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Math Help - How do I solve this linear equation?

  1. #1
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    How do I solve this linear equation?

    How do I solve this linear equation?

    y'+2ytan(x)=sin(x)

    The answer can be found here: y'+2ytan(x)=sin(x) - Wolfram|Alpha

    But I would be really greatful if someone could explain the steps involved.
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by jobloggs View Post
    How do I solve this linear equation?

    y'+2ytan(x)=sin(x)

    The answer can be found here: y'+2ytan(x)=sin(x) - Wolfram|Alpha

    But I would be really greatful if someone could explain the steps involved.

    first thing you need to do is find you integrating factor which will be

    2\tan{x}

    Next

    e^{\int{2\tan{x}}}dx = e^{-2\ln{\cos{x}}} = e^{\ln{\cos^{-2}{x}}} = \cos^{-2}{x} = \sec^2{x}

    so now

    y\sec^2{x} = \int \sin{x}\sec^2{x}dx = \int\sec{x}\tan{x}

    integrate that and you get

    y\sec^2{x} = \sec{x} + C

    divide by \sec^2{x} and get

    y = \cos{x} +C\cos^2{x}
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