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Homework Help: Another Differential Equation

  1. Sep 7, 2014 #1
    1. The problem statement, all variables and given/known data

    [itex]y'sin(2t) = 2(y+cos(t))[/itex]

    [itex] y(\frac{∏}{4}) = 0 [/itex]
    2. Relevant equations

    [itex] \frac{dy}{dx} + p(x)y = q(x) [/itex]

    [itex] y = \frac{\int u(x) q(x) dx + C}{u(x)} [/itex]

    [itex] u(x) = exp(\int p(x)dx) [/itex]

    3. The attempt at a solution

    I've set the equation in the form above, simplified the RHS and solved for u(x):

    [itex] \frac{dy}{dt} - \frac{2}{sin(2t)}y = \frac{2cos(t)}{sin(2t)} [/itex]

    RHS simplification:
    [itex]\frac{2cos(t)}{sin(2t)} = \frac{2cos(t)}{2cos(t)sin(t)} = csc(t)[/itex]

    which gives:

    [itex] \frac{dy}{dt} - \frac{2}{sin(2t)}y = csc(t) [/itex]

    [itex] u(x) = exp(\int -\frac{2}{sin(2t)}dt) [/itex]

    u-sub with u = 2t in the integration gives:

    [itex] u(x) = exp(\int -csc(u)du) = exp(ln|cos(2t)|) = |cos(2t)| [/itex]


    [itex] y = \frac{\int |cos(2t)|csc(t)dt}{|cos(2t)|} + C[/itex]

    And I'm stuck here on the indefinite integral in the numerator:

    [itex] \int |cos(2t)|csc(t)dt [/itex]

    I've tried replacing cos(2t) with [itex]cos^2(t) - sin^2(t) [/itex] but that ends up leaving me with [itex]\int cot(t)cos(t)dt - \int sin(t)dt [/itex] which I find just as hard. Any ideas on where I'm going wrong?
    Last edited: Sep 7, 2014
  2. jcsd
  3. Sep 7, 2014 #2


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    That isn't the correct antiderivative for the csc integral.
  4. Sep 7, 2014 #3
    I did a bit of simplification, but I may have done something wrong in between:

    [itex] \int -csc(u)du = -ln|csc(u) - cot(u)| = ln|csc(2t) - cot(2t)| [/itex]

    [itex] =-ln|\frac{csc(2t)}{cot(2t)}| = -ln|\frac{1}{sin(2t)}*\frac{sin(2t)}{cos(2t)}| [/itex]

    [itex] = -ln|sec(2t)| = ln|sec(2t)|^-1 = ln|cos(2t)| [/itex]

    When exponentiating that, I get to [itex] u(x) = |cos(2t)| [/itex]

    Did I do something wrong along the way?
  5. Sep 7, 2014 #4


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    The correct formula for logarithms is ##\ln a -\ln b = \ln \frac a b##.
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