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    Integrating Factor.

    cos x dy/dx + y sin x = sin x cos x

    dy/dx - y/x = (ln x)^4

    How would i solve for find y in these two problems? I'm looking over a similar example which used substitution but I'm confused on these two.
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    Quote Originally Posted by krzyrice View Post
    cos x dy/dx + y sin x = sin x cos x

    dy/dx - y/x = (ln x)^4

    How would i solve for find y in these two problems? I'm looking over a similar example which used substitution but I'm confused on these two.
    \cos{x}\,\frac{dy}{dx} + y\sin{x} = \sin{x}\cos{x}

    \frac{dy}{dx} + y\tan{x} = \sin{x}


    This is now first order linear, so the integrating factor is

    e^{\int{\tan{x}\,dx}} = e^{\ln{\sec{x}}}= \sec{x}.


    Multiplying through by the integrating factor gives:

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

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

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

    y\,\sec{x} = \ln{|\sec{x}|} + C

    y = \cos{x}\ln{|\sec{x}|} + C\cos{x}.
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    Quote Originally Posted by krzyrice View Post
    cos x dy/dx + y sin x = sin x cos x

    dy/dx - y/x = (ln x)^4

    How would i solve for find y in these two problems? I'm looking over a similar example which used substitution but I'm confused on these two.
    \frac{dy}{dx} - \frac{y}{x} = (\ln{x})^4.

    This is first order linear, so the integrating factor is

    e^{-\frac{1}{x}\,dx} = e^{-\ln{x}} = e^{\ln{(x)^{-1}}} = x^{-1}.


    Multiplying through by the integrating factor gives:

    x^{-1}\,\frac{dy}{dx} - x^{-2}y = x^{-1}(\ln{x})^4

    \frac{d}{dx}(x^{-1}y) = x^{-1}(\ln{x})^4

    x^{-1}y = \int{(\ln{x})^4\,x^{-1}\,dx}.


    Now make the substitution u = \ln{x} so that du = x^{-1}\,dx, the equation becomes

    x^{-1}y = \int{u^4\,du}

    x^{-1}y = \frac{u^5}{5} + C

    x^{-1}y = \frac{(\ln{x})^5}{5} + C

    y = \frac{x(\ln{x})^5}{5} + Cx.
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