1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

This is review for me, but something is going awry

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the solution of the following initial value problems:

    (a) y'' - 2y' + 5y = 0, y(π/2)=0, y'(π/2)=0

    2. Relevant equations

    Just know to "guess" y=Cert, how to bring in Euler's formula, and that sin(x) is odd while cos(x) is even

    3. The attempt at a solution

    Guess: y= ert

    -----> r2ert - 2rert + 5ert = 0
    -----> r2 - 2r + 5 = 0
    -----> (r-1)2 = -5 + 1
    -----> r - 1 = +/- √-4
    -----> r = 1 +/- 2i
    -----> y = et( C1(cos2t+isin2t) + C2(cos(-2t)+isin(-2t)) )
    -----> y = et( C1cos2t + C1sin2t - C2cos2t + C2isin2t)
    -----> y = et(K1cos2t + K2sin2t), where K1, K2 are new constants to represent (C1 - C2), (iC1 + iC2)


    But then come the initial conditions!

    0 = eπ/2K1(-1)
    2 = K1eπ/2(-1)+ 2K2eπ/2(-1)

    It just seems like a ghey problem if K1 disappears
     
    Last edited: Apr 22, 2010
  2. jcsd
  3. Apr 22, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Looks good so far. Did you mean y'(pi/2)=2? What's so hard about the initial conditions? What are sin(pi) and cos(pi)?
     
  4. Apr 22, 2010 #3
    Oh, and another one.

    y'' + 2y' + y = 3e-t

    I'll find the homogenous solution first. The characteristic polynomial is r2 + 2r + 1 = 0 -----> r = -1, a repeated root ----> The homogenous solution is (C1 + C2t)e-t.

    Now I'll guess a second general solution, C3e-t.

    ------> C3et - 2C3e-t + C3e-t = 3e-t
    ------> 0 = 0.

    Fork! So is the homogeneous solution all? Just for the heck of it, I tried y(t)=(At2 + Bt + C)e-t, but to no avail; there doesn't seem to be a polynomial that can be multiplied by e-t without my getting 0=0.

    Where should I go from here?
     
    Last edited: Apr 22, 2010
  5. Apr 22, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    There is an inhomogeneous solution of the form (At^2+Bt+C)*exp(-t). You just missed it.
     
  6. Apr 22, 2010 #5
    Well, yeah.

    y(t) = C1etcos2t + C2etsin2t -----> y'(t) = et(-C12sin2t + C1cos2t + 2C2cost + C2sin2t).

    So,
    0 = C1eπ/2cos(π) + C2eπ/2sin(π)
    2 = -2C1eπ/2sin(π) + C1eπ/2cos(π) + 2C2eπ/2cos(π) + C2eπ/2sin(π)

    ----->

    2 = -C2eπ/2


    ----> y(t) = -e-π/2etcos2t
     
    Last edited: Apr 22, 2010
  7. Apr 22, 2010 #6
    Allow me to demonstrate.

    y(t) = At2e-t + Bte-t + Ce-t
    ----> y'(t)= -At2e-t + 2Ate-t -Bte-t + Be-t - Ce-t

    ----> y''(t)= At2e-t - 2Ate-t - 2Ate-t + 2Ae-t + Bte-t -Be-t - Be-t + Ce-t




    -----> At2e-t - 2Ate-t - 2Ate-t + 2Ae-t + Bte-t -Be-t - Be-t + Ce-t + 2*[ -At2e-t + 2Ate-t -Bte-t + Be-t - Ce-t ] + At2e-t + Bte-t + Ce-t = 3e-t

    Everything on the left side cancels out!
     
  8. Apr 22, 2010 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, what's wrong or 'ghey' about that?
     
  9. Apr 22, 2010 #8
    I'm used to easy system of equations in these problems.

    Something like

    0 = C1 + C2
    2 = 3C1 - C2,

    not something with eπ/2
     
  10. Apr 22, 2010 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Look, you can forget about the B and C parts. You should KNOW they are going to cancel. They are part of the homogenous solution. Are you really going to make me check that? I will ask you, what cancelled the 2*A*exp(-t)?
     
  11. Apr 22, 2010 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    exp(pi/2) is just another constant. Doesn't make it any harder than a constant like '3'.
     
  12. Apr 22, 2010 #11
    Let us see.

    y(t) = At2e-t,
    y'(t) = -At2e-t + 2Ate-t,
    y''(t)=At2e-t - 2Ate-t - 2Ate-t + 2Ae-t

    ------------>

    At2e-t - 2Ate-t - 2Ate-t + 2Ae-t -2At2e-t + 4Ate-t + At2e-t = 3e-t


    ----------> A = 3/2

    I think I finally got it. What a ghey problem.

    WAIT! Now I have the homogeneous solution, in which the coefficients are undetermined, PLUS one another e-t with a determined coefficient. How does that work? What was the point if (3/2 + C1) could just be merged into a new constant?
     
    Last edited: Apr 22, 2010
  13. Apr 22, 2010 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I would stop saying 'ghey'. I don't think it's universally appreciated. You can't absorb the inhomogeneous solution into the homogeneous solution by adjusting a constant. You are just being sloppy again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: This is review for me, but something is going awry
  1. Limits review (Replies: 4)

  2. Midterm Review (Replies: 6)

Loading...