Question

solve (1 + x^2) y'' -2 xy' +3y = e^-x using power of series.

Answer 1

what have you worked out so far?

Answer 2

I am confused about the taylor series for e^(-x)

Answer 3

have you already plugged in your y, y' and y'' series

the rest is just comparison right

Answer 4

yes i have all that. how can i compare to its taylor series(which i dont know) but i do know for e^x

Answer 5

then let me catch up
\[(1 + x^2) y'' -2 xy' +3y = e^{-x}\]
\[(1 + x^2) \sum_2a_n~n(n-1)x^{n-2} -2 x\sum_1a_n~nx^{n-1} +3\sum_0a_nx^n = e^{-x}\]
\[\sum_2a_n~n(n-1)x^{n-2} + \sum_2a_n~n(n-1)x^{n} -2 \sum_1a_n~nx^{n} +3\sum_0a_nx^n = e^{-x}\]
\[\sum_0a_{n+2}~(n+2)(n+1)x^{n} + \sum_2a_n~n(n-1)x^{n} -2 \sum_1a_n~nx^{n} +3\sum_0a_nx^n = e^{-x}\]

\[(3a_0+2a_{2})+(5a_1+6a_{3})x+\\
~~~~
\sum_2 [a_{n+2}~(n+2)(n+1)
+[a_n~n(n-1)-2a_n~n+3a_n]~x^n = e^{-x}\]

so far so good? or have a messed it up along the way

Answer 6

\[e^u = \sum_0 \frac{1}{n!}u^n\]
\[e^{-x} = \sum_0 \frac{1}{n!}(-x)^n\]
\[e^{-x} = \sum_0 \frac{(-1)^nx^n}{n!}\]

Answer 7

\[e^{-x}=\frac{1}{0!}-\frac{1}{1!}x+\sum_2\frac{(-1)^nx^n}{n!}\]

Answer 8

On your last equation above, isnt that supposed to be \[(6a_3+a_1)\]

Answer 9

maybe, its a bit difficult to keep track of when coding :)

Answer 10

it's okay. so now then how do we compare?

Answer 11

\[
a_{0+2}~(0+2)(0+1)x^{0} \\
+a_{1+2}~(1+2)(1+1)x^{1} \\
- 2a_1~1x^{1} \\
+3a_0x^0 \\
+ 3a_1x^1 \\
\]

\[
2a_{2} +6a_{3}~x
- 2a_1~x
+3a_0
+ 3a_1x
\]

Answer 12

well, as long as our first 2 terms are fit to the first 2 terms of e^-x the rest are recurrsive

Answer 13

1 = (3a0 +2a2)
-1x = (6a3+a1)x

\[a_{n+2}~(n+2)(n+1)+a_n~n(n-1)-2a_n~n+3a_n = \frac{(-1)^n}{n!}~:~n\ge2\]
solve for a_n or a_(n+2)

Answer 14

looks like we will be able to define everything we need to in terms of a_0 and a_1

Answer 15

okay this does make sense to me now. i was confused about the taylor series part
Thanks alot for your help

Answer 16

good luck :)