Question

find the series solutions for y''-xy'+2y=0

Answer 1

@Zarkon

Answer 2

@amistre64

Answer 3

@.Sam.

Answer 4

@electrokid

Answer 5

@terenzreignz

Answer 6

this is a second order homogenous ODE
\[
y''-xy'+2y=0
\]
this probably turn out to be a series solution

Answer 7

Yes but I'm not sure what those series are :(

Answer 8

oh duh.. the question itself says about a series solution!!

Answer 9

ok
lets say the solution is:
\[y(x)=\sum_{n=0}^\infty a_nx^n\\
\text{differentiating once,}\\
y'=\sum_{n=1}^\infty na_nx^{n-1}\\
\text{differentiating again,}\\
y''=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}
\]we now plug this in the given DE
\[
\sum_{n=2}^\infty n(n-1)a_nx^{n-2}-x\times \sum_{n=1}^\infty na_nx^{n-1}+2\times\sum_{n=0}^\infty a_nx^n=0\\
\sum_{n=2}^\infty n(n-1)a_nx^{n-2}-\sum_{n=1}^\infty na_nx^n+\sum_{n=0}^\infty 2a_nx^n=0\\
\sum_{\color{red}{n=0}}^\infty (n+2)(n+1)a_{n}x^{n+1}-\sum_{\color{red}{n=0}}^\infty (n+1)a_{n+1}x^{n+1}+\sum_{n=0}^\infty 2a_nx^n=0\\
\sum_{n=0}^\infty\left[(n+2)(n+1)a_{n+2}x^{n}-(n+1)a_{n+1}x^{n+1}+2a_nx^n\right]=0\\
\implies\Large (n+2)(n+1)a_{n+2}-(n+1)a_{n+1}x+2a_n=0\\
\]

Answer 10

yep i've gotten that far

Answer 11

looks like I messed up
there should not be any "x" in the recurrance relation
\[
\sum_{\color{red}{n=1}}^\infty (n+1)(n)a_{n+1}x^{n+1}-\sum_{\color{red}{n=1}}^\infty na_{n}x^{n}+2a_0+\sum_{\color{red}{n=1}}^\infty 2a_nx^n=0\\
\sum_{\color{red}{n=1}}^\infty (n+1)(n)a_{n+1}x^{n+1}-\sum_{\color{red}{n=1}}^\infty na_{n}x^{n}+\sum_{\color{red}{n=1}}^\infty 2a_nx^n=-2a_0\\
\implies n(n+1)a_{n+1}-na_n+2a_n=-2a_0\\
\implies n(n+1)a_{n+1}=(n-2)a_n-2a_0\\
\implies\Large\boxed{a_{n+1}=\frac{(n-1)a_n-2a_0}{n(n+1)}}
\]

Answer 12

how are you combining your series sums when the x terms dont have the same exponents

Answer 13

if a_0=0
\[a_{n+1}=\frac{(n-1)a_n}{n(n+1)}\]

Answer 14

the first term should be x^n it is a type
I am typing in tex and its rediculous to check my syntax!

Answer 15

But I know I am right here

Answer 16

okay its cool.

Answer 17

darn..
\[
a_{n+1}=\frac{(n-2)a_n}{n(n+1)}\qquad\forall \;n\ne0
\]
so, we have, for \[
n=1\quad a_2={-a_1\over2}\\
n=2\quad a_3=0\\
n=3\quad a_4=0\\\implies
a_k=0\quad\forall\quad k\ge3\\
\Large y(x)=a_1x-{a_1\over 2}x^2=a_1x\left(1-{x\over2}\right)\\
y'=a_1-a_1x\\y''=-a_1
\]

Answer 18

Oooooooooh okay I see where I went wrong now! I got it! thank you so much!!!

Answer 19

yw

phew

Answer 20

still made a mistake..
here it is by hand..