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Math Help - Separable Differential Equations

  1. #1
    Junior Member
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    Separable Differential Equations

    So I have been getting caught up on two separable differential equations the first being:

    \frac {3du} {dt} = u^2 with initial conditions: u(0) = 4

    I divided by u^2 and multiplied by dt giving:

    \frac {3du} {u^2} = dt and after integrating I got:

    \frac {3} {u} +C = t + C and inversing it, so that the u was in the numerator and multiplying by 3 I ended up with:

    u = \frac {3} {t+c}

    But I am not sure if this is how the problem is really supposed to be done.

    The other equation I am having trouble with is finding the general term for:

    \frac {dR} {dx} = a(R^2+25)

    where a is some nonzero constant. Would this just be 5tan(ax)+C?

    Thanks ahead of time for the help, I really appreciate it.
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  2. #2
    Member mathemagister's Avatar
    Joined
    Feb 2010
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    Quote Originally Posted by Spudwad View Post
    So I have been getting caught up on two separable differential equations the first being:

    \frac {3du} {dt} = u^2 with initial conditions: u(0) = 4

    I divided by u^2 and multiplied by dt giving:

    \frac {3du} {u^2} = dt and after integrating I got:

    \frac {3} {u} +C = t + C and inversing it, so that the u was in the numerator and multiplying by 3 I ended up with:

    u = \frac {3} {t+c}

    But I am not sure if this is how the problem is really supposed to be done.

    The other equation I am having trouble with is finding the general term for:

    \frac {dR} {dx} = a(R^2+25)

    where a is some nonzero constant. Would this just be 5tan(ax)+C?

    Thanks ahead of time for the help, I really appreciate it.
    3\frac{du}{dt} = u^2

    3u^{-2} du = dt

    Integrate:

    -3u^{-1} + K = t + K

    \frac{-3}{u} = t + C

    u = -\frac{3}{t+C}

    u(0) = 4 \implies -\frac{3}{C}=4 \implies C = -\frac{3}{4}

    u(t) = -\frac{3}{t-\frac{3}{4}}
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