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

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I can't seem to figure out the pattern \[n=0:a_2=-\frac{a_0}{2!}\] \[n=2:a_4=\frac{a_0}{3!}\] \[n=4:a_6=-\frac{a_0}{36}\]
the 36 kinda threw me off
did you forget the 'x' in x*y' ?

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Other answers:

because before the x it was \[y'=\sum_{n=1}^{\infty}a_nnx^{n-1}\]
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"
I can write it all out of you want me to
it's ok...
let me catch up
I don't know anymore
@Outkast3r09 watcha think?
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)
we've got something there!
and a sign changing term...
could also do \[\frac{ 1 }{ 2^{k}k! }\]
but would the k! work?
I think I've done enough math for tonight...g'night!
\[\frac{ 1 }{ 2^{1} 1!}, \frac{ 1 }{ 2^{2} 2!}, \frac{ 1 }{ 2^{3} 3!},\frac{ 1 }{ 2^{4} 4!}\] \[= 1/2, 1/8, 1/48, 1/384\]
definitely an alternating series interp
for the even numbers
this is what I get for the odds \[n=1:a_3=-\frac{a_1}{3}=-\frac{a_1}{3\cdot1}\] \[n=3:a_5=\frac{a_1}{15}=\frac{a_1}{5\cdot3\cdot1}\] \[n=5:a_7=-\frac{a_1}{105}=-\frac{a_1}{7\cdot5\cdot3\cdot1}\]
best way to do this is just to do all these together and then let a1=1 and a2 = 0 \[0+\frac{-1}{3*1}x+0+\frac{1}{5*3*1}x^3+0+\frac{-1}{7*5*3*1}x^5\]
so your x corresponds to \[x^{2k-1}\]
would you agree?
the negatives are constant with each n so you can write that as \[(-1)^k\]
I would agree, but how do we put the denominator in terms of k's? (2k+1)!
\[\sum_{n=0} ^\infty \frac{(-1)^kx^{2k-1}}{1*3*5*7*.....(2k+1)}\]
i believe this is your series =]
This is right lol i overlookd my own self haha
however i'm thinking that your zero term is outside not sure =/
so when we say (2k+1)! Let's try it for k=3 \[\sum_{n=0} ^\infty \frac{(-1)^3x^{2\cdot3-1}}{1*3*5*7*.....(2\cdot3+1)}\]
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
write it as 1*3*5*7
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

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