PDE- method of characteristics, satisfy the given condition How to solve this equation? Should I use method of characteristics? 
Question states: find the solution that satisfies this condition:
\begin{aligned}
xu_{x}-yu_y+u &= x\\
u&= x^2 \ when \ y=x
\end{aligned}
I just plugged "u" and solved accordingly, at the end I got x=0 and x=1/3.
 A: $$xu_x-yu_y=x-u$$
Charpit-Lagrange system of characteristic OEDs :
$$\frac{dx}{x}=\frac{dy}{-y}=\frac{du}{x-u}$$
First characteristic equation from $\frac{dx}{x}=\frac{dy}{-y}$ :
$$xy=c_1$$
Second characteristic equation from $\frac{dx}{x}=\frac{du}{x-u}$ :
$$xu-\frac12 x^2=c_2$$
General solution of the PDE on the form of implicit equation :
$$xu-\frac12 x^2=F(xy)$$
where $F$ is an arbitrary function, to be determined according to the boundary condition.
$$u(x,y)=\frac12 x+\frac{1}{x} F(xy)$$
Condition : $u(x,x)=x^2=\frac12 x+\frac{1}{x} F(x^2)$
$$F(x^2)=x^3-\frac12 x^2$$
This determines the function $F$ :
$$F(X)=X^{3/2}-\frac12 X$$
We put this function into the general solution where $X=xy$ :
$$u(x,y)=\frac12 x+\frac{1}{x}\left((xy)^{3/2}-\frac12 (xy) \right)$$
$$u(x,y)=\frac{x-y}{2}+x^{1/2}y^{3/2}$$
A: Solving the PDE we get
$$
u(x,y) = \frac{x^2+2\phi(x y)}{2x}
$$
and the condition
$$
u(x,x) = \frac{x^2+2\phi(x^2)}{2x} = x^2
$$
gives
$$
\phi(x^2) = x^3-\frac 12 x^2 = x^2\left(\sqrt{x^2}-\frac 12\right)
$$
and finally
$$
\phi(xy) = x y\left(\sqrt{x y}-\frac 12\right)
$$
