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Math Help - First Order Linear Differential Equations Problems

  1. #1
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    First Order Linear Differential Equations Problems

    Hi

    Ok i have attempted the following question, but need help on what should be done next:

    y'+8y=sin(x)

    p(x)=8
    q(x)=sin(x)
    e^{\int p(x)dx} = e^{\int 8xdx} = e^{4x^2}

    e^{4x^2}y'+8ye^{4x^2}=e^{4x^2}sinx

    \frac{d(e^{4x^2}y)}{dx}=e^{4x^2}sin(x)

    e^{4x^2} = \int e^{4x^2} sin(x)
    i used integration by parts

    u=e^{4x^2} du=8xe^{4x^2}
    dv=sin(x) v=-cos(x)

    =-e^{4x^2}cos(x) - \int -8xe^{4x^2}cos(x)

    what should i do now??
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  2. #2
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    Integration by parts again.
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  3. #3
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    after first integration
    =-e^{4x^2}cos(x)+8x-e^{4x^2}sin(x)-8[\int sin(x)8x^2e^{4x^2}-sin(x)e^{4x^2}]
    after two integrations
    =-e^{4x^2}cos(x)+8x-e^{4x^2}sin(x)+8e^{4x^2}sin(x)-64[-x^2cos(x)+2[xsin(x)-sin(x)]]
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  4. #4
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    Can someone show me how this should be done.

    I have tried numerous times and i seem to never stop integrating by parts, therefore it seems to never ends
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  5. #5
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    I found out that your original integrating factor is wrong...

    Your DE is \frac{dy}{dx} + 8y = \sin{x}

    The integrating factor is e^{\int{8\,dx}} = e^{8x}, so multiplying through gives

    e^{8x}\frac{dy}{dx} + 8e^{8x}y = e^{8x}\sin{x}

    \frac{d}{dx}(e^{8x}y) = e^{8x}\sin{x}

    e^{8x}y = \int{e^{8x}\sin{x}\,dx}


    Let u = e^{8x} so that du = 8e^{8x}

    and dv = \sin{x} so that v = -\cos{x} and the integral becomes


    \int{e^{8x}\sin{x}\,dx} = -e^{8x}\cos{x} - \int{-8e^{8x}\cos{x}\,dx}

    \int{e^{8x}\sin{x}\,dx} = -e^{8x}\cos{x} + 8\int{e^{8x}\cos{x}\,dx}.


    Now let u = e^{8x} so that du = 8e^{8x}

    and dv = \cos{x} so that v = \sin{x} and the integral becomes


    \int{e^{8x}\sin{x}\,dx} = -e^{8x}\cos{x} + 8\left(e^{8x}\sin{x} - \int{8e^{8x}\sin{x}\,dx}\right)

    \int{e^{8x}\sin{x}\,dx} = -e^{8x}\cos{x} + 8e^{8x}\sin{x} - 64\int{e^{8x}\sin{x}\,dx}

    65\int{e^{8x}\sin{x}\,dx} = -e^{8x}\cos{x} + 8e^{8x}\sin{x}

    \int{e^{8x}\sin{x}\,dx} = -\frac{1}{65}e^{8x}\cos{x} + \frac{8}{65}e^{8x}\sin{x}.


    So substituting back into the original DE gives

    e^{8x}y = - \frac{1}{65}e^{8x}\cos{x} + \frac{8}{65}e^{8x}\sin{x} + C

    y = -\frac{1}{65}\cos{x} + \frac{8}{65}\sin{x} + Ce^{-8x}.
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  6. #6
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    oh ic now thank you
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