Question

If $\mathit{z}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathit{f}\left(\frac{y}{x}\right)$, prove that $\mathit{x}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\mathit{y}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{y}}\mathbf{+}\mathit{z}\mathbf{=}\mathbf{0}$.

Short answer

It is proved that the equation$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}+z=0$$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}+z=0$.

Step-by-step solution

Given Information

The given information is$z=\frac{1}{x}f\left(\frac{y}{x}\right)$ .

Find the partial differentiation  ∂z∂xand ∂z∂y .

Calculate the partial differentiation with respect to $x$as:

$$\begin{gathered} \frac{\partial z}{\partial x}=\frac{1}{x}f\text{'}\left(\frac{y}{x}\right)\left(\frac{-y}{x^2}\right)-\frac{1}{x^2}f\left(\frac{y}{x}\right) \\ \frac{\partial z}{\partial x}=\frac{-y}{x^3}f\text{'}\left(\frac{y}{x}\right)-\frac{1}{x^2}f\left(\frac{y}{x}\right) \end{gathered}$$

Now calculate the partial differentiation with respect to$y$ as:

$$\begin{gathered} \frac{\partial z}{\partial y}=\frac{1}{x}f\text{'}\left(\frac{y}{x}\right)\left(\frac{1}{x}\right)+f\left(\frac{y}{x}\right)\left(0\right) \\ \frac{\partial z}{\partial y}=\frac{1}{x^2}f\text{'}\left(\frac{y}{x}\right) \end{gathered}$$

Perform the required multiplication.

Multiply $\frac{\partial z}{\partial x}$by $x$as:

$$\begin{gathered} x\frac{\partial z}{\partial x}=x\left(\frac{-y}{x^3}f\text{'}\left(\frac{y}{x}\right)-\frac{1}{x^2}f\left(\frac{y}{x}\right)\right) \\ x\frac{\partial z}{\partial x}=\frac{-y}{x^2}f\text{'}\left(\frac{y}{x}\right)-\frac{1}{x}f\left(\frac{y}{x}\right) \end{gathered}$$

Now multiply $\frac{\partial z}{\partial y}$by as:

$y\frac{\partial z}{\partial y}=\frac{y}{x^2}f\text{'}\left(\frac{y}{x}\right)$

Now adding $x\frac{\partial z}{\partial x}=\frac{-y}{x^2}f\text{'}\left(\frac{y}{x}\right)-\frac{1}{x}f\left(\frac{y}{x}\right)$and $y\frac{\partial z}{\partial y}=\frac{y}{x^2}f\text{'}\left(\frac{y}{x}\right)$as:

$$\begin{gathered} x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}+z=\frac{-y}{x^2}f\text{'}\left(\frac{y}{x}\right)-\frac{1}{x}f\left(\frac{y}{x}\right)+\frac{y}{x^2}f\text{'}\left(\frac{y}{x}\right)+\frac{1}{x}f\left(\frac{y}{x}\right) \\ x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}+z=\frac{-y}{x^2}f\text{'}\left(\frac{y}{x}\right)+\frac{y}{x^2}f\text{'}\left(\frac{y}{x}\right)-\frac{1}{x}f\left(\frac{y}{x}\right)+\frac{1}{x}f\left(\frac{y}{x}\right) \\ x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}+z=0 \end{gathered}$$

Hence, proved that the equation .$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}+z=0$