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Math Help - [SOLVED] Solve for a DE

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Solve for a DE

    I must solve a lot of DE's. I'm finishing my vacations and the upper level undergraduate courses of physics are coming very soon. I just begin with DE's. I hope to improve quickly.

    Find the general solution of this DE:
    y'+3y=x+e^{-2x}.
    I don't expect a full solution. First, I'd be glad to know what kind of DE it is and the way to solve it. Maybe a reference about this thread: http://www.mathhelpforum.com/math-he...-tutorial.html can be nice.
    Thanks for the help.
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  2. #2
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    Quote Originally Posted by arbolis View Post

    Find the general solution of this DE:
    y'+3y=x+e^{-2x}.
    I don't expect a full solution. First, I'd be glad to know what kind of DE it is and the way to solve it.
    It is a first order linear non homogenous D.E. It is not separable.

    You need to employ the integrating factor method.


    Quote Originally Posted by arbolis View Post

    Maybe a reference about this thread: http://www.mathhelpforum.com/math-he...-tutorial.html can be nice.
    Thanks for the help.
    There is a section near the beginning of this thread, let us know how you go.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thank you very much!
    I'm not sure I'm doing well. I reach y=e^{-\int 3 dx} \left [ \int (x+e^{-2x})e^{-\int 3 dx} dx \right ] =e^{-3x+C_0} \left [ \int (x+e^{-2x})e^{-3x+C_1} dx \right ]. Is it right until here?
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  4. #4
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    I get an itegrating factor of e^{\int 3~dx} = e^{ 3x}

    Now you multiply this through both sides of your equation

    y'+3y=x+e^{-2x}

    should give you

    e^{ 3x}y'+3e^{ 3x}y=e^{ 3x}(x+e^{-2x})

    then integrating both sides you get

    e^{ 3x}y=\int e^{ 3x}(x+e^{-2x})~dx

    y=\frac{\int e^{ 3x}(x+e^{-2x})~dx}{e^{ 3x}}

    y=\frac{\int xe^{ 3x}+e^{x}~dx}{e^{ 3x}}

    Use integration by parts for the numerator and you will be finished.
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  5. #5
    MHF Contributor arbolis's Avatar
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    I reached y=\frac{1}{3} \left ( x-1-\frac{e^{2x}}{3}+C \right ) where C is a constant which depends on initial conditions not given here. I don't know if I made an arithmetic error, but at least I've understood this method.
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  6. #6
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  7. #7
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by pickslides View Post
    Ok I made a mistake. I will correct it.
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