Question

I have a problem, Find a power series solution about x=0 for y"+4y=0. I solved it all the way down, but I am not too sure on my answer. Can someone confirm this? http://i42.tinypic.com/30as8ww.png

Answer 1

You can always check your answer by working the "normal" way. You're given a linear homogeneous equation, so it's not too difficult.

You get the characteristic polynomial
\(r^2+4\\
r=\pm2i,\)
which gives you the general solution
\[y=C_1\cos2x+C_2\sin2x\]
Writing each term as their respective power series, you have (I'm using WolframAlpha, by the way)
\[\cos2x=\sum_{k=1}^\infty \frac{(-4)^kx^{2k}}{(2k)!}\\
\sin2x=\sum_{k=1}^\infty \frac{(-1)^k(2)^{2k+1}x^{2k+1}}{(2k+1)!}\]
Now, compared to your series, it looks like you left out the exponent on the \(-1\), which gives the alternating pattern. The same goes for your second series (the \(c_1\) terms you mention), but it also looks like you're missing the odd power of 2.

Answer 2

Ok, Thanks very much!