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Math Help - Nonhomogenious - method of undetermined coefficients

  1. #1
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    Question Nonhomogenious - method of undetermined coefficients

    How to find a particular solution for this using method of undetermined coefficients:
    y'' - y = e^(-x) cos( 4 x )

    I have got the homogenious equation as:
    y = A*e^(1*x) + B*e^(-1*x)

    But now am not sure how to get a particular solution with the right side?

    thanks
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    Quote Originally Posted by taurus View Post
    How to find a particular solution for this using method of undetermined coefficients:
    y'' - y = e^(-x) cos( 4 x )

    I have got the homogenious equation as:
    y = A*e^(1*x) + B*e^(-1*x)

    But now am not sure how to get a particular solution with the right side?

    thanks
    Trial functions:
    e^{-x}cos(4x),~e^{-x}sin(4x)

    -Dan
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    why choose those?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by taurus View Post
    why choose those?
    Because they are similar in form to the non-homogeneous part and have derivatives that are also similar in form to the non-homogeneous part.

    In addition since they are not solutions to the homogeneous equation we don't have to worry about messy terms like
    xe^{-x}cos(4x)
    etc.

    -Dan
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    Question

    could you explain that? How do you choose a particular solution. I am completly confused
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    Sometimes a DE book will have a list of trial particular solutions.

    In this case, we use y_{p}=Ae^{-x}cos(4x)+Be^{-x}sin(4x)

    Because our problem is of that form. Suppose we would've had

    xe^{3x}cos(4x), then we would use

    (ax+B)e^{3x}cos(4x)+(Cx+E)e^{3x}sin(4x)

    But, back to your problem.

    Find the second derivative:

    y''_{p}=(-15A-8B)e^{-x}cos(4x)+(8A-15B)e^{-x}sin(4x)

    Sub them into the equation and get:

    y''_{p}-y_{p}=(-16A-8B)e^{-x}cos(4x)+(8A-16B)e^{-x}sin(4x)=e^{-x}cos(4x)

    Now, equate coefficients:

    There is no sin on the right so that is 0:

    -16A-8B=1

    8A-16B=0

    A=\frac{-1}{20}, \;\ B=\frac{-1}{40}

    A particular solutions is:

    \frac{-1}{20}e^{-x}cos(4x)-\frac{1}{40}e^{-x}sin(4x)

    A general solution is:

    \frac{-1}{20}e^{-x}cos(4x)-\frac{1}{40}e^{-x}sin(4x)+C_{1}e^{x}+C_{2}e^{-x}
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    would you have a list of the trials?
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  8. #8
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    No, I'm sorry. You can probably google some.
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    Quote Originally Posted by galactus View Post
    In this case, we use y_{p}=Ae^{-x}cos(4x)+Be^{-x}sin(4x)
    How would i differentiate that? with respect to what do i differentiate?
    Last edited by taurus; May 21st 2008 at 01:16 PM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by taurus View Post
    How would i differentiate that? with respect to what do i differentiate?
    The derivatives in your differential equation are with respect to x, so take the derivatives with respect to x.
    \frac{dy_p}{dx} = A(-e^{-x})cos(4x) + Ae^{-x}(-4~sin(4x)) + B(-e^{-x})sin(4x) + Be^{-x}(4~cos(x))
    by the usual product rule.

    etc.

    -Dan
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