1 e^{r_1x} + C_2 e^{r_1x}\cdot x$$
2) If $r_1, r_2 \in \mathbb R$ and $r_1 \neq r_2$, then the general solution is: $$u(t) = C_1 e^{r_1x} + C_2 e^{r_2x}$$
3) If $r_1, r_2 = \pm (a + ib)$, then the general solution is: $$u(t) = e^{ax}(C_1 cos(bx) + C_2 sin(bx) )$$
Questions
1) How can I deduce the 3rd formula? I know, that $$ e^{a+i b} = e^{a}e^{ib} = e^{a}(cos(b) + i\cdot sin(b)), $$
but I'm not able to get it to the right form. I've only got this far:
\begin{align} u(t) &= C_1 e^{i\lambda x} + C_2 e^{-i\lambda x} \\ &= C_1 cos(\lambda x) + iC_2 sin(\lambda x) + C_2 cos(\lambda x) - iC_2 sin(\lambda x)\\ &= cos(\lambda x)(C_1 + C_2) + sin(\lambda x)(iC_1 - iC_2) \end{align}
2) I've found in one book, that
$$ u(t) = u(t) = C_1 e^{r_1x} + C_2 e^{r_2x} = C_1 cos(r_1x) + C_2 sinh(r_2x) $$
And I'm also not able to "prove" it.
