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First order linear partial differential equation 
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#1
Feb2609, 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(xy)e^y BUT if we took first and last terms: z=Be^x z=f(xy)e^x....... 


#2
Feb2609, 09:16 PM

Math
Emeritus
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Thanks
PF Gold
P: 39,338

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



#3
Feb2709, 05:02 AM

P: 167

You are quite the genius! Thanks



#4
Apr209, 06:20 PM

P: 4

First order linear partial differential equation
First solution [tex]z = e^xf(xy)[/tex] second solution [tex]z = e^y g(xy)[/tex]. Since [tex]f[/tex] is arbitrary the set [tex]f(xy) = e^{(xy)} g(xy)[/tex] and the first becomes the second. 


#5
Apr309, 07:56 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,338

What, you mean I'm NOT a genius?



#6
Apr309, 07:59 AM

P: 4




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