Computing the flows $\theta, \Psi$ of $X$ and $Y,$ and verifying that the flows do not commute. This is Problem 9-18 from Professor Lee's Intro to Smooth Manifolds (2nd edition). I believe I have most of the proof done, however, I am unsure how to finish it from where I currently am at. Can someone please help me? Thank you so much!
Define vector fields $X$ and $Y$ on the plane by $$X = x\frac{\partial}{\partial x} - y\frac{\partial}{\partial y}, \hspace{20 pt} Y = x\frac{\partial}{\partial y} + y \frac{\partial}{\partial x}.$$ Compute the flows $\theta, \Psi$ of $X$ and $Y,$ and verify that the flows do not commute by finding explicit open intervals $J$ and $K$ containing $0$ such that $\theta_s \circ \Psi_t$ and $\Psi_t \circ \theta_s$ are both defined for all $(s,t) \in J\times K,$ but they are unequal for some such $(s,t).$
$\textit{Proof.}$ Recall that the flow is a family of integral curves for a specific vector field.
Let $\gamma(t) = (x(t),y(t))$ be the candidate for the flow. We must solve $$X(\gamma(t)) = \gamma'(t).$$ We need to delineate $\gamma'(t).$ It must be an element of the tangent space at $\gamma(t).$ So, now we verify that. $$\gamma'(t) = x'(t)\partial_{x(t)} + y'(t)\partial_{y(t)}.$$ On the other side we have $X(\gamma(t)) = x(t)\partial_{x(t)} -y(t)\partial_{y(t)}.$ Once $X(\gamma(t)) = \gamma'(t)$ it assigns us $$x'(t) = x(t), y'(t) = -y(t).$$ These equations have the solutions $$x(t) = a\cos t = b\sin t, \hspace{20 pt} y(t) = a\sin t + b \cos t,$$ where $a,b$ are arbitrary constants.
 A: You have the right idea by computing the flows, but you did a mistake while solving the differential equation. The solutions of $x' = x$ are $t\mapsto ae^t$, and the solutions of $y'=-y$ are $t\mapsto be^{-t}$.
For the flow of $Y$, the solutions of $x'= y$ and $y' = x$ are $x(t) = x_0\cosh t + y_0 \sinh t$ and $y(t) = y_0\cosh t + x_0 \sinh t$.
Thus the flows of $X$ and $Y$ are
\begin{align}
\varphi_X(t,(x_0,y_0)) &= (x_0e^t,y_0e^{-t}) \\ \varphi_Y(s,(x_0,y_0)) &= (x_0\cosh s + y_0 \sinh s,y_0\cosh s + x_0 \sinh s)
\end{align}
Try now to find $t$ and $s$, and $x_0$ and $y_0$ such that $\varphi_X\left(t,\varphi_Y(s,(x_0,y_0)) \right)\neq \varphi_Y\left(s,\varphi_X(t,(x_0,y_0))\right)$.
I would add two things to help you understand your mistakes:

*

*when you solved the differential equations, you had two parameters for each solution, that is uncoherent with the fact that you tried to solve a first order linear equation

*if you know the formulation of Lie brackets, the flows of two vector fields $X$ and $Y$ commute if and only if $[X,Y]=0$. Consequently, if you compute $[X,Y]$, and it is non zero, you know that the flows do not commute. But you know more than that: if you identify the non-zero component of $[X,Y]$, you can identify how they do not commute, and that can help you find $s,t,x_0$ and $y_0$
Edit
Here is a precise computation.
\begin{align}
\varphi_X\left(t,\varphi_Y(s,(x_0,y_0)) \right) &= \varphi_X\left(t,(x_0\cosh s + y_0 \sinh s,y_0\cosh s + x_0 \sinh s)\right)\\ 
&=\left( x_0(\cosh s) e^{t} + y_0(\sinh s) e^t,y_0(\cosh s) e^{-t} + x_0(\sinh s)e^{-t}\right) \\
\varphi_Y(s,\varphi_X(t,x_0,y_0))) &= \varphi_Y\left(s,(x_0e^{t},y_0e^{-t} )\right) \\
&= \left(x_0e^t(\cosh s) +y_0e^{-t}(\sinh s), y_0e^{-t}(\cosh s) + x_0e^t (\sinh s) \right)
\end{align}
