# 2nd order PDE- general solution

• November 9th 2009, 05:17 AM
ArTiCK
2nd order PDE- general solution
Hi all,

I am trying to determine the general solution for the following equation:

http://www.mathhelpforum.com/math-he...1&d=1257775597

First of all i factorised it and determined the 1st order ODEs to be:

http://www.mathhelpforum.com/math-he...1&d=1257775955

For equation (1) there are no characterisitcs. While, for equation (2), i determined the characterisitics to be c_2 = x - t

The answer is.... the general solution,
phi(x,t) = F(t) + G(x -t)

Could some explain to me why there is F(t) in the general solution and why is it a function of t?

ArTiCk
• November 9th 2009, 08:00 AM
Aliquantus
Well, if $\phi(x,t)=F(t)$, then $\frac{\partial \phi}{\partial x}=0$.

$\frac{\partial \phi}{\partial x}$ kills of all functions of t.
• November 9th 2009, 08:02 AM
HallsofIvy
Quote:

Originally Posted by ArTiCK
Hi all,

I am trying to determine the general solution for the following equation:

http://www.mathhelpforum.com/math-he...1&d=1257775597

First of all i factorised it and determined the 1st order ODEs to be:

http://www.mathhelpforum.com/math-he...1&d=1257775955

This is simply wrong. You cannot "factor" derivatives like that.
You can write it as $\frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial x}+ \frac{\partial \phi}{\partial t}\right)= 0$ which says that $\frac{\partial \phi}{\partial x}+ \frac{\partial \phi}{\partial t}$ is independent of x. That is, that $\frac{\partial \phi}{\partial x}+ \frac{\partial \phi}{\partial t}= F(t)$ where F(t) can be any function of t only.

Quote:

For equation (1) there are no characterisitcs. While, for equation (2), i determined the characterisitics to be c_2 = x - t

The answer is.... the general solution,
phi(x,t) = F(t) + G(x -t)

Could some explain to me why there is F(t) in the general solution and why is it a function of t?