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

**dwsmith** · December 25th 2010, 03:25 PM

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.

**zzzoak** · December 25th 2010, 04:02 PM

Let
$ \displaystyle z=x-\frac{3}{4}y $
There is a mistake. The chain rule
$ \displaystyle g_y(z)=g_z \; z_y \; = \; - \; \frac{3}{4} \; g_z. $

Also

$ \displaystyle g_x(z)=g_z \; z_x \; = \; g_z. $

**dwsmith** · December 25th 2010, 04:04 PM

Originally Posted by zzzoak

Let
$ \displaystyle z=x-\frac{3}{4}y $
There is a mistake. The chain rule
$ \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 $\displaystyle \frac{-3}{4}$ from $g_y(x-3/4y)\mbox{?}$

**zzzoak** · December 25th 2010, 04:28 PM

For example consider the function

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

and find

$ g_y \; . $

**dwsmith** · December 25th 2010, 04:31 PM

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?

**zzzoak** · December 25th 2010, 04:37 PM

Please look post #2.
Instead of
$ g_x \; and \; g_y $

there must be

$ g_z \; . $

**dwsmith** · December 25th 2010, 04:48 PM

$\displaystyle u_x=e^{y/2}g(z)_x$

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

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

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

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

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

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

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

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

**zzzoak** · December 25th 2010, 05:06 PM

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

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

$ \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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