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: $$u(t) = e^{ax}(C_1 cos(bx) + C_2 sin(bx) )$$


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.

  • 1
    $\begingroup$ Take the solution you have derived. $C_1,C_2$ are arbitrary. So take $C_1=\frac{1}{2}B_1-\frac{1}{2}iB_2,C_2=\frac{1}{2}B_1+\frac{1}{2}iB_2$, then your solution has the required form. Now try something similar for your Q 2). $\endgroup$
    – almagest
    Apr 13, 2016 at 11:54
  • $\begingroup$ @almagest Oh, thank you! Could you write it as an answer, so I could accept it? $\endgroup$
    – Eenoku
    Apr 13, 2016 at 12:32

1 Answer 1


1) you correctly got as far as $e^{ax}((C_1+C_2)\cos bx+i(C_1-C_2)\sin bx)$. But $C_1,C_2$ are arbitrary. So take $C_1=\frac{1}{2}B_1-\frac{1}{2}iB_2,C_2=\frac{1}{2}B_1+\frac{1}{2}iB_2$. Then $C_1+C_2=B_1,i(C_1-C_2)=B_2$, so your solution becomes $e^{ax}(B_1\cos bx+B_2\sin bx)$, which has the required form.


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