Use power series to solve for y'' + y = 0.

Question

kc_kennylau · Answer

"Use power series to..."
What if no.

kc_kennylau · Answer

xD

shamil98 · Answer

It's quite long, and i'm actually pondering on how it's done...
It's the next section after nonhomogenous equations ( which I have yet to learn to xD)

kainui · Answer

$$y=\sum_{n=0}^{\infty}a_nx^n$$ $$y''=\sum_{n=0}^{\infty}(n+1)(n+2)a_{n+2}x^n$$ $$y''+y=0=\sum_{n=0}^{\infty}x^n[a_n+(n+1)(n+2)a_{n+2}]$$

since x is variable, that part can't be assumed to be 0. So we have to assume the sum of the other stuff is 0.

$$0=a_n+(n+1)(n+2)a_{n+2}$$

This gives you a recursion relation which amounts to finding even and odds which are your two constants for your fundamental solutions which are determined by a_0 and a_1.

kc_kennylau · Answer

wait, isn't y'' [SUM] n(n-1)x^(n-2)?

kc_kennylau · Answer

http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx Just googled this and learnt it by myself xD (I always do this when I meet new terms)

kainui · Answer

See, when you take the derivative of a power series, just increase the index from n=0 to n=1 since you know your n=0 term will go to zero since it's a constant term. That way when you get to the next thing you have the same index.

kc_kennylau · Answer

Oh I see I get it

kainui · Answer

$$y=\sum_{n=0}^{\infty}a_nx^n=a_0+\sum_{n=1}^{\infty}a_{n+1}x^{n+1}$$

kc_kennylau · Answer

@kainui Thanks for teaching me :D

kainui · Answer

No problem, that's why I come here. =P

kc_kennylau · Answer

but @shamil98 is gone :/

shamil98 · Answer

Oh, I get it, over thought it a bit at first.. (sleep-deprivation kickin in xD)

shamil98 · Answer

What is the next step to solve for it? or is that it?

kc_kennylau · Answer

I think @Kainui means that there has to be more information

kc_kennylau · Answer

I'm new in this matter

anonymous · Answer

a good practice...$$y''+xy'+(x^2+2)y=0 $$ around $$x_0=0$$

kc_kennylau · Answer

Using whatever approach? @Jonask

anonymous · Answer

yes...it was explained above very well

kainui · Answer

So yeah, we are definitely not done, we need to find a recursion relation. I just got lazy cause it's x-mas and I figured that's pretty much the easy part. 

$$0=a_n+(n+2)(n+1)a_{n+2}$$
$$-a_n/(n+2)(n+1)=a_{n+2}$$

So now you plug in basically n=0 to find the recursion relation for the even terms, n=1 for the odd.

Notice that we're going to obviously have a (-1)^n since it oscillates between positive and negative terms.

Since each one up we go, we'll be continually multiplying (n+1)(n+2)(n+3)(n+4) etc... the bottom part will be a factorial.

So then we can pretty much say after some mathmagic... (actually I just wrote in the power series for sine and cosine from memory, so they might be slightly wrong.)

$$y=a_0\sum_{n=0}^{\infty}\frac{ (-1)^nx^{2n} }{ (2n)! }+a_1\sum_{n=0}^{\infty}\frac{ (-1)^nx^{2n+1} }{ (2n+1)! }$$

shamil98 · Answer

thanks! I think i understood it. (i'll re-check this when I do have some sleep.)

solomonzelman · Answer

you forgot the 1st quotation mark.

solomonzelman · Answer

$$"why"+why=0$$This?

shamil98 · Answer

y'' means the second derivative , solomon.

kc_kennylau · Answer

nope it's y differentiated twice

kc_kennylau · Answer

@SolomonZelman

solomonzelman · Answer

Yeah I looked it up before typing....

kainui · Answer

The good thing about power series solutions to differential equations is that you can easily make up your own problems for yourself to solve. Obviously the answer to this one is sinx and cosx since their second derivatives are just the negative versions of themselves, but as long as its any differential equation you've found the answer to by some other method you should be fine if you're practicing the method.