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Math Help - c1 and c2

  1. #1
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    c1 and c2

    I need your help with particular solution of the following DE

    y'' + 2y'+5y=e^x , y(0)=0.125, y'(0)=0.125

    The solution I get is:
    y=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)+0.125 e^x



    To get C1 and C2 I do this:
    y_1=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)
    C_1=0.125

    y_1^{'}=-C_1 e^{-x} cos2x-2C_1 e^{-x} sin2x-C_2 e^{-x} sin2x+2C_2 e^{-x} cos2x

    C_2=0.125

    I know that the particular solution should be: y=0.125e^{x}
    Can you tell me what I'm doing wrong?
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  2. #2
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    Quote Originally Posted by hekt View Post
    I need your help with particular solution of the following DE

    y'' + 2y'+5y=e^x , y(0)=0.125, y'(0)=0.125

    The solution I get is:
    y=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)+0.125 e^x



    To get C1 and C2 I do this:
    y_1=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)
    C_1=0.125

    y_1^{'}=-C_1 e^{-x} cos2x-2C_1 e^{-x} sin2x-C_2 e^{-x} sin2x+2C_2 e^{-x} cos2x

    C_2=0.125

    I know that the particular solution should be: y=0.125e^{x}
    Can you tell me what I'm doing wrong?
    I haven't checked your general solution but you should realise that to find C1 and C2 you substitute the given boundary conditions into y=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)+0.125 e^x not y_1=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x).
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