Finding values $a$ and $b$ which transforms a differential equation Given the function $u = u(\xi, \eta)$ where $\xi = x + ay$ and $\eta = x + by$, find the values of $a$ and $b$ such that they transform the equation $$\frac{\partial^{2}u}{\partial x^{2
}} + 4\frac{\partial^{2}u}{\partial x\partial y}+3\frac{\partial^{2}u}{\partial y^{2}} = 0$$
into
$$\frac{\partial^{2}u}{\partial\xi\partial\eta} = 0$$

I have managed to get the differential equation down to: $$(1+4a+3a^{2})\frac{\partial^{2}u}{\partial\xi^{2}} + (1 + 4b + 3ab)\frac{\partial^{2}u}{\partial\xi\partial\eta} + (1+4a+3ab)\frac{\partial^{2}u}{\partial\eta\partial\xi}+(1+4b+3b^{2})\frac{\partial^{2}u}{\partial\eta^{2}} = 0$$

Now I am not told about the continuity of this function so I assume I can't use clairaut's theorem hence I must solve the following system:$$\begin{cases} 1 + 4a + 3a^{2} = 0 &amp; (1)\\ 1 + 4b + 3ab = 1 &amp; (2)\\ 1 + 4a + 3ab = 0 &amp; (3) \\ 1 + 4b + 3b^{2} = 0 &amp; (4)\end{cases}$$

From $(1)$: $1 + 4a + 3a^{2} = 0 \iff a = -1 \text{ or } a = -\frac{1}{3}$.

If $a = -1$, $(2) \implies b = 0$ whereas $(3)\implies b = -1$. So clearly I have done something wrong since they do not agree. What have I done wrong?
Perhaps the mistake is,
$$
\frac{\partial^2 u}{\partial \xi \partial \eta}
 = \frac{\partial^2 u}{\partial \eta \partial \xi}
$$
and therefore you only get 3 equations: top and bottom as you wrote, and the thirs equation comes from adding LHS and RHS of (2) and (3) together to form one equation.
Perhaps the mistake is,
$$
\frac{\partial^2 u}{\partial \xi \partial \eta}
 = \frac{\partial^2 u}{\partial \eta \partial \xi}
$$
and therefore you only get 3 equations: top and bottom as you wrote, and the thirs equation comes from adding LHS and RHS of (2) and (3) together to form one equation.