First order linear partial differential equation


by coverband
Tags: differential, equation, linear, order, partial
coverband
coverband is offline
#1
Feb26-09, 05:38 PM
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.......
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HallsofIvy
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#2
Feb26-09, 09:16 PM
Math
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PF Gold
P: 38,877
Neither of those is the "general" solution. z= f(x-y)ex+ g(x-y)ey is the general solution.
coverband
coverband is offline
#3
Feb27-09, 05:02 AM
P: 167
You are quite the genius! Thanks

Mathwebster
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#4
Apr2-09, 06:20 PM
P: 4

First order linear partial differential equation


Quote Quote by HallsofIvy View Post
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 Quote by coverband View Post
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 [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
HallsofIvy is offline
#5
Apr3-09, 07:56 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877
What, you mean I'm NOT a genius?
Mathwebster
Mathwebster is offline
#6
Apr3-09, 07:59 AM
P: 4
Quote Quote by HallsofIvy View Post
What, you mean I'm NOT a genius?
I've never met you so I really don't know


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