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Thread: c1 and c2

  1. #1
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    c1 and c2

    I need your help with particular solution of the following DE

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

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



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

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

    $\displaystyle C_2=0.125$

    I know that the particular solution should be: $\displaystyle y=0.125e^{x}$
    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

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

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



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

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

    $\displaystyle C_2=0.125$

    I know that the particular solution should be: $\displaystyle y=0.125e^{x}$
    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 $\displaystyle y=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)+0.125 e^x$ not $\displaystyle y_1=C_1 (e^{-x} cos2x)+C_2 (e^{-x} sin2x)$.
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