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Thread: 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 $\displaystyle e^{\int{\tan{x}\,dx}} = e^{\ln{\sec{x}}} = \sec{x}$.

    So multiplying through gives

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

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

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

    $\displaystyle y\sec{x} = x + C$

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


    Now substituting the boundary condition

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

    $\displaystyle 0 = -\pi - C$

    $\displaystyle C = -\pi$.


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