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Math Help - differential equation, solution check

  1. #1
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    differential equation, solution check

    can someone take a look at this
    y^{'}(t)+y(t)=sint
    y(0)=0
    y^{'}e^x+ye^x=e^xsint

    \int y^{'}e^x=\int e^xsintdx
    ye^x=\frac{e^x}{2}(sinx-cosx)+C
    y=\frac{1}{2}(sinx-cosx)+Ce^{-x}
    0=\frac{1}{2}(0-1)+C
    C=\frac{1}{2}

    y=\frac{1}{2}(sinx-cosx)+\frac{1}{2}e^{-x}
    oh yes it was a stupid mistake I have corrected it thank you
    Last edited by rayman; October 13th 2010 at 11:48 AM.
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  2. #2
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    Quote Originally Posted by rayman View Post
    can someone take a look at this
    y^{'}(t)+y(t)=sint
    y(0)=0
    y^{'}e^x+ye^x=e^xsint

    \int y^{'}e^x=\int e^xsintdx
    ye^x=\frac{e^x}{2}(sinx-cosx)+C
    y=\frac{1}{2}(sinx-cosx)+Ce^{-x}
    0=\frac{1}{2}(0-1)+C
    C=\frac{3}{2}

    y=\frac{1}{2}(sinx-cosx)+\frac{3}{2}e^{-x}


    With your solution I get y(0)=-\frac{1}{2}+\frac{3}{2}=1\neq 0 ...

    The mistake is a rather minor one and it is in the one before last line, the value of C.

    Tonio
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