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

• December 25th 2010, 03:25 PM
dwsmith
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.
• December 25th 2010, 04:02 PM
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.
$

Also

$
\displaystyle g_x(z)=g_z \; z_x \; = \; g_z.
$
• December 25th 2010, 04:04 PM
dwsmith
Quote:

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 $g_y(x-3/4y)\mbox{?}$
• December 25th 2010, 04:28 PM
zzzoak
For example consider the function

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

and find

$
g_y \; .
$
• December 25th 2010, 04:31 PM
dwsmith
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?
• December 25th 2010, 04:37 PM
zzzoak
$
g_x \; and \; g_y
$

there must be

$
g_z \; .
$
• December 25th 2010, 04:48 PM
dwsmith
$\displaystyle u_x=e^{y/2}g(z)_x$

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

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

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

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

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

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

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

$\displaystyle g(z)_x=g(z)_y$
• December 25th 2010, 05:06 PM
zzzoak
$
\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 \: .
$