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Math Help - Exactness Test of differential equations

  1. #1
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    Exactness Test of differential equations

    I have the problem of (ylny + e^x)dx + (x+e^-y + xlny)dy. I have determined that they are exact. However, I am running into a problem when I am trying to solve the equation. I have done the following and got stuck!

    1. partial F with respect to x = ylny + e^x and
    2. partial F with respect to y = x + e^-y + xlny

    When I integrated #1 with respect to x, I came up with: xylny + e^x + g(y) and then when I integrated #2 with respect to y, I came up with:
    xy - e^-y + xylny -x + h(x). From here I am stuck, therefore I think I made a mistake somewhere and I can't seem to find it. Can somebody help?
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  2. #2
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    (y\ln{y} + e^x)dx + (x+e^{-y} + x\ln{y})dy=0

    Set M(x,y)=y\ln{y} + e^x and N(x,y)=x+e^{-y} + x\ln{y}

    You found an exact ODE, therefore, exists a function f(x,y) such that

    \frac{\partial{f}}{\partial{x}}=y\ln{y} + e^x and \frac{\partial{f}}{\partial{y}}=x+e^{-y} + x\ln{y}

    Integrate the first equation with respect to x, after that find the partial derivative with respect to y.
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  3. #3
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    Hello, graceofayak!

    I have the problem of: . (y\ln y + e^x)dx + (x + e^{-y} + x\ln y)dy \:=\:0

    I have determined that they are exact.
    I have done the following and got stuck:

    (1)\;\frac{\partial f}{\partial x} \:=\:y\ln y + e^x .and . (2)\;\frac{\partial f}{\partial y} \:=\: x + e^{-y} + x\ln y

    When I integrated #1 with respect to x, I came up with: xy\ln y + e^x + g(y) . Yes!

    Then when I integrated #2 with respect to y, I came up with: . xy - e^{-y} + xy\ln y -x + h(x) . no

    Your error is in integrating the third term: . x\int \ln y\,dy

    By parts: . \begin{Bmatrix}u & = & \ln y & \quad & dv & = & dy \\ du & = & \frac{dy}{y} & \quad & v & = & y\end{Bmatrix}

    Then: . x\left(y\ln y - \int dy\right) \;=\;x\left(y\ln y - y\right)\;=\;xy\ln y - xy


    So the integral of #2 is: . xy - e^{-y} + xy\ln y - xy + h(x)

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