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

Callisto Group Title

Given that \(y=x^2cosx\) satisfies \(x^2y^{(2)} - 4xy^{(1)} + (x^2 +6)y =0\) , where \( y^{(n)} = \frac{d^ny}{dx^n}\). (a) Show that when x=0, \((n-2)(n-3)y^{(n)} = -n(n-1)y^{(n-2)}\). (b) If we use the notation \(\bar y^{(n)} = \frac{d^ny}{dx^n}|_{x=0}\), find \(\bar y^{(2m)}\) and \(\bar y^{(2m+1)}\). How to do part (b)?

  • one year ago
  • one year ago

  • This Question is Closed
  1. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\bar y^{(n)} = -(\frac{n(n-1)}{(n-2)(n-3)})\bar y^{(n-2)}\]When n=2m \[\bar y^{(2m)} = -(\frac{2m(2m-1)}{(2m-2)(2m-3)})\bar y^{(2m-2)}\]\[\bar y^{(2m)} = -\frac{m(2m-1)}{(m-1)(2m-3)}\bar y^{(2m-2)}\]Then... I don't know how to continue.. When m=0, \[\bar y^{(0)} = 0\] When m=1, \[\bar y^{(2)} = \frac{d^2y}{dx^2}(x^2cosx)|_{x=0} = 2\] When m=2, \[\bar y^{(4)} = \frac{2(4-1)}{(2-1)(4-3)}\bar y^{(4-2)} =2\times 3 \times (2) = 12\]Hmm...

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

    Oh wait. It should be -12. My bad!

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

    Well, I guess 0 is not in the domain of m, so just ignore it...

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

    Hmm.... By observing the pattern, I got \[\bar y^{(2m)} = (-1)^{m-1}(2m-1)(2m)\]

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

    So far so good? Or...?

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

    Not that good. 0 is not a domain when n=2m. Since in part a, domain of n is positive integers. So, for m, it can't be 0. Hmm...

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

    *

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

    But for n=2m+1, m can be 0 Since the domain of n is positive integers, When m=0, 2m+1 = 1 >0 => it's in the domain of n, so for n=2m+1, domain of m is integers greater than or equal to 0.

    • 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.