$$u_x+u_y=2xu $$
The characteristic curves are solution of the differential equations :
$$\frac{dx}{1}=\frac{dy}{1}=\frac{du}{2xu}$$
From $dx=dy$ a first family of characteristic curves is $\quad y-x=c_1$
From $dx=\frac{du}{2xu}$ a second family of characteristic curves is $\quad ue^{-x^2}=c_2$
The general solution of the PDE expressed on the form of implicit equation is :
$$\Phi\left((y-x)\:,\:(ue^{-x^2})\right)=0$$
or, on explicit form :
$$ue^{-x^2}=f(y-x) \quad\to\quad u=e^{x^2}f(y-x)$$
where $f$ is any differentiable function.
With the condition $u(x,x)=e^{x^2}=e^{x^2}f(x-x)=e^{x^2}f(0)\quad\implies\quad f(0)=1$
The solutions are :
$$u(x,y)=e^{x^2}f(y-x)\quad \text{any function }f \text{ having the property }f(0)=1$$
Since they are an infinity of functions which have the property $f(0)=1$, this proves that they are an infinity of solutions for the PDE with condition
$\begin{cases}
u_x+u_y=2xu \\
u(x,x)=e^{x^2}
\end{cases}
$
EXAMPLE of solutions :
With $f(X)=C\quad$ a set of solutions is : $\quad u(x,y)=C\:e^{x^2}$
With $f(X)=CX\quad$ a set of solutions is : $\quad u(x,y)=C\:e^{x^2}(y-x)$
With $f(X)=CX^b\quad$ a set of solutions is : $\quad u(x,y)=C\:e^{x^2}(y-x)^b$
With $f(X)=C\sin(X)\quad$ a set of solutions is : $\quad u(x,y)=C\:e^{x^2}\sin(y-x)$
With $f(X)=Ce^{-bX^2}\quad$ a set of solutions is : $\quad u(x,y)=C\:e^{x^2}e^{-b(y-x)^2}$
An so on ...
One see that they are an infinity of examples, many are easy to find. And all linear combinations of those solutions.
First, we can drop $x_0$ as it doesn't entail the graph of the curves to change, then apply the initial conditions.
$$\quad x(t) = t; \quad y(t) = \frac{1}{2}t^2+y_0; \quad u(t)=t+u_0$$
$$y= \frac{1}{2}x^2+y_0; \quad u=x+u_0$$
Now $u(3,y)=y^2$, so is $3+u_0=\left(\frac{1}{2}3^2+y_0\right)^2$, bringing us the desired relation between $y_0$ and $u_0$... to get rid of them!
$y_0=y-\frac{1}{2}x^2; \quad u_0=\left(\frac{1}{2}3^2+y_0\right)^2-3$
$u=x+\left(\frac{1}{2}3^2-\frac{1}{2}x^2+y\right)^2-3$
Best Answer
We apply the method of characteristics, and we use the notation $u(x,0)=f(x)$ for the initial values. The characteristic curves satisfy
$y'(t) = 1$, letting $y(0)=0$, we know $y(t)=t$.
$x'(t) = a(x(t),y(t))$, letting $x(0)=x_0$, we know $x(t)=x_0+\int_0^t a(x(s),s)\, \text d s$.
$u'(t) = 0$, letting $u(0)=f(x_0)$, we know $u(t)=f(x_0)$.
Therefore, the expression of $u$ relies on the fact that $x$ solves $x'(y) = a(x(y),y)$ with $x(0)=x_0$. This type of differential equation may not have a solution over the entire plane $\Bbb R^2$. Consider for instance $$a(x,y)= x^2 .$$ Then we have $x(y) = x_0/(1-x_0 y)$ and $x_0 = x(y)/(1+x(y) y)$, so that finally, $$ u(x,y) = f\left(\frac{x}{1+xy}\right) . $$ If $f$ is not trivial, then the above expression is valid for $xy>{-1}$.