Pde

moonta33

New member
Joined
Dec 13, 2012
Messages
4
  1. look at the partial differential equation
    xux - tut= u

    show that u( x , t) = xf( xt) is a general solution , where f is any differentiable function?
 
Observe that:

\(\displaystyle xu_x(x,t)=x\left(xf'(xt)\cdot t+f(xt) \right)=x^2tf'(xt)+xf(xt)=x^2tf'(xt)+u(x,t)\)

\(\displaystyle tu_t(x,t)=t\left(xf'(xt)\cdot x \right)=x^2tf'(xt)\)

So, substituting into the equation, what do we find?
 
Top