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

    is a solution of

    3u_x+4u_y-2u=1

    Attempt:

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

    \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 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 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  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  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 e^{y/2}\neq 0

    How am I suppose to show this:

    \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; December 25th 2010 at 04:12 PM. Reason: Adjusted derivative
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  2. #2
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    Let
    <br />
\displaystyle z=x-\frac{3}{4}y<br />
    There is a mistake. The chain rule
    <br />
\displaystyle g_y(z)=g_z \; z_y \; = \; - \; \frac{3}{4} \; g_z.<br />

    Also

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

    <br />
g=sin(x-\frac{3}{4}y)<br />

    and find

    <br />
g_y \; .<br />
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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
    <br />
g_x \; and \; g_y<br />

    there must be

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

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

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

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

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

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

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

    <br />
\displaystyle u_x=e^{y/2}g_z z_x=e^{y/2}g_z<br />

    <br />
 \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 \: .<br />
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