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Math Help - Nonhomogenious Equation

  1. #1
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    Question Nonhomogenious Equation

    I am a bit confused on how to solve the following using the method of undetermined coefficients:

    equation: y'' + y - 2y = x + sin2x

    So what i have done so far is:
    roots are 1, -2 so
    Yc = Ae^(x) + Be^(-2x)

    But from the homogenious equation above i dont know how to go on. can someone help me solve this?

    Thanks
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  2. #2
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    Quote Originally Posted by taurus View Post
    I am a bit confused on how to solve the following using the method of undetermined coefficients:

    equation: y'' + y - 2y = x + sin2x

    So what i have done so far is:
    roots are 1, -2 so
    Yc = Ae^(x) + Be^(-2x)

    But from the homogenious equation above i dont know how to go on. can someone help me solve this?

    Thanks
    the particular solution will be of the form
    y_p(x)=Ax^2+Bx+C+D\sin(2x)+E\cos(2x)

    y'_p=2Ax+B+2D\cos(2x)-2E\sin(2x)

    y''_p=2A-4D\sin(2x)-4E\cos(2x)

    just sub these back into the non homogeonious equation and solve for the constants.

    I hope this helps.
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  3. #3
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    Dont quite get it, which equation to you sub back?
    and is A=1, B, 1, C=-2 So whats D and E?

    thanks
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  4. #4
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    Quote Originally Posted by taurus View Post
    Dont quite get it, which equation to you sub back?
    and is A=1, B, 1, C=-2 So whats D and E?

    thanks
    y'' + y - 2y = x + sin2x

    y_p=D\sin(2x)+E\cos(2x)

    y_p=2D\cos(2x)-2E\sin(2x)

    y_p=-4D\sin(2x)-4E\cos(2x)

    =-4D\sin(2x)-4E\cos(2x)+2D\cos(2x)-2E\sin(2x)-2[D\sin(2x)+E\cos(2x)]
    =\sin(2x)

    [-6D-2E]\sin(2x)+[-6E+2D]\cos(2x)=\sin(2x)

    so we get

    -6D-2E=1
    2D-6E=0

    Solve this system gives E=-\frac{1}{20},D=-\frac{3}{20}

    I hope this clears it up.
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  5. #5
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    Wait so now i put what back into the nonhomogenious equation?

    so it becomes:
    y = e^x + e^(-2x) + x^2 + -2 + (-3/20)sin(2x) + (-1/20)cos(2x) ?
    Is this correct?
    Last edited by taurus; May 14th 2008 at 08:03 AM.
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  6. #6
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    Question

    Actually I think its this:

    y = Ae^x + Be^(-2x) + x^2 + -2 + (-3/20)sin(2x) + (-1/20)cos(2x)

    and then i can solve for A and B using y(0)=1 and y'(0)=0

    is this correct?
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