Question

Solve power series equation y" + xy =0

Answer 1

that looks like a differential equation

Answer 2

Yes it is differential equation but I need to solve it using a power series or frobenius method

Answer 3

not my area ... doesn't have an elementary solution

Answer 4

here is a reference:
http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx

Answer 5

hmm

Answer 6

\[\Large\rm y''+xy=0\]
So we're assuming our solution is of the form:\[\Large\rm y=\sum_{k=0}^{\infty} a_k x^k\]Then:\[\Large\rm xy=x\sum_{k=0}^\infty a_k x^k=\sum_{k=0}^\infty a_k x^{k+1}\]We want to shift our k's so the power on x isn't +1.
Let, \(\Large\rm k+1=i\)
Then,\[\Large\rm xy=\sum_{i=1}^{\infty} a_{i-1}x^i\]Ok that takes care of one part.

Answer 7

\[\Large\rm y=\sum_{k=0}^{\infty} a_k x^k\]Taking derivative twice gives us our y'',\[\Large\rm y''=\sum_{k=2}^{\infty}k(k-1) a_k x^{k-2}\]Power rule applied twice ^.
Again, we want the exponent on our x to be without the -2.
Let, \(\Large\rm k-2=i\)
\(\Large\rm \qquad\qquad k=i+2\)
Then we have,\[\Large\rm y''=\sum_{i=0}^{\infty} (i+2)(i+1)a_{i+2}x^i\]

Answer 8

So we've got:\[\Large\rm \color{royalblue}{xy=\sum_{i=1}^{\infty} a_{i-1}x^i}\]\[\Large\rm \color{royalblue}{y''=\sum_{i=0}^{\infty} (i+2)(i+1)a_{i+2}x^i}\]We would like to combine these and write them as a single summation but they don't have the same starting point!

So we'll have to peel one off of the y'' term.
We'll let i=0, and pull that term out of the sum.\[\Large\rm y''=(2)(1)a_0 x^0+\sum_{i=1}^{\infty} (i+2)(i+1)a_{i+2}x^i\]\[\Large\rm y''=2a_0+\sum_{i=1}^{\infty} (i+2)(i+1)a_{i+2}x^i\]Ok great now they start from the same location! i=1!

Answer 9

\[\Large\rm y''+xy=0\]We'll write as:\[\Large\rm 2a_0+\sum_{i=1}^{\infty} x^i\left[(i+2)(i+1)a_{i+2}+a_{i-1}\right]=0\]So I wrote it as one sum, and factored out the x^i.

Answer 10

Then we set our coefficients equal to zero,
(I can't seem to remember why... )

Answer 11

\[\Large\rm a_0=0\]\[\Large\rm (i+2)(i+1)a_{i+2}+a_{i-1}=0\]

Answer 12

So we have a starting value and a recursive relationship.
In the recursive relationship, we'll solve for the larger indexed value.\[\Large\rm a_{i+2}=\frac{-a_{i-1}}{(i+2)(i+1)}\]

Answer 13

Ah ok I see a little mistake I made, the term we peeled off corresponded to i=0, which should have given us our a2 term, not a0.\[\Large\rm a_2=0\]

Answer 14

So then start plugging in some values to get an idea of what is going on, look for a pattern.
a3 comes next, let's see what's going on,\[\Large\rm a_{1+2}=\frac{-a_{1-1}}{(1+2)(1+1)}\]\[\Large\rm a_3=\frac{-a_0}{3\cdot 2}\]Then the a4 term,\[\Large\rm a_{2+2}=\frac{-a_{2-1}}{(2+2)(2+1)}\]\[\Large\rm a_4=\frac{-a_1}{4\cdot 3}\]Then our a5 term will relate back to the a2 term, so it is also zero.\[\Large\rm a_5=0\]

Answer 15

bahhh im too tired right now >.<