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Math Help - Particular solution to this non-homogen differential equation

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    Particular solution to this non-homogen differential equation

    Find the particular solution to y''+9y=2 \cos (2x)+3 \sin (3x)

    I have y=A \cos (2x)+B \sin (3x), but for some reason when I plug in at the end everything cancels, what was wrong? Thanks
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    Quote Originally Posted by tttcomrader View Post
    Find the particular solution to y''+9y=2 \cos (2x)+3 \sin (3x)

    I have y=A \cos (2x)+B \sin (3x), but for some reason when I plug in at the end everything cancels, what was wrong? Thanks
    THis is happening because your complimentry solution is

    y_c=c_1\cos(3x)+c_2\sin(3x)

    Since you have repeated factors of \sin(3x) your sine term will look like

    y_p=(Ax^2+Bx+C)\sin(3x)

    Be careful when your complimentry solution and particular have the same factors. Good luck.
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    Quote Originally Posted by tttcomrader View Post
    Find the particular solution to y''+9y=2 \cos (2x)+3 \sin (3x)

    I have y=A \cos (2x)+B \sin (3x), but for some reason when I plug in at the end everything cancels, what was wrong? Thanks
    Hi

    The term in B \sin (3x) effectively cancels but not the one in A \cos (2x)

    First derivative
    y'=-2A \sin (2x)+3B \cos (3x)

    Second derivative
    y''=-4A \cos (2x)-9B \sin (3x)

    y''+9y=5A \cos (2x)

    This means that y=\frac25 \cos (2x) is a particular solution to y''+9y=2 \cos (2x)

    If you manage to find a particular solution to y''+9y=3 \sin (3x) then by addition you will get a particular solution to y''+9y=2 \cos (2x)+3 \sin (3x)

    Try y=Cx \cos (3x)

    EDIT : sorry, too late !
    Last edited by running-gag; March 27th 2009 at 08:35 AM. Reason: Too late
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    Hello, tttcomrader!

    Find the particular solution to: y''+9y\:=\:2 \cos (2x)+3 \sin (3x)

    I have y=A \cos (2x)+B \sin (3x) .??
    but for some reason when I plug in at the end everything cancels.
    What was wrong?
    Of course, everything cancels out . . .

    To get \cos(2x) in the solution, you need: A\sin(2x) + B\cos(2x)

    And \sin(3x) is a solution to the homogeneous equation.
    . . You need: . Cx\sin(3x) + Dx\cos(3x)

    Try it again with: . y \;=\;A\sin(2x) + B\cos(2x) + Cx\sin(3x) + Dx\cos(3x)


    You should get: . A = 0,\;B = \tfrac{2}{5},\;C = 0,\;D = -\tfrac{1}{2}

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    Thank you so much, I was actually tutoring this course, and embarrassingly don't know how to do this.

    Is there a proof for this on the internet? I really want to understand why I do these process. Thanks!

    P.S. Why is one of the terms is A \sin (2x) , I didn't start with that term in the particular solution, and I didn't find that in the homogeneous solution.
    Last edited by tttcomrader; March 28th 2009 at 09:05 AM.
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  6. #6
    Behold, the power of SARDINES!
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    Quote Originally Posted by tttcomrader View Post
    Thank you so much, I was actually tutoring this course, and embarrassingly don't know how to do this.

    Is there a proof for this on the internet? I really want to understand why I do these process. Thanks!

    P.S. Why is one of the terms is A \sin (2x) , I didn't start with that term in the particular solution, and I didn't find that in the homogeneous solution.
    Chris has a section on this here in MHF here is the link

    http://www.mathhelpforum.com/math-he...-tutorial.html

    The term A \sin (2x) comes from the original driving function and needs to be included in the particular solution.
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