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Math Help - Solve the Differential Equation

  1. #1
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    Solve the Differential Equation

    sec(x) dy/dx = e^(y + sin(x))

    So far I have :

    ln (secx) dy/dx = y + sin x

    ln (secx) dy = y + sinx dx

    Not sure what to do now... I don't think I can just subtract the y?
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  2. #2
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    you should be able to solve this using implicit differentiation.

    so you have \frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} = -\frac{F_x}{F_y}

    so then

    \sec(x) dy/dx = e^{y + \sin(x)}

    = \frac{1}{\cos(x)} \frac{dy}{dx} = e^{y + \sin(x)}

    \frac{dy}{dx}= e^{y + \sin(x)} \times \cos(x)

    \frac{\partial F}{\partial y} = e^{y + \sin(x)} \times \cos(x)

    \frac{\partial F}{\partial x} = \cos(x) \times e^{y + \sin(x)} \times \cos(x) -sin(x) \times e^{y + \sin(x)} = e^{y + \sin(x)}(\cos^2(x) -\sin(x))

    now just combine both function...
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by veronicak5678 View Post
    sec(x) dy/dx = e^(y + sin(x))

    So far I have :

    ln (secx) dy/dx = y + sin x

    ln (secx) dy = y + sinx dx

    Not sure what to do now... I don't think I can just subtract the y?
    Note that \sec(x)\frac{\,dy}{\,dx}=e^{y+\sin x}\implies\sec(x)\frac{\,dy}{\,dx}=e^{y}e^{\sin x}\implies e^{-y}\,dy=\frac{e^{\sin x}}{\sec x}\,dx \implies e^{-y}\,dy=e^{\sin x}\cos x \,dx

    Can you take it from here?

    --Chris
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  4. #4
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    I see. Thanks!
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by veronicak5678 View Post
    I see. Thanks!
    e^{-y}, because in simplifying, I had \frac{1}{e^y}\,dy=e^{\sin x}\cos x\,dx, but \frac{1}{e^y}=e^{-y}

    Does this clarify things?

    --Chris
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