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Math Help - Integration of a differential equation

  1. #1
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    Integration of a differential equation

    Hi,

    I am unsure if I am getting the correct answer to my problem, I have to find y

    \frac{dy}{dx} = y\cos(x) which goes to

    \int\frac{1}{\cos(x)\cdot dx} = \int\frac{y}{dy} then

    ln(\cos(x)) + c = y^2 therefore

    y = \sqrt{ln(\cos(x)) + c}

    Is this correct, if not where am i going wrong?

    Thanks
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  2. #2
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    Quote Originally Posted by Beard View Post
    Hi,

    I am unsure if I am getting the correct answer to my problem, I have to find y

    \frac{dy}{dx} = y\cos(x) which goes to

    \int\frac{1}{\cos(x)\cdot dx} = \int\frac{y}{dy} then

    ln(\cos(x)) + c = y^2 therefore

    y = \sqrt{ln(\cos(x)) + c}

    Is this correct, if not where am i going wrong?

    Thanks
    Dear Beard,

    \frac{dy}{dx} = y\cos~x

    \frac{1}{y}\frac{dy}{dx}=cos~x

    Now integrate both sides with respect to x,

    \int{\frac{1}{y}\frac{dy}{dx}}dx=\int{cos~x}~dx

    \int{\frac{1}{y}}dy=sin~x+C ; Cis an arbitary constant.

    ln\mid{y}\mid=sin~x+C

    Hope this will help you.
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  3. #3
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    Nice and clear to follow thankyou
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  4. #4
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    And never, ever again write " \int \frac{f(x)}{dx}"!!!
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