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Math Help - 2nd ODE problem

  1. #1
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    2nd ODE problem

    find the general solution:

    y" + 4y = 3sin(2x)

    I'm able to solve for the homogeneous by:

    y" + 4y = 0

    y(x) = Asin(2x) + Bcos(2x)

    however I'm unsure about the particular solution for 3sin(2x).

    any hints to point me in the right direction would be much appreciated. Thank you
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  2. #2
    Member Mauritzvdworm's Avatar
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    consider the homogenous equation

    \frac{d^{2}y}{dx^{2}}+4y=0

    this has roots repeated at -2 since

    r^{2}+4=(r+2)^{2}

    the solution to the homogenous equation will then be of the form

    y=(c_{1}+c_{2}x)e^{-2x}

    now we consider the particular solution, since we are differentiating twice if you try the following solution

    y_{particular}=(c_{1}+c_{2}x)e^{-2x}+c_{3}\sin (2x)

    you should find the correct answer.

    Plug it into your DE and solve for the constant c_{3}
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  3. #3
    MHF Contributor Calculus26's Avatar
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    r^2 +4 is not (r+2)^2


    Tasukete you are correct

    y(x) = Asin(2x) + Bcos(2x)


    is the correct homogeneous solution

    For the particular solution if you use undetermined cofficients
    use yp = Axsin(2x) + Bxcos(2x)

    multiply by x since sin(2x) and cos(2x) are homogeneous solutions.

    Personally I would use variation of parameters if you have learned that method if not see Laplace Transforms for the notes on var of parameters
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  4. #4
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    Quote Originally Posted by Mauritzvdworm View Post
    consider the homogenous equation

    \frac{d^{2}y}{dx^{2}}+4y=0

    this has roots repeated at -2 since

    r^{2}+4=(r+2)^{2}

    the solution to the homogenous equation will then be of the form

    y=(c_{1}+c_{2}x)e^{-2x}

    now we consider the particular solution, since we are differentiating twice if you try the following solution

    y_{particular}=(c_{1}+c_{2}x)e^{-2x}+c_{3}\sin (2x)

    you should find the correct answer.

    Plug it into your DE and solve for the constant c_{3}
    Sorry but this is completely wrong.

    1. The roots of r^{2}+4 = 0 are r = \pm 2i. Therefore the homogenous solution is y_h = A \cos (2x) + B \sin (2x).

    @OP: The difficulty in getting a particular solution is that the term on the right hand side of the differential equation is part of the homogenous solution .... Therefore try this one:

    y_p = (ax + b) \sin (2x) + (cx + d) \cos (2x)

    Your job is to determine the value of the coefficients a, b, c and d.
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  5. #5
    MHF Contributor Calculus26's Avatar
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    Mr fantastic you don't need the b and d in



    to find the particular solution.

    though in the general solution you will have yc(x) = Asin(2x) + Bcos(2x)

    as part of the general solution
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  6. #6
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    Quote Originally Posted by Calculus26 View Post
    Mr fantastic you don't need the b and d in



    to find the particular solution.

    though in the general solution you will have yc(x) = Asin(2x) + Bcos(2x)

    as part of the general solution
    Yes.

    When f(x) is part of the homogeneous solution, the rule of thumb is to multiply the 'usual form' of the particular solution by x^m, where m is the smallest positive integer needed to ensure that the 'new' particular solution is not part of the homogeneous solution.
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  7. #7
    Member Mauritzvdworm's Avatar
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    sorry, that was a bit careless of me. we will have imaginary roots
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