Question

In the partial differential equation

$\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\mathbf{5}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }\mathbf{x}\mathbf{\partial }\mathbf{y}}\mathbf{+}\mathbf{6}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }{\mathbf{y}}^{\mathbf{2}}}\mathbf{=}\mathbf{0}$

put $\mathit{s}\mathbf{=}\mathit{y}\mathbf{+}\mathbf{2}\mathit{x}\mathbf{,}\mathit{t}\mathbf{=}\mathit{y}\mathbf{+}\mathbf{3}\mathit{x}$ and show that the equation becomes role="math" localid="1665051128513" $\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }\mathbf{s}\mathbf{\partial }\mathbf{t}}\mathbf{=}\mathbf{0}$. Following the method of solving (11.6), solve the equation.

Short answer

Hence, $\frac{{\partial }^{2}z}{\partial s\partial t}=0$ and the solution is of the form: $F=f\left(y+2x\right)+g\left(y+3x\right)$ .

Step-by-step solution

Determine the Chain Rule

For function $v=v\left(x,y\right)$has independent variables that are also functions of some independent variables $x=x\left(s,t\right)$and $y=y\left(s,t\right)$, the chain rule for the function is represented as follows:

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

Find the differentials

The given differential equation is:

$\frac{{\partial }^{2}z}{\partial x^2}-5\frac{{\partial }^{2}z}{\partial x\partial y}+6\frac{{\partial }^{2}z}{\partial y^2}=0$

Consider the equations:

$$\begin{gathered} s=y+2x \\ t=y+3x \end{gathered}$$

From this equation, solve as:

$$\begin{gathered} \frac{\partial z}{\partial x}=\frac{\partial z}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial z}{\partial t}\frac{\partial t}{\partial x} \\ \frac{\partial z}{\partial x}=2\frac{\partial z}{\partial s}+3\frac{\partial z}{\partial t} \\ \frac{\partial }{\partial x}=2\frac{\partial }{\partial s}+3\frac{\partial }{\partial t} \end{gathered}$$

And similarly,

$\frac{\partial }{\partial y}=\frac{\partial }{\partial s}+\frac{\partial }{\partial t}$

Find the second derivative

Now, we have:

$$\begin{gathered} \frac{{\partial }^{2}z}{\partial x^2}=\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial x}\right) \\ =\left(2\frac{\partial }{\partial s}+3\frac{\partial }{\partial t}\right)\left(2\frac{\partial z}{\partial s}+3\frac{\partial z}{\partial t}\right) \\ =4\frac{{\partial }^{2}z}{\partial s^2}+12\frac{{\partial }^{2}z}{\partial s\partial t}+9\frac{{\partial }^{2}z}{\partial t^2} \end{gathered}$$

Similarly,

$$\begin{gathered} \frac{{\partial }^{2}z}{\partial y^2}=\frac{\partial }{\partial y}\left(\frac{\partial z}{\partial y}\right) \\ =\left(\frac{\partial }{\partial s}+\frac{\partial }{\partial t}\right)\left(\frac{\partial z}{\partial s}+\frac{\partial z}{\partial t}\right) \\ =\frac{{\partial }^{2}z}{\partial s^2}+2\frac{{\partial }^{2}z}{\partial s\partial t}+\frac{{\partial }^{2}z}{\partial t^2} \end{gathered}$$

Also,

$$\begin{gathered} \frac{{\partial }^{2}z}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial y}\right) \\ =\left(2\frac{\partial }{\partial s}+3\frac{\partial }{\partial t}\right)\left(\frac{\partial z}{\partial s}+\frac{\partial z}{\partial t}\right) \\ =2\frac{{\partial }^{2}z}{\partial s^2}+5\frac{{\partial }^{2}z}{\partial s\partial t}+3\frac{{\partial }^{2}z}{\partial t^2} \end{gathered}$$

Solve for the solution of the differential equation as:

Substitute all obtained expression in $\frac{{\partial }^{2}z}{\partial x^2}-5\frac{{\partial }^{2}z}{\partial x\partial y}+6\frac{{\partial }^{2}z}{\partial y^2}=0$and solve as:

$$\begin{gathered} \frac{{\partial }^{2}z}{\partial x^2}-5\frac{{\partial }^{2}z}{\partial x\partial y}+6\frac{{\partial }^{2}z}{\partial y^2}=0 \\ \left(4\frac{{\partial }^{2}z}{\partial s^2}+12\frac{{\partial }^{2}z}{\partial s\partial t}+9\frac{{\partial }^{2}z}{\partial t^2}\right)-5\left(2\frac{{\partial }^{2}z}{\partial s^2}+5\frac{{\partial }^{2}z}{\partial s\partial t}+3\frac{{\partial }^{2}z}{\partial t^2}\right)+6\left(\frac{{\partial }^{2}z}{\partial s^2}+2\frac{{\partial }^{2}z}{\partial s\partial t}+\frac{{\partial }^{2}z}{\partial t^2}\right)=0 \\ \left(4-10+6\right)\frac{{\partial }^{2}z}{\partial s^2}+\left(12-25+12\right)\frac{{\partial }^{2}z}{\partial s\partial t}+\left(9-15+6\right)\frac{{\partial }^{2}z}{\partial t^2}=0 \\ \frac{{\partial }^{2}z}{\partial s\partial t}=0 \end{gathered}$$

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

$F=f\left(y+2x\right)+g\left(y+3x\right)$.