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Help with ODE

  1. Jul 10, 2005 #1

    cronxeh

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    Ok I'm stuck on this problem - and it didnt particularly look hard when I started doing it, oh say about a good 2 hours ago.. hmm

    Solve the IVP
    [tex]2y'' + 3y' + y = t^2 + 85 sin(2t), y(0) = -2, y'(0) = 0.[/tex]

    My homogeneous solution is [tex]c1y1 + c2y2[/tex]

    [tex]y1 = e^\frac{-t}{2}, y2 = e^-^t, Wronskian = e^\frac{-3t}{2}+e^\frac{-t}{2} [/tex]

    So for a particular solution I've tried 2 different methods

    1. Undetermined Coefficients
    Assumed solution was [tex]At^2 + Bt + C + Dcos(2t) + Esin(2t)[/tex]

    [tex]yp' = 2At + B - 2Dsin(2t) + 2Ecos(2t)[/tex]
    [tex]yp'' = 2A - 4Dcos(2t) -4Esin(2t)[/tex]

    After pluggin this back into original equation [tex]2y'' + 3y' + y[/tex] I got the following equalities:
    [tex]At^2 + Bt + C + 6At + 3B + 4A = t^2[/tex]
    [tex]cos(2t)(-7D + 6F) = 0[/tex]
    [tex]sin(2t)(-7F - 6D) = 85sin(2t)[/tex]

    Fiddling around with the equalities I got C=0, D=-6, F=-7

    [tex]A=\frac{3t}{3t+14}, B=\frac{-4t}{3t+14}[/tex]

    After assembling the whole thing together I dont get the right hand side after integrating to check the solutions, and so even when I derive the y=yp+yh solution to get my constants for homogeneous equation I get wrong coefficients c1 and c2 for yh - an infinite number of them, so I'm pretty sure my solution is wrong

    Should I have multiplied the assumed solution by t anywhere?

    2. Variation of Parameters

    [tex]Yp = -y1 \int \frac{y2 g(t) dt}{Wronskian} + y2 \int \frac{y1 g(t) dt}{Wronskian}[/tex]
    where [tex]g(t) =\frac{ t^2 + 85sin(2t) } {2}[/tex]
    The result was extremely long and contained a lot of exponentials, and I assumed it to be incorrect, although any comment on the correct answer you got would be appreciated
     
    Last edited: Jul 10, 2005
  2. jcsd
  3. Jul 10, 2005 #2

    James R

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    In your undetermined coefficients solution, why do A and B depend on t? A and B are supposed to be constants, right?

    The particular solution I get is:

    [tex]y = t^2 - 6t + 14 -6\cos 2t -7 \sin 2t[/tex]
     
  4. Jul 10, 2005 #3

    saltydog

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    I get the following equations:

    4A+3B+C=0
    6A+B=0
    A=1
    6E-7D=0
    -7E-6D=85

    Edit: Suppose I should compare this to the results reported by James . . .
    Ok, I'm cool.
     
    Last edited: Jul 10, 2005
  5. Jul 10, 2005 #4

    cronxeh

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    Doh.. of course! So since [tex]At^2 = t^2 [/tex] then [tex]A=1[/tex] and I simply equate the other terms to 0. Thanks.
     
  6. Jul 11, 2005 #5

    GCT

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    With variation of parameters you typically want to work out the entire step, especially if you're new to it, I prefer the practical variation of parameters approach v.s. plugging in to the wronskian derivation. Reduction of order should work with this also, except try using your y1 intstead of the simpler y2. Have you gotten to the laplace transform yet?
     
  7. Jul 11, 2005 #6

    GCT

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    I think that you will have more cancellations with the y1, however if you need to go with y2 (using reduction of order), you'll probably need to use euler's form of sin(2t) to solve for your first integral
     
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