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Homework Help: 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

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    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

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    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

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    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

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    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

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    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

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    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.
     
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