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

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

  • 10 months ago
  • 10 months ago

  • This Question is Closed
  1. oldrin.bataku
    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)

    • 10 months ago
  2. goalie2012
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 10 months ago
  3. oldrin.bataku
    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 $$y'=\sum_{n=0}^\infty (n+1)a_{n+1}x^n\\y''=\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n$$

    • 10 months ago
  4. oldrin.bataku
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 10 months ago
  5. oldrin.bataku
    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:$$x\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty a_nx^{n+1}=\sum_{n=1}^\infty a_{n-1}x^n$$

    • 10 months ago
  6. oldrin.bataku
    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:$$\begin{align*}(1-2x)&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-2x\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=0}^\infty2(n+1)(n+2)a_{n+2}x^{n+1}\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=1}^\infty2n(n+1)a_{n+1}x^n\end{align*}$$

    • 10 months ago
  7. oldrin.bataku
    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!

    • 10 months ago
  8. oldrin.bataku
    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...

    • 10 months ago
  9. goalie2012
    Best Response
    You've already chosen the best response.
    Medals 0

    ya. that's where I got a little lost

    • 10 months ago
  10. oldrin.bataku
    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 \(2(0)(1)a_1x^0=0\):$$\begin{align*}&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=1}^\infty2n(n+1)a_{n+1}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=0}^\infty2n(n+1)a_{n+1}x^n\\=&\sum_{n=0}^\infty[(n+2)a_{n+2}-2na_{n+1}](n+1)x^n\end{align*}$$

    • 10 months ago
  11. oldrin.bataku
    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.

    • 10 months ago
  12. oldrin.bataku
    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 \(0\):$$\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}=0$$

    • 10 months ago
  13. oldrin.bataku
    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 \(n=0\):$$a_0x^0=0\\a_0=0$$

    • 10 months ago
  14. oldrin.bataku
    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, \(a_0=0,a_1=1\).

    • 10 months ago
  15. oldrin.bataku
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 10 months ago
  16. oldrin.bataku
    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:$$(2a_2-a_1)+a_{-1}=0\\2a_2-1=0\\a_2=\frac12$$

    • 10 months ago
  17. oldrin.bataku
    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 \(a_2=\frac12\):$$6a_3-3=0\\6a_3=3\\a_3=\frac12$$

    • 10 months ago
  18. oldrin.bataku
    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 \(a_3=\frac12\):$$12a_4-\frac{15}2+1=0\\24a_4-15+2=0\\24a_4-13=0\\a_4=\frac{13}{24}$$

    • 10 months ago
  19. oldrin.bataku
    Best Response
    You've already chosen the best response.
    Medals 1

    Can you take it from here? ;-)

    • 10 months ago
  20. goalie2012
    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?

    • 10 months ago
  21. oldrin.bataku
    Best Response
    You've already chosen the best response.
    Medals 1

    I can try!

    • 10 months ago
  22. goalie2012
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 10 months ago
  23. goalie2012
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 10 months 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.