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.