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Thread: u(x,y) is a solution to 3u_x+4u_y-2u=1

  1. #1
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    u(x,y) is a solution to 3u_x+4u_y-2u=1

    If g is an arbitrary differentiable function of one variable, then

    $\displaystyle \displaystyle u(x,y)=\frac{-1}{2}+e^{y/2}g\left(x-\frac{3}{4}y\right)$

    is a solution of

    $\displaystyle 3u_x+4u_y-2u=1$

    Attempt:

    $\displaystyle \displaystyle u_x=e^{y/2}g_x\left(x-\frac{3}{4}y\right)$

    $\displaystyle \displaystyle u_y=\frac{e^{y/2}}{2}g\left(x-\frac{3}{4}y\right)-\frac{3e^{y/2}}{4}g_y\left(x-\frac{3}{4}y\right)$

    $\displaystyle \displaystyle 3\left[e^{y/2}g_x\left(x-\frac{3}{4}y\right)\right]+4\left[\frac{e^{y/2}}{2}g\left(x-\frac{3}{4}y\right)-\frac{3e^{y/2}}{4}g_y\left(x-\frac{3}{4}y\right)\right]-2\left[\frac{-1}{2}+e^{y/2}g\left(x-\frac{3}{4}y\right)\right]=1$

    $\displaystyle \displaystyle 3e^{y/2}g_x\left(x-\frac{3}{4}y\right)-3e^{y/2}g\left(x-\frac{3}{4}y\right)+4e^{y/2}g_y\left(x-\frac{3}{4}y\right)+1-2e^{y/2}g\left(x-\frac{3}{4}y\right)=1$

    $\displaystyle \displaystyle 3e^{y/2}g_x\left(x-\frac{3}{4}y\right)-3e^{y/2}g_y\left(x-\frac{3}{4}y\right)+1=1$

    $\displaystyle \displaystyle 3e^{y/2}\left[g_x\left(x-\frac{3}{4}y\right)-g_y\left(x-\frac{3}{4}y\right)\right]=0$

    $\displaystyle \displaystyle e^{y/2}\neq 0$

    How am I suppose to show this:

    $\displaystyle \displaystyle g_x\left(x-\frac{3}{4}y\right)-g_y\left(x-\frac{3}{4}y\right)=0$

    Thanks.
    Last edited by dwsmith; Dec 25th 2010 at 04:12 PM. Reason: Adjusted derivative
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  2. #2
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    Let
    $\displaystyle
    \displaystyle z=x-\frac{3}{4}y
    $
    There is a mistake. The chain rule
    $\displaystyle
    \displaystyle g_y(z)=g_z \; z_y \; = \; - \; \frac{3}{4} \; g_z.
    $

    Also

    $\displaystyle
    \displaystyle g_x(z)=g_z \; z_x \; = \; g_z.
    $
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  3. #3
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    Quote Originally Posted by zzzoak View Post
    Let
    $\displaystyle
    \displaystyle z=x-\frac{3}{4}y
    $
    There is a mistake. The chain rule
    $\displaystyle
    \displaystyle g_y(z)=g_z \; z_y \; = \; - \; \frac{3}{4} \; g_z.
    $
    Since g is function of one variable, how can we be sure that the derivative will produce $\displaystyle \displaystyle \frac{-3}{4}$ from $\displaystyle g_y(x-3/4y)\mbox{?}$
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  4. #4
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    For example consider the function

    $\displaystyle
    g=sin(x-\frac{3}{4}y)
    $

    and find

    $\displaystyle
    g_y \; .
    $
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  5. #5
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    I believe you. I was just asking.

    I edited the original post. How can I show g_y=g_x because that is the only way it will be 0?
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  6. #6
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    Please look post #2.
    Instead of
    $\displaystyle
    g_x \; and \; g_y
    $

    there must be

    $\displaystyle
    g_z \; .
    $
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  7. #7
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    $\displaystyle \displaystyle u_x=e^{y/2}g(z)_x$

    $\displaystyle \displaystyle u_y=\frac{e^{y/2}}{2}g(z)-\frac{3e^{y/2}}{4}g(z)_y$

    $\displaystyle \displaystyle 3\left[e^{y/2}g(z)_x\right]+4\left[\frac{e^{y/2}}{2}g(z)-\frac{3e^{y/2}}{4}g(z)_y\right]-2\left[\frac{-1}{2}+e^{y/2}g(z)\right]=1$

    $\displaystyle \displaystyle 3e^{y/2}g(z)_x-3e^{y/2}g(z)+4e^{y/2}g(z)_y+1-2e^{y/2}g(z)=1$

    $\displaystyle \displaystyle 3e^{y/2}g(z)_x-3e^{y/2}g(z)_y+1=1$

    $\displaystyle \displaystyle 3e^{y/2}\left[g(z)_x-g(z)_y\right]=0$

    $\displaystyle \displaystyle e^{y/2}\neq 0$

    $\displaystyle \displaystyle g(z)_x-g(z)_y=0$

    $\displaystyle \displaystyle g(z)_x=g(z)_y$
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  8. #8
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    $\displaystyle
    \displaystyle u=\frac{-1}{2}+e^{y/2}g(z)
    $

    $\displaystyle
    \displaystyle u_x=e^{y/2}g_z z_x=e^{y/2}g_z
    $

    $\displaystyle
    \displaystyle u_y=\frac{e^{y/2}}{2}g(z)+e^{y/2}g_z z_y=\frac{e^{y/2}}{2}g(z)-\frac{3}{4}e^{y/2}g_z \: .
    $
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