How to find other equations of lagrange for the Initial Value Problem

Question

Find the solution of the Initial Value Problem

$(x-y)\dfrac{\partial u}{\partial x}+(y-x-u)\dfrac{\partial u}{\partial y}=u$

where $u(x,0)=1$ .

My try:

$\dfrac{dx}{x-y}=\dfrac{dy}{y-x-u}=\dfrac{du}{u}$

Now $\dfrac{dx+dy}{-u}=\dfrac{du}{u}\implies x+y+u=c_1$

But I am not getting any other conditions to work on. Please help.

Answer 1

Hint: Use multipliers $1,-1,1$ and then equate to the third term.

Edit: Using above you will get, $$\frac{dx-dy+du}{2(x-y+u)}=\frac{du}{u}$$ Now solve, and you will get $\frac{x-y+u}{u^2}=c_2$. Then combine with the one you have already found and use the initial condition.