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Math Help - DE help

  1. #1
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    DE help

    y' + ytanx = cosx; y(x) = (x+C)cosx; y(pi) = 0

    Please help!
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  2. #2
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    This is first order linear, so use the integrating factor method.

    The integrating factor is e^{\int{\tan{x}\,dx}} = e^{\ln{\sec{x}}} = \sec{x}.

    So multiplying through gives

    \sec{x}\,\frac{dy}{dx} + y\sec{x}\tan{x} = 1

    \frac{d}{dx}(y\sec{x}) = 1

    y\sec{x} = \int{1\,dx}

    y\sec{x} = x + C

    y = (x + C)\cos{x}.


    Now substituting the boundary condition

    0 = (\pi + C)\cos{\pi}

    0 = -\pi - C

    C = -\pi.


    Therefore y = (x - \pi)\cos{x}.
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