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Math Help - [SOLVED] Initial value problem

  1. #1
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    [SOLVED] Initial value problem

    I need some help with this question.

    Show that the equation (1+2xe^{-y})dx-(1+x^2)e^{-y}dy=0 is exact and hence find the solution of the initial value problem dy/dx=\frac{2x+e^y}{1+x^2} on the interval 0<_x<_1


    The first part i have done.
    (1+2xe^{-y})dx-(1+x^2)e^{-y}dy=0
    Use Mdx+Ndy=0

    dM/dy = (1+2xe^{-y})dx
    0+(0.e^{-y})+(2xe^{-y})
    2xe^{-y}

    dN/dx = (1+x^2)e^{-y}dy
    0+(2x.e^{-y})+(x^2.0)
    2xe^{-y}
    therefore the eqn. is exact.

    Its just the next part that i need some help with, a nudge in the right direction would be great!
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  2. #2
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    Quote Originally Posted by sterps View Post
    I need some help with this question.

    Show that the equation (1+2xe^{-y})dx-(1+x^2)e^{-y}dy=0 is exact and hence find the solution of the initial value problem dy/dx=\frac{2x+e^y}{1+x^2} on the interval 0<_x<_1


    The first part i have done.
    (1+2xe^{-y})dx-(1+x^2)e^{-y}dy=0
    Use Mdx+Ndy=0

    dM/dy = (1+2xe^{-y})dx
    0+(0.e^{-y})+(2xe^{-y})
    2xe^{-y}

    dN/dx = (1+x^2)e^{-y}dy
    0+(2x.e^{-y})+(x^2.0)
    2xe^{-y}
    therefore the eqn. is exact.

    Its just the next part that i need some help with, a nudge in the right direction would be great!
    The solution is M(x, y) = C where

    \frac{\partial M}{\partial x} = 1 + 2x e^{-y} .... (1)

    \frac{\partial M}{\partial y} = -(1 + x^2) e^{-y} = -e^{-y} - x^2 e^{-y} .... (2)

    Solve equations (1) and (2) simultaneously for M.
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  3. #3
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    thanks again for your help, i managed to figure out the answer with your help
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