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
Take the partial derivative of $u$ with respect to $t$.
$u_{t}(x,t)=f'(x+at)[\dfrac{\partial }{\partial t}(x+at)]+g'(x-at)[\dfrac{\partial }{\partial t}(x-at)]$
or, $=af'(x+at)-ag'(x-at)$
$u_{tt}(x,t)=a^2f''(x+at)+a^2g''(x-at)$
Take the partial derivative of $u$ with respect to $x$.
$u_{x}(x,t)=f'(x+at)+g'(x-at)$
and
Take the second derivative of $u$ with respect to $x$.
$u_{xx}(x,t)=f''(x+at)+g''(x-at)$
Hence, it has been proved that the function $u_{x}=f(x+at)+g(x-at)$ is a solution of the equation $u_{tt}(x,t)=a^2 u_{xx}(x,t)$
