Question

Determine 2 linearly independent power series solutions to the following differential equation:
y'' - 2xy' - 2y = 0

I'm having trouble rewriting the diff. eqn. as a single sum of index 0.

Answer 1

what class is it for?

Answer 2

Linear Algebra/Diff Eqns

Answer 3

oh , I only know the latter

Answer 4

Oh, there's no linear algebra involved in solving this. I just need help rewriting the diff equation in terms of a single sum. Whenever I try to combine the 3 sums you end up with, I keep getting and excessive (x) term. Messes up the whole thing

Answer 5

so do you solve it normal way first then write the soloution as power series?

Answer 6

Here's how they do it in the book. I've highlighted the part that I'm having trouble with.

Answer 7

i have no idea what this is about, but i can follow the algebra ok.
what is the question?

Answer 8

Here's how they do it in the book. I've highlighted the part that I'm having trouble with.

Answer 9

Combining the 3 sums they come up with is not making sense to me. In particular the 2nd sum, when re-indexed about 0 should have an x^(k+1) term in it, but that disappears in the simplificaiton

Answer 10

let me take a closer look because it seems ok to me.

Answer 11

oh ok i see what the problem is, but i don't think they are reindexing (if that is a word) the second sum. it is already in terms of k and it is
\[-2\sum_{k=1}^{\infty}ka_kx^k\] so at
\[k=0\] this term vanishes

Answer 12

clear right? because of the
\[ka_k\]

Answer 13

so you might as well start at 0, because at 0 this term is gone. then the algebra should make sense

Answer 14

forgetting about the unimportant -2 for a second what this is saying is that
\[\sum_{k=1}^{\infty}ka_kx^k+\sum_{k=0}^{\infty}a_kx^k
\]
\[=\sum_{k=0}^{\infty}(k+1)a_kx^k\] and you can check that it is right by replacing k by 0

Answer 15

i can see why this is confusing because they did reindex the first term making
\[k=n-2\]

Answer 16

Here's how they do it in the book. I've highlighted the part that I'm having trouble with.

Answer 17

Oh! Ha! That works! My only hiccup is that shouldn't I still be able to re-index the sum and end up with the same answer?

Answer 18

you are just trying to make the exponents match up right? they match in the second and third summation, just not the first

Answer 19

Here's how they do it in the book. I've highlighted the part that I'm having trouble with.

Answer 20

Right, and by adding zero to the second sum, we can start at k = 0 instead of k = 1.

I guess I'm just curious as to why we can't substitute k = x + 1 and end up with the exact same answer. (ie: the method we used in the first sum to end up with the proper exponents)

Answer 21

i guess the best way to see that is to make the replacement and see if you get the exact same thing. i assume you mean substitute
\[k=j+1\] or something right?

Answer 22

Here's how they do it in the book. I've highlighted the part that I'm having trouble with.

Answer 23

Right, I mean right off the bat you can see that you'll end up with an x^(k+1) if you do that. I guess it's just a matter of recognizing that starting that particular sum at zero is exactly the same as starting it at 1.

Answer 24

the first one starts at n = 2 and the exponent is n - 2 so the natural thing to do is shift up by two so the exponent starts at 0 and the index starts at 0 as well