Thread: solution of simple partial differential equation

1. solution of simple partial differential equation

If $u(x,t)$ satisfy the partial differential equation $\frac{\partial^2 u}{\partial t^2}=4\frac{\partial^2 u}{\partial x^2}$, then how
$u(x,t)=f(e^{x-2t})+g(x+2t)$?

2. Re: solution of simple partial differential equation

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