$$u_x+u_y=2xu $$
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.
Your Cauchy problem has no solution, indeed
if there were a solution $\;u(x,y)\;,\;$ then $\;f(x)=u\left(x,\dfrac{3x}2\right)=1\;$ for all $\;x\in\mathbb{R}\;.$
Consequently,
$f’(x)=u_x+\dfrac32u_y=0\;$ for all $\;x\in\mathbb{R}\;,\;$ that is
$2u_x+3u_y=0\;$ which is a contradiction.
Best Answer
Using the method of characteristics we obtain that $u$ is constant along the parametrised by $s$ curves: $$ (t,x)=\left(\frac{s^2}{2}+c,s\right), $$ and $u$ is of the form $u=f(x^2-2t)$.
This means that the initial data cover only the region: $$ \{(t,x): 0\le t\le x^2/2\}. $$ In other words, the characteristics which start from the $x-$axis never arrive in the region. $$ \{(t,x): t> x^2/2\}. $$ In particular, if $$ f(x)=\left\{ \begin{array}{lll} \cos(x^{1/2}) & \text{if} & x\ge 0, \\ 1+xg(x) & \text{if} & x< 0, \end{array} \right. $$ where $g$ is an arbitrary continuously differentiable function, then $u(t,x)=f(x^2-2t)$ satisfies the given initial value problem.