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

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MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I can't seem to figure out the pattern \[n=0:a_2=\frac{a_0}{2!}\] \[n=2:a_4=\frac{a_0}{3!}\] \[n=4:a_6=\frac{a_0}{36}\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
the 36 kinda threw me off
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
did you forget the 'x' in x*y' ?
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
because before the x it was \[y'=\sum_{n=1}^{\infty}a_nnx^{n1}\]
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
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"
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I can write it all out of you want me to
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
it's ok...
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
let me catch up
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I don't know anymore
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
@Outkast3r09 watcha think?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
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)
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
we've got something there!
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
and a sign changing term...
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
could also do \[\frac{ 1 }{ 2^{k}k! }\]
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
k=1,2,3...
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
but would the k! work?
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I think I've done enough math for tonight...g'night!
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ 1 }{ 2^{1} 1!}, \frac{ 1 }{ 2^{2} 2!}, \frac{ 1 }{ 2^{3} 3!},\frac{ 1 }{ 2^{4} 4!}\] \[= 1/2, 1/8, 1/48, 1/384\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
definitely an alternating series interp
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[y=\sum_{n=0}^{\infty}a_nx^n=a_0\sum_{n=0}^{\infty}(1)^{2k}\frac{x^{2k}}{2^kk!}\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
@Outkast3r09 @Algebraic!
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
for the even numbers
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
this is what I get for the odds \[n=1:a_3=\frac{a_1}{3}=\frac{a_1}{3\cdot1}\] \[n=3:a_5=\frac{a_1}{15}=\frac{a_1}{5\cdot3\cdot1}\] \[n=5:a_7=\frac{a_1}{105}=\frac{a_1}{7\cdot5\cdot3\cdot1}\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
best way to do this is just to do all these together and then let a1=1 and a2 = 0 \[0+\frac{1}{3*1}x+0+\frac{1}{5*3*1}x^3+0+\frac{1}{7*5*3*1}x^5\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
so your x corresponds to \[x^{2k1}\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
would you agree?
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
the negatives are constant with each n so you can write that as \[(1)^k\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I would agree, but how do we put the denominator in terms of k's? (2k+1)!
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_{n=0} ^\infty \frac{(1)^kx^{2k1}}{1*3*5*7*.....(2k+1)}\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
i believe this is your series =]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
This is right lol i overlookd my own self haha
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
however i'm thinking that your zero term is outside not sure =/
 one year ago

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

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{x^5}{7!}\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
write it as 1*3*5*7
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago
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