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Math Help - solving diffyQ for calc2

  1. #1
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    solving diffyQ for calc2

    I am having trouble towards the end of this problem:
    dy/dx=(e^(y)sin^(2)x)/ysecx

    towards the end i get (y^(2)/2e^(y))=sin^(3)x/3

    i am not sure really how to isolate y

    Can anyone please help?
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  2. #2
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    Quote Originally Posted by RhotelidlewildR View Post
    I am having trouble towards the end of this problem:
    dy/dx=(e^(y)sin^(2)x)/ysecx

    towards the end i get (y^(2)/2e^(y))=sin^(3)x/3

    i am not sure really how to isolate y

    Can anyone please help?
    You can't! If you write it as y^2 e^{-y} you can integrate by parts.
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  3. #3
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    my only problem with that is that i already integrated. That is how i ended up in that dilemma
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  4. #4
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    Quote Originally Posted by RhotelidlewildR View Post
    I am having trouble towards the end of this problem:
    dy/dx=(e^(y)sin^(2)x)/ysecx

    towards the end i get (y^(2)/2e^(y))=sin^(3)x/3

    i am not sure really how to isolate y

    Can anyone please help?
    your solution is incorrect ...

    \frac{dy}{dx} = \frac{e^y \cdot \sin^2{x}}{y \cdot \sec{x}}

    \int y \cdot e^{-y} \, dy = \sin^2{x} \cdot \cos{x} \, dx

    using parts on the left side ...

    -(y+1)e^{-y} = \frac{\sin^3{x}}{3} + C

    you're stuck here as far as solving for y is concerned ... unless you are skilled in using the Lambert W function.
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