Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

goalie2012 Group Title

solve the differential equation with initial conditions; compute the first 6 coefficients(a0-a5); find the general pattern: (1-2x)y''-y'+xy=0 y(0)=0, y'(0)=1

  • one year ago
  • one year ago

  • This Question is Closed
  1. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Have you read the FAQ? It's generally frowned upon to post the same question multiple times... additionally, you are expected to reward helpful answers with medals (re: your previous question)

    • one year ago
  2. goalie2012 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry about that. Kind of new. thanks for the tip

    • one year ago
  3. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Anyways, again consider an analytic solution \(y=\sum\limits_{n=0}^\infty a_nx^n\) with derivatives $$y'=\sum_{n=0}^\infty (n+1)a_{n+1}x^n\\y''=\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n$$

    • one year ago
  4. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Plugging into our equation we have:$$(1-2x)\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=0}^\infty(n+1)a_{n+1}x^n+x\sum_{n=0}^\infty a_nx^n=0$$

    • one year ago
  5. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    For our last term, note we can multiply into our sum \(x\) and shift our indices of summation:$$x\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty a_nx^{n+1}=\sum_{n=1}^\infty a_{n-1}x^n$$

    • one year ago
  6. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    For our first term, observe we can distribute \(1-2x\) and do something similar:$$\begin{align*}(1-2x)&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-2x\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=0}^\infty2(n+1)(n+2)a_{n+2}x^{n+1}\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=1}^\infty2n(n+1)a_{n+1}x^n\end{align*}$$

    • one year ago
  7. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @goalie2012 you should award the medal on the other question... I haven't answered this one yet!

    • one year ago
  8. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Now, I haven't combined them into 1 yet because dealing with the starting indices would be gross...

    • one year ago
  9. goalie2012 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ya. that's where I got a little lost

    • one year ago
  10. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Well, we can safely rewrite that summation again with 'nicer' indices since starting at \(n=0\) will just add \(2(0)(1)a_1x^0=0\):$$\begin{align*}&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=1}^\infty2n(n+1)a_{n+1}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=0}^\infty2n(n+1)a_{n+1}x^n\\=&\sum_{n=0}^\infty[(n+2)a_{n+2}-2na_{n+1}](n+1)x^n\end{align*}$$

    • one year ago
  11. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    We can also extend our previous summation:$$\sum_{n=1}^\infty a_{n-1}x^n=\sum_{n=0}^\infty a_{n-1}x^n$$where we can impose \(a_m=0\) for \(m<0\) (this way we don't change the value of the summation); this is okay because any \(a_m\) for \(m<0\) do not appear in our solution.

    • one year ago
  12. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Anyways, let's put it all together:$$\sum_{n=0}^\infty[(n+2)a_{n+2}-2na_{n+1}](n+1)x^n-\sum_{n=0}^\infty (n+1)a_{n+1}x^n+\sum_{n=0}^\infty a_{n-1}x^n=0\\\sum_{n=0}^\infty\left(\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}\right)x^n=0$$Because our right must be identically \(0\), we require the coefficient of each power of \(x\) to be \(0\):$$\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}=0$$

    • one year ago
  13. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Now, consider our initial conditions \(y(0)=0,y'(0)=1\):$$y(0)=0\\\sum_{n=0}^\infty a_nx^n=0$$Since \(x^n=0\) for \(n>0\), we're left with only the term for \(n=0\):$$a_0x^0=0\\a_0=0$$

    • one year ago
  14. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Moving on to our second condition \(y'(0)=1\), we find:$$\sum_{n=0}^\infty(n+1)a_{n+1}x^n=1$$Again, \(x=0\) leaves all but our \(n=0\) term:$$a_1x^0=1\\a_1=1$$... so we've solved for first two coefficients, \(a_0=0,a_1=1\).

    • one year ago
  15. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    For our last three, we go back to that expression!

    • one year ago
  16. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    $$\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}=0$$Letting \(n=0\) we know \(a_{-1}=0,a_0=0,a_1=1\) so:$$(2a_2-a_1)+a_{-1}=0\\2a_2-1=0\\a_2=\frac12$$

    • one year ago
  17. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    For \(n=1\) we find:$$2(3a_3-2a_2-a_2)+a_0=0\\6a_3-6a_2+a_0=0$$Knowing \(a_0=0,a_1=1\) and having just concluded \(a_2=\frac12\):$$6a_3-3=0\\6a_3=3\\a_3=\frac12$$

    • one year ago
  18. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    $$\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}=0$$ For \(n=2\) we find:$$3(4a_4-2(2)a_3-a_3)+a_1=0\\12a_4-15a_3+a_1=0$$Again, knowing \(a_1=1\) and having concluded \(a_3=\frac12\):$$12a_4-\frac{15}2+1=0\\24a_4-15+2=0\\24a_4-13=0\\a_4=\frac{13}{24}$$

    • one year ago
  19. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Can you take it from here? ;-)

    • one year ago
  20. goalie2012 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yep. thank you very much. I have two more. would you be willing to help?

    • one year ago
  21. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I can try!

    • one year ago
  22. goalie2012 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    one won't take long and the other I'm not sure about

    • one year ago
  23. goalie2012 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I have to close this question to write another one, right?

    • one year ago
    • Attachments:

See more questions >>>

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.