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Math Help - diff eqn using integrating factor

  1. #1
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    Red face diff eqn using integrating factor

    Want to know if my answer is correct;

    solve the diff eqn \left (u^2 + 1 \right) \, \frac{dv}{du} + 4uv = 3u

    \frac{dv}{du} + \frac{4uv}{u^2 + 1} = \frac{3u}{u^2 + 1}

    Compare \, \, \frac{dy}{dx} + Py = Q

    \therefore \, P = \frac{4u}{u^2 + 1} \, \, \, and \, \, \, Q = \frac{3u}{u^2 + 1}

    Integrating \, \, Factor = e^{\int Pdu} = e^{\int \frac{4u}{u^2 + 1}} = e^{2 \int \frac{2u}{u^2 + 1}} = e^{2ln(u^2 + 1)} = e^{ln(u^2 + 1)^2}

     = (u^2 + 1)^2

    v \cdot Int \, Factor \, = \int Q \cdot Int \, Factor \, du

    \therefore \, \, v(u^2 + 1)^2 = \int \frac{3u}{u^2 + 1} \cdot (u^2 + 1)^2 du

    v(u^2 + 1)^2 = \int 3u(u^2 + 1) du

    v(u^2 + 1)^2 = \int (3u^3 + 3u) du

    = \frac{3u^4}{4} + \frac{3u^2}{2} + C

    = \frac{3u^4 + 6u^2}{4} + C

     = \frac{3u^2(u^2 + 2)}{4} + C

    v = \frac{3u^2(u^2 + 2)}{4(u^2 +1 )^2} + \frac{C}{(u^2 + 1) ^2}

    v = \frac{3u^2(u^2 + 2) + 4C}{4(u^2 +1 )^2} \, \,ans
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  2. #2
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    Quote Originally Posted by ashura View Post

    v = \frac{3u^2(u^2 + 2) + 4C}{4(u^2 +1 )^2} \, \,ans
    Good job!

    Just a comment.
    Instead of,
    4C you can write C.
    Because it is a konstant also.
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