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Math Help - seperable equation

  1. #1
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    seperable equation

    ive got a differential equation

    y'-y=xy to the power 5,

    so have let (ie. .)

    Then

    Turning this around we get:


    So the differential equation becomes:


    <-- Multiply both sides by .





    and g(x)= exp^(intergral of 4 dx)

    so




    finally giving y= -x + 1/4 -c/e^{4x}

    is this the final answer, and do i need to bring back in the substitution i did earlier?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Student_richard View Post
    ive got a differential equation

    y'-y=xy to the power 5,

    so have let (ie. .)

    Then

    Turning this around we get:


    So the differential equation becomes:


    <-- Multiply both sides by .





    and g(x)= exp^(intergral of 4 dx)

    so
    You are good up to here.

    And remember, you are currently solving the u(x) equation! Now we have that
    u(x) = \frac{\int e^{4x} \cdot (-4x) dx + C}{e^{4x}}

    u(x) = \frac{-4 \int xe^{4x}dx + C}{e^{4x}}

    u(x) = \frac{-4 \left ( \frac{1}{4}xe^{4x}- \frac{1}{16}e^{4x} \right ) + C}{e^{4x}}

    u(x) = -x + \frac{1}{4} + Ce^{-4x}

    Now, y = u^{-1/4}, so

    y(x) = \frac{1}{\sqrt[4]{-x + \frac{1}{4} + Ce^{-4x}}}

    -Dan
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  3. #3
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