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anonymous
 4 years ago
Use power series to solve the differential equation
\[y''+xy'+y=0\]
\[y=\sum_{n=0}^{\infty}\]
\[y'=\sum_{n=1}^{\infty}a_nnx^{n1}\]
\[y''=\sum{n=2}^{\infty}n(n1)a_nx^{2}\]
\[\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^{\infty}a_nx^n=0\]
\[\sum_{n=0}^{\infty}x^n\left[(n+2)(n+1)a_{n+2}+a_nn+a_n\right]=0\]
\[a_{n+2}=\frac{a_n}{n+2}\]
anonymous
 4 years ago
Use power series to solve the differential equation \[y''+xy'+y=0\] \[y=\sum_{n=0}^{\infty}\] \[y'=\sum_{n=1}^{\infty}a_nnx^{n1}\] \[y''=\sum{n=2}^{\infty}n(n1)a_nx^{2}\] \[\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^{\infty}a_nx^n=0\] \[\sum_{n=0}^{\infty}x^n\left[(n+2)(n+1)a_{n+2}+a_nn+a_n\right]=0\] \[a_{n+2}=\frac{a_n}{n+2}\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can't seem to figure out the pattern \[n=0:a_2=\frac{a_0}{2!}\] \[n=2:a_4=\frac{a_0}{3!}\] \[n=4:a_6=\frac{a_0}{36}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the 36 kinda threw me off

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did you forget the 'x' in x*y' ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because before the x it was \[y'=\sum_{n=1}^{\infty}a_nnx^{n1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0bit 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"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can write it all out of you want me to

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Outkast3r09 watcha think?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0factorial 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we've got something there!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and a sign changing term...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0could also do \[\frac{ 1 }{ 2^{k}k! }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but would the k! work?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think I've done enough math for tonight...g'night!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ 2^{1} 1!}, \frac{ 1 }{ 2^{2} 2!}, \frac{ 1 }{ 2^{3} 3!},\frac{ 1 }{ 2^{4} 4!}\] \[= 1/2, 1/8, 1/48, 1/384\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0definitely an alternating series interp

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y=\sum_{n=0}^{\infty}a_nx^n=a_0\sum_{n=0}^{\infty}(1)^{2k}\frac{x^{2k}}{2^kk!}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Outkast3r09 @Algebraic!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is what I get for the odds \[n=1:a_3=\frac{a_1}{3}=\frac{a_1}{3\cdot1}\] \[n=3:a_5=\frac{a_1}{15}=\frac{a_1}{5\cdot3\cdot1}\] \[n=5:a_7=\frac{a_1}{105}=\frac{a_1}{7\cdot5\cdot3\cdot1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0best way to do this is just to do all these together and then let a1=1 and a2 = 0 \[0+\frac{1}{3*1}x+0+\frac{1}{5*3*1}x^3+0+\frac{1}{7*5*3*1}x^5\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so your x corresponds to \[x^{2k1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the negatives are constant with each n so you can write that as \[(1)^k\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I would agree, but how do we put the denominator in terms of k's? (2k+1)!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=0} ^\infty \frac{(1)^kx^{2k1}}{1*3*5*7*.....(2k+1)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i believe this is your series =]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is right lol i overlookd my own self haha

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0however i'm thinking that your zero term is outside not sure =/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so when we say (2k+1)! Let's try it for k=3 \[\sum_{n=0} ^\infty \frac{(1)^3x^{2\cdot31}}{1*3*5*7*.....(2\cdot3+1)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh 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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