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Math Help - solve the differential equation by separation of variables dy/dx = (cos x)e^(y+sinx)

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    solve the differential equation by separation of variables dy/dx = (cos x)e^(y+sinx)

    dy/dx = (cos x)e^(y+sinx)

    i mean i put the y's and x's on the same side but couldn't proceed from it. guidance?
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    e^{y+\sin x}=e^{y}e^{\sin x}.
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    Quote Originally Posted by -DQ- View Post
    dy/dx = (cos x)e^(y+sinx)

    i mean i put the y's and x's on the same side but couldn't proceed from it. guidance?
    \frac{dy}{dx} = \cos{(x)}e^{y + \sin{(x)}}

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

    e^{-y}\,\frac{dy}{dx} = \cos{(x)}e^{\sin{(x)}}

    \int{e^{-y}\,\frac{dy}{dx}\,dx} = \int{\cos{(x)}e^{\sin{(x)}}\,dx}

    \int{e^{-y}\,dy} = \int{e^u\,du} upon making the substitution u = \sin{(x)}.
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    Quote Originally Posted by Prove It View Post
    \frac{dy}{dx} = \cos{(x)}e^{y + \sin{(x)}}

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

    e^{-y}\,\frac{dy}{dx} = \cos{(x)}e^{\sin{(x)}}

    \int{e^{-y}\,\frac{dy}{dx}\,dx} = \int{\cos{(x)}e^{\sin{(x)}}\,dx}

    \int{e^{-y}\,dy} = \int{e^u\,du} upon making the substitution u = \sin{(x)}.
    I did this then i did the integral. i got

    ln(e^y) + C = e^(sinx) + C

    y = e^(sinx) + C [not same C but its still a constant)

    however my answer sheet says im wrong
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  5. #5
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    Quote Originally Posted by -DQ- View Post
    I did this then i did the integral. i got

    ln(e^y) + C = e^(sinx) + C

    y = e^(sinx) + C [not same C but its still a constant)

    however my answer sheet says im wrong
    \int{e^{-y}\,dy} \neq \ln{e^y}.


    Think about \int{e^{ax}\,dx}. What does that equal? What is a in this case?
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