Is my solution to the following problem correct?
Solve $4u_x+3u_y=0$ subject to $u(0,y)=y^3$
Changing coordinates so that we have: $\displaystyle \frac{\partial x}{\partial \alpha}=4$ and $\displaystyle \frac{\partial y}{\partial \beta}=3$, so take $x=4\alpha$ and $y=3\beta+\gamma$, we then get:
$\displaystyle \frac{\partial u}{\partial \alpha}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \alpha} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial \beta}=0$
so we have that $u=f(\gamma)$ and as $\gamma=y-\frac{3}{4}x$ we get $f(y-\frac{3}{4}\gamma)$, letting $x=0$, $f(y)=y^3$ which gives $u(x,y)=f(y-\frac{3}{4}x)=(y-\frac{3}{4}x)^3$
Thanks very much for any help