If $\displaystyle u(x,t)$ satisfy the partial differential equation $\displaystyle \frac{\partial^2 u}{\partial t^2}=4\frac{\partial^2 u}{\partial x^2}$, then how
$\displaystyle u(x,t)=f(e^{x-2t})+g(x+2t)$?
If $\displaystyle u(x,t)$ satisfy the partial differential equation $\displaystyle \frac{\partial^2 u}{\partial t^2}=4\frac{\partial^2 u}{\partial x^2}$, then how
$\displaystyle u(x,t)=f(e^{x-2t})+g(x+2t)$?
$\displaystyle \frac{\partial^2 u}{\partial t^2}-4\frac{\partial^2 u}{\partial x^2}=(\frac{\partial u}{\partial t}-2\frac{\partial u}{\partial x})}(\frac{\partial u}{\partial t}+2\frac{\partial u}{\partial x})=0$
$\displaystyle \frac{\partial u}{\partial t}-2\frac{\partial u}{\partial x}=0 $ OR $\displaystyle \frac{\partial u}{\partial t}+2\frac{\partial u}{\partial x}=0$
$\displaystyle u(x,t)=f(x-2t)+g(x+2t)$
which is the same as $\displaystyle u(x,t)=F(e^{x-2t})+g(x+2t) $ because $\displaystyle F(e^{x-2t})=f(x-2t)$