I have a first order PDE: x u x </msub> + ( x + y )

res2bfitjq

res2bfitjq

Answered question

2022-05-23

I have a first order PDE:
x u x + ( x + y ) u y = 1
With the initial condition:
I have calculated result in Mathematica: u ( x , y ) = y x , but I am trying to solve the equation myself, but I had no luck so far. I tried with method of characteristics, but I could not get the correct results. I would appreciate any help or maybe even whole procedure.

Answer & Explanation

Hailee Henderson

Hailee Henderson

Beginner2022-05-24Added 12 answers

x u x + ( x + y ) u y = 1
u x + ( y x + 1 ) u y = 1 x
d x d t = 1, letting x(1)=1 , we have x=t
d y d t = y x + 1 = y t + 1, letting y ( 1 ) = y 0 , we have
y ( 1 ) = y 0
d u d t = 1 x = 1 t , letting u ( 1 ) = f ( y 0 ), we have
u ( x , y ) = ln t + f ( y 0 ) = ln x + f ( y x ln x )
u ( 1 , y ) = y:
f ( y ) = y
u ( x , y ) = ln x + y x ln x = y x

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