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Thread: help with integrals needed

  1. #1
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    help with integrals needed

    I have to solve the following:

    $\displaystyle
    dy=\frac{du}{x-u}$, where $\displaystyle u=u(x, y)$.

    How do I solve this?
    Thank you so much for all your help!
    Last edited by georgel; Sep 6th 2008 at 11:08 AM.
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  2. #2
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    Quote Originally Posted by georgel View Post
    I have to solve the following:

    $\displaystyle
    dy=\frac{du}{x-u}$, where $\displaystyle u=u(x, y)$.

    How do I solve this?
    Thank you so much for all your help!
    Is this exactly as the original question is stated?
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Is this exactly as the original question is stated?
    No. I have to solve a quasilinear equation:

    $\displaystyle x u_x + y u_y= xy-yu$

    I'm trying to find the general soultion so I do the standard procedure from my textbook:

    $\displaystyle \frac{dx}{x}=\frac{dy}{y}=\frac{du}{y(x-u)}$


    $\displaystyle \frac{dx}{x}=\frac{dy}{y}$ gives me
    $\displaystyle ln x=ln y + ln c$, $\displaystyle \phi (x, y)=c=\frac{x}{y}$

    And now I try to do the same with $\displaystyle \frac{dy}{y}=\frac{du}{y(x-u)}$ to get $\displaystyle \psi(x, y, u)$, but don't know how.
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  4. #4
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    I've had a similar problem:

    $\displaystyle \frac{dy}{y+u}=\frac{du}{y-u}$, and this is what I did:

    $\displaystyle ydy-udy=ydu+udu$
    $\displaystyle (y-u)dy-(y-u)du=2udu$
    Then I substituted $\displaystyle z=y-u$ and got $\displaystyle zdz=2udu$.

    But I don't know what to do with the problem I stated above..
    Please, could anyone help?

    I'm stuck and unfortunately, I'm running out of time to solve it.
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