91Ó°ÊÓ

Given$u=f(x-ct)+g(x+ct)$

Partial differentiation is defined as the process, in which find the partial derivative of a function.

In Partial differentiation, the function has more than one variable and find the partial derivative of a function with respect to one variable and keeping the other variable constant.

$\frac{{∂}^{2}u}{∂x^2}=\frac{1}{c^2}\frac{{∂}^{2}u}{∂t^2}$Find the partial differentiation of function,$u=f(x-ct)+g(x+ct)$ with respect to x.

$\frac{∂u}{∂x}=\frac{∂f(x-ct)}{∂x}+\frac{∂g(x+ct)}{∂x}.........\left(1\right)$

Find the partial differentiation of equation (1) with respect to x.

$\frac{{∂}^{2}u}{∂x^2}=\frac{{∂}^{2}f(x-ct)}{∂x^2}+\frac{{∂}^{2}g(x+ct)}{∂x^2}.........\left(2\right)$

Find the partial differentiation of function,$u=f(x-ct)+g(x+ct)$ with respect to t.

$\frac{∂u}{∂t}=-c\frac{∂f(x-ct)}{∂t}+c\frac{∂g(x+ct)}{∂t}.........\left(3\right)$

Find the partial differentiation of equation (3) with respect to x.

$$\begin{gathered} \frac{{∂}^{2}u}{∂t^2}=c^2\frac{∂f(x-ct)}{∂t}+c^2\frac{∂g(x+ct)}{∂t} \\ =c^2\left[\frac{{∂}^{2}f(x-ct)}{∂t^2}+\frac{{∂}^{2}g(x+ct)}{∂t^2}\right].........\left(4\right) \end{gathered}$$

From equation (2),

localid="1659151087874" $\frac{{∂}^{2}u}{∂t^2}=c^2\frac{{∂}^{2}u}{∂x^2}$

Or it can be simplified as:

$\frac{{∂}^{2}u}{∂x^2}=\frac{1}{c^2}\frac{{∂}^{2}u}{∂t^2}$

Hence it is proved.