Answer
The function $u_{x}=f(x+at)+g(x-at)$ is a solution of the equation $u_{tt}(x,t)=a^2 u_{xx}(x,t)$
Work Step by Step
Need to take the partial derivative of $u$ with respect to $t$.Thus, we have
$u_{t}(x,t)=f'(x+at)[\dfrac{\partial }{\partial t}(x+at)]+g'(x-at)[\dfrac{\partial }{\partial t}(x-at)]=af'(x+at)-ag'(x-at)$
and $u_{tt}(x,t)=a^2 \times f''(x+at)+a^2 \times g''(x-at)$
Now, take the partial derivative of $u$ with respect to $x$.Thus, we have
$u_{x}(x,t)=f'(x+at)+g'(x-at)$ ...(1)
Need to take the second derivative of $u$ with respect to $x$.Thus, we get
$u_{xx}(x,t)=f''(x+at)+g''(x-at)$ ....(2)
This has been verified that the function $u_{x}=f(x+at)+g(x-at)$ is a solution of the equation $u_{tt}(x,t)=a^2 u_{xx}(x,t)$
