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 '16 at 11:54
  • $\begingroup$ @almagest Oh, thank you! Could you write it as an answer, so I could accept it? $\endgroup$ – Eenoku Apr 13 '16 at 12:32

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.