3
$\begingroup$

Just doing some revision for ODEs and came across this problem. Find the general solution to $$u''+4u=0.$$

So far I've applied the characteristic polynomial: $$\begin{array}{r c l} \lambda^2 +4 & = & 0 \\ \lambda^2 & = & -4 \\ \lambda & = & i\sqrt{4} \\ \lambda & = & 2i, -2i. \\ \end{array}$$

So the general solution should be: $$\begin{array}{l c l} u_H & = & Ae^{2ix}+Be^{-2ix} \\ & = & A(\cos{2x}+i\sin{2x})+B(\cos{(-2x)}+i\sin{(-2x)}) \\ & = & A\cos{2x}+iA\sin{2x}+B\cos{2x}-iB\sin{2x} \\ & = & (A+B)\cos{2x}+i(A-B)\sin{2x} \\ & = & C_1\cos{2x}+iC_2\sin{2x}. \\ \end{array}$$

The answers have $u=C_1\cos{2x}+C_2\sin{2x}$, and my question is "what happened to the $i$?" Does it drop out somewhere or is there an error in the answers?

Many thanks for a quick explanation/link to the appropriate website explaining this. :)

$\endgroup$
8
$\begingroup$

$i$ is a constant, and so is included in $C_2$. So you're both right! :)

$\endgroup$
3
$\begingroup$

If we wrote:

$$\tag 1 e^{a+ 2i} = e^{a t}(\cos 2t + i \sin 2t)v_1$$

where $v_1$ is the eigenvector, and for your problem $a = 0$.

When we expand $(\cos 2t + i \sin 2t)v_1$, we get an expression of the form:

$$(\alpha) + i(\beta)$$

Because we know that the real imaginary parts are both solutions, we have:

$$c_1(\alpha) + c_2(\beta).$$

$\endgroup$
  • $\begingroup$ Nice observations my dear friend+ $\endgroup$ – mrs Aug 3 '13 at 9:00
  • $\begingroup$ @BabakS.: Even though the answer is a bit cryptic, it won't be appreciated until the OP has to use eigenvalues/vectors to write out solutions. Hope all is well. Regards! $\endgroup$ – Amzoti Aug 3 '13 at 12:27
  • $\begingroup$ @BabakS.: This was an interesting problem math.stackexchange.com/questions/458105/… $\endgroup$ – Amzoti Aug 3 '13 at 13:15
  • $\begingroup$ Hello, Amzoti! (+1) ;-) $\endgroup$ – Namaste Aug 4 '13 at 1:11

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.