Question

Use power series to solve the differential equation
\[y''+xy'+y=0\]
\[y=\sum_{n=0}^{\infty}\]
\[y'=\sum_{n=1}^{\infty}a_nnx^{n-1}\]
\[y''=\sum{n=2}^{\infty}n(n-1)a_nx^{-2}\]
\[\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^{\infty}a_nx^n=0\]
\[\sum_{n=0}^{\infty}x^n\left[(n+2)(n+1)a_{n+2}+a_nn+a_n\right]=0\]
\[a_{n+2}=-\frac{a_n}{n+2}\]

Answer 1

I can't seem to figure out the pattern
\[n=0:a_2=-\frac{a_0}{2!}\]
\[n=2:a_4=\frac{a_0}{3!}\]
\[n=4:a_6=-\frac{a_0}{36}\]

Answer 2

the 36 kinda threw me off

Answer 3

did you forget the 'x' in x*y' ?

Answer 4

no

Answer 5

because before the x it was
\[y'=\sum_{n=1}^{\infty}a_nnx^{n-1}\]

Answer 6

bit hard to follow your work since things are missing. I honor you for putting so much info into your question statement though:) refreshing change from " solve y=4x +5"

Answer 7

I can write it all out of you want me to

Answer 8

it's ok...

Answer 9

let me catch up

Answer 10

ok

Answer 11

I don't know anymore

Answer 12

@Outkast3r09 watcha think?

Answer 13

factorial of the evens, not sure how to express that...
n=0 1/2
n=2 1/(2*4)
n=4 1/(2*4*6)
n=6 1/(2*4*6*8)

Answer 14

we've got something there!

Answer 15

and a sign changing term...

Answer 16

could also do
\[\frac{ 1 }{ 2^{k}k! }\]

Answer 17

k=1,2,3...

Answer 18

but would the k! work?

Answer 19

I think I've done enough math for tonight...g'night!

Answer 20

\[\frac{ 1 }{ 2^{1} 1!}, \frac{ 1 }{ 2^{2} 2!}, \frac{ 1 }{ 2^{3} 3!},\frac{ 1 }{ 2^{4} 4!}\]
\[= 1/2, 1/8, 1/48, 1/384\]

Answer 21

definitely an alternating series interp

Answer 22

\[y=\sum_{n=0}^{\infty}a_nx^n=a_0\sum_{n=0}^{\infty}(-1)^{2k}\frac{x^{2k}}{2^kk!}\]

Answer 23

@Outkast3r09 @Algebraic!

Answer 24

for the even numbers

Answer 25

this is what I get for the odds
\[n=1:a_3=-\frac{a_1}{3}=-\frac{a_1}{3\cdot1}\]
\[n=3:a_5=\frac{a_1}{15}=\frac{a_1}{5\cdot3\cdot1}\]
\[n=5:a_7=-\frac{a_1}{105}=-\frac{a_1}{7\cdot5\cdot3\cdot1}\]

Answer 26

best way to do this is just to do all these together
and then let a1=1 and a2 = 0

\[0+\frac{-1}{3*1}x+0+\frac{1}{5*3*1}x^3+0+\frac{-1}{7*5*3*1}x^5\]

Answer 27

so your x corresponds to \[x^{2k-1}\]

Answer 28

would you agree?

Answer 29

the negatives are constant with each n so you can write that as

\[(-1)^k\]

Answer 30

I would agree, but how do we put the denominator in terms of k's?
(2k+1)!

Answer 31

\[\sum_{n=0} ^\infty \frac{(-1)^kx^{2k-1}}{1*3*5*7*.....(2k+1)}\]

Answer 32

i believe this is your series =]

Answer 33

This is right lol i overlookd my own self haha

Answer 34

however i'm thinking that your zero term is outside not sure =/

Answer 35

so when we say (2k+1)!
Let's try it for k=3
\[\sum_{n=0} ^\infty \frac{(-1)^3x^{2\cdot3-1}}{1*3*5*7*.....(2\cdot3+1)}\]

Answer 36

\[\frac{-x^5}{7!}\]

Answer 37

what 1*3*5*7*.....(2k+1) means is that you continuously keep the one before it so

for n=0 you get 2(0)+1 = 1, next do n=1 1*(2*1+1)=3

so now you have for n=1 , 1*3

I wouldn't write it as 7! because that is 7*6*5*4*3*2*1

Answer 38

write it as 1*3*5*7

Answer 39

oh ok, so I'm allowed to write it as \[1*3*5*....(2k+1)\]
I thought I had to write the whole denominator in terms of k, but that makes sense now