Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

MathSofiya

  • 2 years ago

Use power series to solve the differential equation \[(x-1)y''+y'=0\] \[y=\sum_{n=0}^\infty a_nx^n\] \[y'=\sum_{n=1}^\infty na_nx^{n-1}\] \[y''=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}\] \[(x-1)\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}=0\]

  • This Question is Closed
  1. MathSofiya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    allow me to complete this problem as far as I can, give me like 5 mins. Thanks!

  2. MathSofiya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\sum_{n=2}^{\infty}n(n-1)a_nx^{n-1}-\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}=0\]

  3. MathSofiya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\sum_{n=1}^{\infty}n(n+1)a_{n+1}x^n-\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^\infty (n+1)a_{n+1}x^n=0\]

  4. MathSofiya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so when we strip out a term on \[\sum_{n=1}^{\infty}n(n+1)a_{n+1}x^n\] for the sum to start at n=0 would I have... \[a_0+ \sum_{n=0}^\infty n(n+1)a_{n+1}x^n \]

  5. MathSofiya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no that would be \[a_0+ \sum_{n=1}^\infty n(n+1)a_{n+1}x^n\]

  6. mahmit2012
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cof of x^n (left)=cof of x^n (right) don't use sigmas.

  7. mahmit2012
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    already I have showed you ! Take a look to my solution.

  8. mahmit2012
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    y=an y'=(n+1)an+1 xy'=nan y''=(n+2)(n+1)an+2 xy''=(n+1)nan+1 x2y=n(n-1)an and so on.

  9. mahmit2012
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (n+1)nan+1-(n+2)(n+1)an+2+(n+1)an+1=0 an+2=(n+1)/(n+2) an+1 an+2=a1/(n+2) one answer is just a0(check it out) y=a0+a1sigma(n=1 to inf)(x^n/n)

  10. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find 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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.