A community for students.
Here's the question you clicked on:
 0 viewing
MathSofiya
 3 years ago
Use power series to solve the differential equation
\[y''+xy'+y=0\]
\[y=\sum_{n=0}^{\infty}\]
\[y'=\sum_{n=1}^{\infty}a_nnx^{n1}\]
\[y''=\sum{n=2}^{\infty}n(n1)a_nx^{2}\]
\[\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^{\infty}a_nx^n=0\]
\[\sum_{n=0}^{\infty}x^n\left[(n+2)(n+1)a_{n+2}+a_nn+a_n\right]=0\]
\[a_{n+2}=\frac{a_n}{n+2}\]
MathSofiya
 3 years ago
Use power series to solve the differential equation \[y''+xy'+y=0\] \[y=\sum_{n=0}^{\infty}\] \[y'=\sum_{n=1}^{\infty}a_nnx^{n1}\] \[y''=\sum{n=2}^{\infty}n(n1)a_nx^{2}\] \[\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^{\infty}a_nx^n=0\] \[\sum_{n=0}^{\infty}x^n\left[(n+2)(n+1)a_{n+2}+a_nn+a_n\right]=0\] \[a_{n+2}=\frac{a_n}{n+2}\]

This Question is Closed

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

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1the 36 kinda threw me off

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

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

Algebraic!
 3 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"

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1I can write it all out of you want me to

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1I don't know anymore

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1@Outkast3r09 watcha think?

Algebraic!
 3 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)

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1we've got something there!

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.0and a sign changing term...

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

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1but would the k! work?

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1I think I've done enough math for tonight...g'night!

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

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.1definitely an alternating series interp

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

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1@Outkast3r09 @Algebraic!

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1for the even numbers

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

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

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.1so your x corresponds to \[x^{2k1}\]

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

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

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

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.1i believe this is your series =]

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.1This is right lol i overlookd my own self haha

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.1however i'm thinking that your zero term is outside not sure =/

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

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.1what 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

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.1write it as 1*3*5*7

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.1oh 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.