# Solution of ODE $u''+4u=0$

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. :)

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

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:

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

• Nice observations my dear friend+ – mrs Aug 3 '13 at 9:00
• @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! – Amzoti Aug 3 '13 at 12:27
• @BabakS.: This was an interesting problem math.stackexchange.com/questions/458105/… – Amzoti Aug 3 '13 at 13:15
• Hello, Amzoti! (+1) ;-) – Namaste Aug 4 '13 at 1:11