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

$A=1,B=y,d=-1,f=y{e}^{-x}\phantom{\rule{0ex}{0ex}}{\displaystyle \frac{dy}{dx}}={\displaystyle \frac{b}{a}}=y\phantom{\rule{0ex}{0ex}}dy=ydx\phantom{\rule{0ex}{0ex}}\int {\displaystyle \frac{1}{y}}dy=\int dx\phantom{\rule{0ex}{0ex}}\mathrm{ln}y=x+c\phantom{\rule{0ex}{0ex}}\eta (x,y)=c\phantom{\rule{0ex}{0ex}}\text{let}\xi =x$

Then

$\eta =\mathrm{ln}y-x$

$\xi =x$

those are my characteristics curves,

when I put it into $[a{\xi}_{x}+b{\xi}_{y}]{\omega}_{\xi}+d\omega =F$

I get the equation ${\omega}_{\xi}-\omega ={e}^{\eta}$

Is this all correct?

${U}_{x}+y{U}_{y}-y=y{e}^{-x}$

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

$A=1,B=y,d=-1,f=y{e}^{-x}\phantom{\rule{0ex}{0ex}}{\displaystyle \frac{dy}{dx}}={\displaystyle \frac{b}{a}}=y\phantom{\rule{0ex}{0ex}}dy=ydx\phantom{\rule{0ex}{0ex}}\int {\displaystyle \frac{1}{y}}dy=\int dx\phantom{\rule{0ex}{0ex}}\mathrm{ln}y=x+c\phantom{\rule{0ex}{0ex}}\eta (x,y)=c\phantom{\rule{0ex}{0ex}}\text{let}\xi =x$

Then

$\eta =\mathrm{ln}y-x$

$\xi =x$

those are my characteristics curves,

when I put it into $[a{\xi}_{x}+b{\xi}_{y}]{\omega}_{\xi}+d\omega =F$

I get the equation ${\omega}_{\xi}-\omega ={e}^{\eta}$

Is this all correct?