$\displaystyle

\displaystyle (x+u)u_x+yu_y=u+y^2

$

Asked to derive the general solution in the given form which I have done and is as attached...

$\displaystyle

x=y_0^2(s)e^{2t}+te^t (u_0(s)-y_0^2(s))+(x_0(s)-y_0^2(s)e^t

$

$\displaystyle

y=y_0(s)e^t

$

$\displaystyle

u=(u_0(s)-y_0^2(s))e^t+y_0^2(s)e^{2t}

$

The intial data curve is $\displaystyle x=x_0(s), y=y_0(s), u=u_0(s)$

For clarity the last line of attachment is as above which I know is correct...

How do I show that $\displaystyle \displaystyle u(x,y)=\frac{x-y^2}{1+ln(y)} +y^2$ for the special case u=x when y=1?

I believe this requires getting u in terms of x and y but I dont know how to do that with variables $\displaystyle x_0(s), y_0(s), u_0(s)$ in the above equations...