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Math Help - solution of simple partial differential equation

  1. #1
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    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)?
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  2. #2
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    Re: solution of simple partial differential equation

    Quote Originally Posted by Suvadip View Post
    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)
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