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]

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

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

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.

 

[HallsofIvy]

What, you mean I'm NOT a genius?

 

[Mathwebster]

What, you mean I'm NOT a genius?

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