Infinitely many solutions for PDE $u_x+u_y=2xu$ I want to show that the following PDE of the function $u(x,y)$ has infinitely many solutions:
$$
\left\{ 
\begin{array}{c}
u_x+u_y=2xu \\ 
u(x,x)=e^{x^2} \\ 
\end{array}
\right. 
$$
By using the method of characteristics and choosing a curve $\Gamma(r,r,e^{r^2})$ in $u(x(r,s),y(r,s))$, I get the characteristic curve $(x(r,s),y(r,s),z(r,s))=(s+r,s+r,e^{(s+r)^2})$
I notice: $x(r,s)=y(r,s)$ $\implies$ $u(x,y)=z(r(x,y),s(x,y))=e^{x^2}=e^{y^2}$. However, this solution that I find is unique. I seem to be missing something to show that there are infinite solutions. Any tips?

Consider the Ansatz $u(x,y) = \exp\left(ax^2 + bxy + cy^2\right)$, then $a,b,c$ must satisfy
\begin{align}
u_x + u_y &= \left( 2ax + by \right) \exp\left( a^2 + bxy + cy^2 \right) + \left( bx + 2cy \right) \exp\left( ax^2 + bxy + cy^2 \right) \\
&=\left( 2a + b \right)x u(x,y) + \left( 2c +b \right) y u(x,y)
\end{align}
Thus $2a + b = 2$ and $2c + b = 0$, so $a = 1 - \frac{b}{2}$ and $  c =- \frac{b}{2}$.  
This also gives $a + b + c = 1 - \frac{b}{2} + b - \frac{b}{2} = 1$, or
\begin{align}
u(x,x) = \exp\left(x^2 \right)
\end{align}
So for all $b \in \mathbb{R}$ another solution is
\begin{align}
u(x,y) = \exp\left(x^2 \right) \exp\left(- \frac{b}{2}x^2 + b xy - \frac{b}{2} y^2\right) = \exp\left(x^2 \right) \exp\left(- \frac{b}{2} \left(x - y \right)^2\right)
\end{align}
