As in Problem 1, solve

$\mathbf{2}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }\mathbf{x}\mathbf{\partial }\mathbf{y}}\mathbf{-}\mathbf{10}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }{\mathbf{y}}^{\mathbf{2}}}\mathbf{=}\mathbf{0}$

by making the change of variables $\mathit{u}\mathbf{=}\mathbf{5}\mathit{x}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{,}\mathit{v}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{+}\mathit{y}$.

Consider $v=v\left(x,y\right)$ has independent variables and the independent variables such that $x=x\left(s,t\right)$and $y=y\left(s,t\right)$are also the variables of the function.

Consider the equation for the chain rule for the function is:

$$\begin{gathered} \frac{\partial v}{\partial s}=\frac{\partial v}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial s} \\ \frac{\partial v}{\partial t}=\frac{\partial v}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial t} \end{gathered}$$

The given differential equation is:

$2\frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial x\partial y}-10\frac{{\partial }^{2}z}{\partial y^2}=0$

And we have:

$$\begin{gathered} u=5x-2y \\ v=2x+y \end{gathered}$$

From these equations, we get:

$$\begin{gathered} \frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial x}=5\frac{\partial z}{\partial u}+2\frac{\partial z}{\partial v} \\ \frac{\partial }{\partial x}=5\frac{\partial }{\partial u}+2\frac{\partial }{\partial v} \end{gathered}$$

And similarly,

$\frac{\partial }{\partial y}=-2\frac{\partial }{\partial u}+\frac{\partial }{\partial v}$

Now, we have:

$$\begin{gathered} \frac{{\partial }^{2}z}{\partial x^2}=\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial x}\right) \\ =\left(5\frac{\partial }{\partial u}+2\frac{\partial }{\partial v}\right)\left(5\frac{\partial z}{\partial u}+2\frac{\partial z}{\partial v}\right) \\ =25\frac{{\partial }^{2}z}{\partial u^2}+20\frac{{\partial }^{2}z}{\partial u\partial v}+4\frac{{\partial }^{2}z}{\partial v^2} \end{gathered}$$

Similarly,

$$\begin{gathered} \frac{{\partial }^{2}z}{\partial y^2}=\frac{\partial }{\partial y}\left(\frac{\partial z}{\partial y}\right) \\ =\left(-2\frac{\partial }{\partial u}+\frac{\partial }{\partial v}\right)\left(-2\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}\right) \\ =4\frac{{\partial }^{2}z}{\partial u^2}-4\frac{{\partial }^{2}z}{\partial u\partial v}+\frac{{\partial }^{2}z}{\partial v^2} \end{gathered}$$

Also,

$$\begin{gathered} \frac{{\partial }^{2}z}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial y}\right) \\ =\left(5\frac{\partial }{\partial u}+2\frac{\partial }{\partial v}\right)\left(-2\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}\right) \\ =-10\frac{{\partial }^{2}z}{\partial u^2}+\frac{{\partial }^{2}z}{\partial u\partial v}+2\frac{{\partial }^{2}z}{\partial v^2} \end{gathered}$$

Substitute all obtained expression in $2\frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial x\partial y}-10\frac{{\partial }^{2}z}{\partial y^2}=0$, solve as:

$$\begin{gathered} 2\frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial x\partial y}-10\frac{{\partial }^{2}z}{\partial y^2}=0 \\ 2\left(25\frac{{\partial }^{2}z}{\partial u^2}+20\frac{{\partial }^{2}z}{\partial u\partial v}+4\frac{{\partial }^{2}z}{\partial v^2}\right)+\left(-10\frac{{\partial }^{2}z}{\partial u^2}+\frac{{\partial }^{2}z}{\partial u\partial v}+2\frac{{\partial }^{2}z}{\partial v^2}\right)-10\left(4\frac{{\partial }^{2}z}{\partial u^2}-4\frac{{\partial }^{2}z}{\partial u\partial v}+\frac{{\partial }^{2}z}{\partial v^2}\right)=0 \\ \left(50-10-10\right)\frac{{\partial }^{2}z}{\partial u^2}+\left(40+1+40\right)\frac{{\partial }^{2}z}{\partial u\partial v}+\left(8+2-10\right)\frac{{\partial }^{2}z}{\partial v^2}=0 \\ \frac{{\partial }^{2}z}{\partial u\partial v}=0 \end{gathered}$$

Clearly, $\frac{\partial z}{\partial v}=0$ means $\frac{\partial z}{\partial v}$ is independent of u. Thus, the solution is of the form:

$F=f\left(5x-2y\right)+g\left(2x+y\right)$.