Solving : $2xy \cdot u_x + u_y - u = 0$ Exercise :

Solve the Initial Value Problem (IVP) :
  $$2xy \cdot u_x + u_y - u = 0$$
  $$u(x,0) = x$$

Attempt :
The functions $a(x,y) = 2xy$, $c(x,y) = 1$, $f(x,y) = 0$ are $C^1$ in an open set including the random point $(x_0,0)$ of the $x-$axis, while the function $\phi(x) = x$ is also $C^1$ in an open interval including the point $x_0$. Thus, via a known lemma, there exists a unique solution for the IVP in a neighborhood of $(x_0,0)$.
Yielding the Lagrange problem :
$$\frac{\mathrm{d}x}{2xy} = \frac{\mathrm{d}y}{1} = \frac{\mathrm{d}u}{u}$$
Question : Moving from here on, how would one yield the results that Wolfram Alpha calculates here ?
 A: You have that $\frac{dx}{x} = 2ydy$. Integrating you get $\ln(x) = y^2 + C$
By integrating $\frac{du}{u} = dy$ we get $\ln(u) = y + f(C)$ and so $\ln(u) = y + f(\ln(x) - y^2)$ is your general solution for some real-valued differentiable function $f$.
To find the particular solution evaluate at $(x,0)$ to get:
$$\ln(x) = f(\ln(x)) \implies f(t) = t$$
Finally the wanted solution is: 
$$\ln(u) = y - y^2 + \ln(x) \implies \boxed{u(x,y) = xe^{y-y^2}}$$

To get the general solution that Wolfram Alpha gives just exponentiate both sides and the set $F = e^f$ in my solution. Also the factor of $\frac 12$ is irrelevant as we take $f$ to be any real-valued differentiable function $f$.
I'm not sure why Wolfram takes $\pm$ in front of $y$. I mean the $-$ sign doesn't give you a solution. Take $F \equiv 1$ then $u=e^{-y}$, assuming $y$ is positive. However then we have:
$$2xyu_x + u_y - u = -e^{-y} - e^{-y} = -2e^{-y} \not = 0$$ 
A: An alternate solution, without use of WA:
From
$$\frac{\mathrm{d}x}{2xy} = \frac{\mathrm{d}y}{1} = \frac{\mathrm{d}u}{u}$$
it can be seen that
$$\frac{dx}{2 x y} = \frac{dy}{1} \hspace{5mm} \to \hspace{5mm} \frac{dx}{x} = 2 y \, dy$$
leads to $\ln x - y^{2} = c_{1}$ and 
$$ \frac{du}{u} = \frac{dy}{1}$$
leads to $\ln u - y = c_{2}$. A solution of the form $ u = e^{y + c_{2}} = e^{y} \, e^{f(\ln x - y^{2})}$ can be obtained. Now using $u(x,0) = x$ provides
$$ x = e^{0} \, e^{f(\ln x - 0^{2})} = e^{f(\ln x)}$$
or $f(t) = t$ and leads to the solution
$$u(x,y) = x \, e^{y - y^{2}}. $$
As a check of the solution:
\begin{align}
u_{x} &= e^{y - y^{2}} = \frac{u}{x} \\
u_{y} &= x \, (1 - 2 y) \, e^{y - y^{2}} = (1 -2 y) \, u
\end{align}
then $2 x y \, u_{x} + u_{y} = 2 y u + (1 - 2 y ) u = u$ as required.
A minor extension can be seen as:
$$u = a x y \, u_{x} + b \, u_{y} \, \hspace{5mm} u(x,0) = x$$
has the solution
$$u(x,y) = x \, e^{-\frac{a}{2 b} \, (y^2 - 2 y/a)}$$
A: $$2xyu_x+u_y=u$$
$\frac{dx}{2xy}=\frac{dy}{1}=\frac{du}{u}$
First characteristics, from $\quad\frac{dx}{2xy}=\frac{dy}{1}$ :
$$xe^{-y^2}=c_1$$
Second characteristics, from $\quad \frac{dy}{1}=\frac{du}{u}$ :
$$ue^{-y}=c_2$$
General solution : $\quad ue^{-y}=F\left(xe^{-y^2}\right)$
$$u(x,y)=e^yF\left(xe^{-y^2}\right)$$
$F$ is an arbitrary function, to be determined according to the boundary condition.
$$ $$
Condition : $\quad u(x,0)=x=e^0F\left(xe^{-0^2}\right)=F(x)$
Now the function is determined : $\quad F(X)=X\quad$ any $X$.
We put it into the above general solution where $X=xe^{-y^2}\quad$ thus $\quad F\left(xe^{-y^2}\right) =xe^{-y^2}$.
The particular solution fitting to the boundary condition is :
$\quad u(x,y)=e^y\left(xe^{-y^2}\right)$
$$u(x,y)=xe^{y-y^2}$$
