How to find $u(x,y)$? I have the second-order PDE $6u_{xx}+u_{xy}-u_{yy}=0$ .
The change of variables I'm given is $s=x+2y$ and $j=x-y$ .
I used the chain rule and solved $u_{xx}$ and $u_{xy}$ and $u_{yy}$ in terms of $s$ and $t$ , then replaced the original pde, and I got:
$$4u_{ss}+17u_{sj}+4u_{jj}=0$$
Things are supposed to cancel out and I should be left with just one term but they're not and I don't know how to solve for $u(x,y)$ . Please help?
 A: $\because6\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial x\partial y}-\dfrac{\partial^2}{\partial y^2}=\left(3\dfrac{\partial}{\partial x}-\dfrac{\partial}{\partial y}\right)\left(2\dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial y}\right)$
$\therefore$ Let $\begin{cases}\zeta=x+3y\\\eta=x-2y\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial\zeta}\dfrac{\partial\zeta}{\partial x}+\dfrac{\partial u}{\partial\eta}\dfrac{\partial\eta}{\partial x}=\dfrac{\partial u}{\partial\zeta}+\dfrac{\partial u}{\partial\eta}$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial\zeta}+\dfrac{\partial u}{\partial\eta}\right)=\dfrac{\partial}{\partial\zeta}\left(\dfrac{\partial u}{\partial\zeta}+\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\zeta}{\partial x}+\dfrac{\partial}{\partial\eta}\left(\dfrac{\partial u}{\partial\zeta}+\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\eta}{\partial x}=\dfrac{\partial^2u}{\partial\zeta^2}+\dfrac{\partial^2u}{\partial\zeta\partial\eta}+\dfrac{\partial^2u}{\partial\zeta\partial\eta}+\dfrac{\partial^2u}{\partial\eta^2}=\dfrac{\partial^2u}{\partial\zeta^2}+2\dfrac{\partial^2u}{\partial\zeta\partial\eta}+\dfrac{\partial^2u}{\partial\eta^2}$
$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial\zeta}\dfrac{\partial\zeta}{\partial y}+\dfrac{\partial u}{\partial\eta}\dfrac{\partial\eta}{\partial y}=3\dfrac{\partial u}{\partial\zeta}-2\dfrac{\partial u}{\partial\eta}$
$\dfrac{\partial^2u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(3\dfrac{\partial u}{\partial\zeta}-2\dfrac{\partial u}{\partial\eta}\right)=\dfrac{\partial}{\partial\zeta}\left(3\dfrac{\partial u}{\partial\zeta}-2\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\zeta}{\partial y}+\dfrac{\partial}{\partial\eta}\left(3\dfrac{\partial u}{\partial\zeta}-2\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\eta}{\partial y}=3\left(3\dfrac{\partial^2u}{\partial\zeta^2}-2\dfrac{\partial^2u}{\partial\zeta\partial\eta}\right)-2\left(3\dfrac{\partial^2u}{\partial\zeta\partial\eta}-2\dfrac{\partial^2u}{\partial\eta^2}\right)=9\dfrac{\partial^2u}{\partial\zeta^2}-12\dfrac{\partial^2u}{\partial\zeta\partial\eta}+4\dfrac{\partial^2u}{\partial\eta^2}$
$\dfrac{\partial^2u}{\partial x\partial y}=\dfrac{\partial}{\partial x}\left(3\dfrac{\partial u}{\partial\zeta}-2\dfrac{\partial u}{\partial\eta}\right)=\dfrac{\partial}{\partial\zeta}\left(3\dfrac{\partial u}{\partial\zeta}-2\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\zeta}{\partial x}+\dfrac{\partial}{\partial\eta}\left(3\dfrac{\partial u}{\partial\zeta}-2\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\eta}{\partial x}=3\dfrac{\partial^2u}{\partial\zeta^2}-2\dfrac{\partial^2u}{\partial\zeta\partial\eta}+3\dfrac{\partial^2u}{\partial\zeta\partial\eta}-2\dfrac{\partial^2u}{\partial\eta^2}=3\dfrac{\partial^2u}{\partial\zeta^2}+\dfrac{\partial^2u}{\partial\zeta\partial\eta}-2\dfrac{\partial^2u}{\partial\eta^2}$
$\therefore6\left(\dfrac{\partial^2u}{\partial\zeta^2}+2\dfrac{\partial^2u}{\partial\zeta\partial\eta}+\dfrac{\partial^2u}{\partial\eta^2}\right)+3\dfrac{\partial^2u}{\partial\zeta^2}+\dfrac{\partial^2u}{\partial\zeta\partial\eta}-2\dfrac{\partial^2u}{\partial\eta^2}-\left(9\dfrac{\partial^2u}{\partial\zeta^2}-12\dfrac{\partial^2u}{\partial\zeta\partial\eta}+4\dfrac{\partial^2u}{\partial\eta^2}\right)=0$
$25\dfrac{\partial^2u}{\partial\zeta\partial\eta}=0$
$\dfrac{\partial^2u}{\partial\zeta\partial\eta}=0$
$u=F(\zeta)+G(\eta)$
$u=F(x+3y)+G(x-2y)$
