# First order linear partial differential equation

by coverband
Tags: differential, equation, linear, order, partial
 P: 167 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.......
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,338 Neither of those is the "general" solution. z= f(x-y)ex+ g(x-y)ey is the general solution.
 P: 167 You are quite the genius! Thanks
P: 4
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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,338 What, you mean I'm NOT a genius?
P: 4
 Quote by HallsofIvy What, you mean I'm NOT a genius?
I've never met you so I really don't know

 Related Discussions Calculus & Beyond Homework 2 Calculus & Beyond Homework 9 Calculus & Beyond Homework 3 Differential Equations 0 Calculus & Beyond Homework 1