What to do with the $y$ term when solving $yu_{y} + uu_{x} =u-y$

Question

I have an initial value problem that looks like this

$$yu_{y} + uu_{x} =u-y; \qquad u(x,1)=x$$

I think I can just use the method of characteristics to solve it, which I would be perfectly capable of doing, except I am not sure how to treat the $-y$

Perhaps I am missing something obvious, but it has me stuck.

Answer 1

The equation of the characteristics is $$ \frac{dy}{y}=\frac{dx}{u}=\frac{du}{u-y}. $$

Answer 2

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=y$ , letting $x(0)=1$ , we have $y=e^t$

$\dfrac{du}{dt}=u-y=u-e^t$ , we have $u=e^t(u_0-t)=y(u_0-\ln y)$

$\dfrac{dx}{dt}=u=e^t(u_0-t)$ , we have $x=f(u_0)+e^t(u_0-t+1)=f\left(\dfrac{u}{y}+\ln y\right)+u+y$

$u(x,1)=x$ :

$f(x)+x+1=x$

$f(x)=-1$

$\therefore x=u+y-1$

$u(x,y)=x-y+1$