Find a solution of the first order PDE $\pa_tu=y\pa_xu-x\pa_y$ on $M=\R^2$ satisfying $u(0,x,y)=x^2-y^2$.
Put $X=y\pa_x-x\pa_y$, then its flow $\theta(t,x,y)=\colon(x(t),y(t))$ is given by the ODE:
$$
x^\prime(t)=y(t),\quad
y^\prime(t)=-x(t)
\quad\mbox{and}\quad
x(0)=x, y(0)=y~.
$$
Thus $x(t)=x\cos(t)+y\sin(t)$ and $y(t)=-x\sin(t)+y\cos(t)$, i.e.
$$
\theta(t,x,y)=(x\cos(t)+y\sin(t),-x\sin(t)+y\cos(t))
$$
and $u(t,x,y)=P_tf(x,y)=f(\theta(t,x,y))$ for $f=x^2-y^2$, i.e.
\begin{eqnarray*}
u(t,x,y)&=&f(x\cos(t)+y\sin(t),-x\sin(t)+y\cos(t))\\
&=&(x\cos(t)+y\sin(t))^2-(-x\sin(t)+y\cos(t))^2\\
&=&(x^2-y^2)\cos(2t)+2xy\sin(2t)~.
\end{eqnarray*}