The function $u(x, y)$ satisfies the partial differential equation

$$a \frac{\partial^{2} u}{\partial x^{2}}+b \frac{\partial^{2} u}{\partial x \partial y}+c \frac{\partial^{2} u}{\partial y^{2}}=0$$

where $a, b$ and $c$ are non-zero constants.

Defining the variables $\xi=\alpha x+y$ and $\eta=\beta x+y$, where $\alpha$ and $\beta$ are constants, and writing $v(\xi, \eta)=u(x, y)$ show that

$$a \frac{\partial^{2} u}{\partial x^{2}}+b \frac{\partial^{2} u}{\partial x \partial y}+c \frac{\partial^{2} u}{\partial y^{2}}=A(\alpha, \beta) \frac{\partial^{2} v}{\partial \xi^{2}}+B(\alpha, \beta) \frac{\partial^{2} v}{\partial \xi \partial \eta}+C(\alpha, \beta) \frac{\partial^{2} v}{\partial \eta^{2}},$$

where you should determine the functions $A(\alpha, \beta), B(\alpha, \beta)$ and $C(\alpha, \beta)$.

If the quadratic $a s^{2}+b s+c=0$ has distinct real roots then show that $\alpha$ and $\beta$ can be chosen such that $A(\alpha, \beta)=C(\alpha, \beta)=0$ and $B(\alpha, \beta) \neq 0$.

If the quadratic $a s^{2}+b s+c=0$ has a repeated root then show that $\alpha$ and $\beta$ can be chosen such that $A(\alpha, \beta)=B(\alpha, \beta)=0$ and $C(\alpha, \beta) \neq 0$.

Hence find the general solutions of the equations

$$\frac{\partial^{2} u}{\partial x^{2}}+3 \frac{\partial^{2} u}{\partial x \partial y}+2 \frac{\partial^{2} u}{\partial y^{2}}=0$$

and

$$\frac{\partial^{2} u}{\partial x^{2}}+2 \frac{\partial^{2} u}{\partial x \partial y}+\frac{\partial^{2} u}{\partial y^{2}}=0$$