Consider the linear first order non-homogeneous partial differential equation U x

Holetaug 2022-07-01 Answered
Consider the linear first order non-homogeneous partial differential equation
U x + y U y y = y e x
By using the method of characteristics show that its general solution is given by
u ( x , y ) = y e x + e x g ( y e x )
where g is any differentiable function of its argument
A = 1 , B = y , d = 1 , f = y e x d y d x = b a = y d y = y d x 1 y d y = d x ln y = x + c η ( x , y ) = c let  ξ = x
Then
η = ln y x
ξ = x
those are my characteristics curves,
when I put it into [ a ξ x + b ξ y ] ω ξ + d ω = F
I get the equation ω ξ ω = e η
Is this all correct?
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Answers (1)

Camron Herrera
Answered 2022-07-02 Author has 16 answers
U x + y U y y = y e x
U x + y U y = y ( e x + 1 )
d x d t = 1, letting x(0)=0 , we have x=t
d y d t = y, letting y ( 0 ) = y 0 , we have y = y 0 e t = y 0 e x
d U d t = y ( e x + 1 ) = y 0 ( e t + 1 ), we have U ( x , y ) = y 0 ( e t + t ) + g ( y 0 ) = y ( x e x + 1 ) + g ( y e x )

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