# 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
$u\left(x,y\right)=-y{e}^{-x}+{e}^{x}g\left(y{e}^{-x}\right)$
where g is any differentiable function of its argument

Then
$\eta =\mathrm{ln}y-x$
$\xi =x$
those are my characteristics curves,
when I put it into $\left[a{\xi }_{x}+b{\xi }_{y}\right]{\omega }_{\xi }+d\omega =F$
I get the equation ${\omega }_{\xi }-\omega ={e}^{\eta }$
Is this all correct?
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## Answers (1)

Camron Herrera
Answered 2022-07-02 Author has 16 answers
${U}_{x}+y{U}_{y}-y=y{e}^{-x}$
${U}_{x}+y{U}_{y}=y\left({e}^{-x}+1\right)$
$\frac{dx}{dt}=1$, letting x(0)=0 , we have x=t
$\frac{dy}{dt}=y$, letting $y\left(0\right)={y}_{0}$, we have $y={y}_{0}{e}^{t}={y}_{0}{e}^{x}$
$\frac{dU}{dt}=y\left({e}^{-x}+1\right)={y}_{0}\left({e}^{t}+1\right)$, we have $U\left(x,y\right)={y}_{0}\left({e}^{t}+t\right)+g\left({y}_{0}\right)=y\left(x{e}^{-x}+1\right)+g\left(y{e}^{-x}\right)$

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