Solve first-order pde $x^2u_x + uu_y = 1$ with $u(x,1-x) = 0$ I am solving a first order pde:
$$x^2u_x + uu_y = 1 \,\,,\,\,\, u = u(x,y) $$
$$u(x,1-x) = 0$$
After the calculations I end up with the following expression:
$$y-\frac{u^2}{2} = 1 - \frac{1}{u+\frac{1}{x}} \,\,\,\,\,(1)$$
In order to finish the question I have to solve for $u(x,y)$ in (1) but I can't find a way to do so. 
Any help is appreciated.
 A: $$x^2u_x+uu_y=1$$
System of characteristic ODEs :
$$\frac{dx}{x^2}=\frac{dy}{u}=\frac{du}{1}$$
A first characteristic equation comes from $\frac{dx}{x^2}=\frac{du}{1}$
$$u+\frac{1}{x}=c_1$$
A second characteristic equation comes from $\frac{dy}{u}=\frac{du}{1}$
$$u^2-2y=c_2$$
General solution of the PDE on implicit form $c_2=F(c_1)$ :
$$u^2-2y=F(u+\frac{1}{x})$$
$F$ is an arbitrary function (to be determined according to the bondary condition).
Condition : $u(x,1-x)=0$
$$-2(1-x)=F(\frac{1}{x})$$
Let $X=\frac{1}{x}\quad;\quad x=\frac{1}{X}$
$$F(X)=-2+2x=-2+\frac{2}{X}$$
Now the function $F=-2+\frac{2}{X}$ is determined. We put it into the above general solution where $X=u+\frac{1}{x}\quad\implies\quad F(X)=-2+\frac{2}{u+\frac{1}{x}}$
$$u^2-2y=-2+\frac{2}{u+\frac{1}{x}}$$
The particular solution of the PDE which satisfies the boundary condition is found on the form of the implicit equation :
$$u^2-2y+2-\frac{2}{u+\frac{1}{x}}=0$$
This is a cubic equation which could be solved for $u(x,y)$. It is simpler to accept the solution on this implicit form.
