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Math Help - Initial Value Problem

  1. #1
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    Initial Value Problem

    Hi All,

    Can somebody please help me solve this IVP? Or simply explain ivp's to me so I know what I'm supposed to do with this?

    (1+t^2)dy/dt + 2ty = -t sin(t), y(pi) = 0

    Thank you
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  2. #2
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    Quote Originally Posted by woody198403 View Post
    Hi All,

    Can somebody please help me solve this IVP? Or simply explain ivp's to me so I know what I'm supposed to do with this?

    (1+t^2)dy/dt + 2ty = -t sin(t), y(pi) = 0

    Thank you
    It is a first order linear equation.
    If you put it in standard form and solve for the integrating factor you get the same equation you start with.

    so it is the same as

    \frac{d}{dt}\left[ (1+t^2)y\right]=-t\sin(t)

    Now we integrate both sides to get

    (1+t^2)y=t\cos(t)-\sin(t)+C

    You should be able to finish from here.

    Good luck.
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  3. #3
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    Okay. Here is what ive done, and now Im stuck on a part.

    (1+t^2)dy/dt + 2ty = -t sin(t), y(pi) = 0

    dy/dt + (2ty/(1+t^2))*y = (-t sin(t))/(1+t^2)

    Then I find the integrating factor, which I think is

    =e^(y*ln(1+t^2))

    = (1 +t^2)^y

    Multiplying this through, you get:

    [((1+t^2)^y) * y] ' = [-(((1+t^2)^y)*t*sin(t))/(1+t^2)]

    Then I think I'm supposed to integrate both sides. I know the l.h.s. is just


    ((1+t^2)^y)*y

    but im having trouble integrating [-(((1+t^2)^y)*t*sin(t))/(1+t^2)]

    Have I done something wrong here? Why does your answer look so neat compared to mine?

    I know that once I have that I can apply the initial condition and workout the particular solution.
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by woody198403 View Post

    Multiplying this through, you get:

    [((1+t^2)^y) * y] ' = [-(((1+t^2)^y)*t*sin(t))/(1+t^2)]

    Then I think I'm supposed to integrate both sides. I know the l.h.s. is just


    ((1+t^2)^y)*y

    but im having trouble integrating [-(((1+t^2)^y)*t*sin(t))/(1+t^2)]

    Have I done something wrong here? Why does your answer look so neat compared to mine?


    I know that once I have that I can apply the initial condition and workout the particular solution.
    Reduce your fractions

    \frac{-(1+t^2)(t\sin(t)}{1+t^2}=-t\sin(t)

    Edit: You need to use integration by parts twice from here
    Last edited by TheEmptySet; May 20th 2008 at 08:31 AM. Reason: To be clearer
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    Reduce your fractions

    \frac{-(1+t^2)(t\sin(t)}{1+t^2}=-t\sin(t)

    Edit: You need to use integration by parts twice from here
    Of course. I cant believe I didnt see that.
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