First order linear partial differential equation

coverband

Do these equations have two general solutions!?

e.g. z_x + z_y -z = 0

Using the method of characteristics

a=1
b=1
c=-1
d=0

Therefore dx/1=dy/1=dz/z

Taking first two terms: x = y + A
*Taking last two terms: z = Be^y
So general solution is z = f(x-y)e^y

BUT if we took first and last terms: z=Be^x
z=f(x-y)e^x.......

HallsofIvy

Neither of those is the "general" solution. z= f(x-y)e^x+ g(x-y)e^y is the general solution.

coverband

You are quite the genius! Thanks

Mathwebster

I disagree - first order PDE's don't have two arbitrary functions in their solutions!

Actually, they're both right.

First solution [tex]z = e^xf(x-y)[/tex] second solution [tex]z = e^y g(x-y)[/tex]. Since [tex]f[/tex] is arbitrary the set [tex]f(x-y) = e^{-(x-y)} g(x-y)[/tex] and the first becomes the second.

HallsofIvy

What, you mean I'm NOT a genius?

Mathwebster

I've never met you so I really don't know