## First order linear partial differential equation

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.......
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 Recognitions: Gold Member Science Advisor Staff Emeritus Neither of those is the "general" solution. z= f(x-y)ex+ g(x-y)ey is the general solution.
 You are quite the genius! Thanks

## First order linear partial differential equation

 Quote by HallsofIvy Neither of those is the "general" solution. z= f(x-y)ex+ g(x-y)ey is the general solution.
I disagree - first order PDE's don't have two arbitrary functions in their solutions!

 Quote by 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.......
Actually, they're both right.

First solution $$z = e^xf(x-y)$$ second solution $$z = e^y g(x-y)$$. Since $$f$$ is arbitrary the set $$f(x-y) = e^{-(x-y)} g(x-y)$$ and the first becomes the second.
 Recognitions: Gold Member Science Advisor Staff Emeritus What, you mean I'm NOT a genius?

 Quote by HallsofIvy What, you mean I'm NOT a genius?
I've never met you so I really don't know