Compute the flow $\Phi_t$ of $\mathbb{X}(x,y)=(y,x)$ Let $\mathbb{X}$ be the vector field on $\mathbb{R}^2$ given by  $$ \mathbb{X}(x,y) = (y,x). $$ Compute the flow $\Phi_t$ of
$\mathbb{X}$
I was reading through an answer on math.stackexchange entitled "Finding the flow of a pushforward of vector field" and the first part of his problem was this question.
The solution he gave was $\Phi_t(x,y)=(\frac{x+y}{2}e^t+\frac{y-x}{2}e^{-t},\frac{x+y}{2}e^t +\frac{x-y}{2}e^{-t})$
My understanding is that to calculate the flow, you consider: $\Phi_t(x,y)=(\dot{x},\dot{y})=(y,x)$ and try to solve. 
Why cant the flow be $\Phi(x,y)=((y-x)e^{-t},(x-y)e^{-t})$?
It is probably a problem with definitions, but any help would be greatly appreciated.
 A: No, it is not a problem with definition.
Hint. The flow of the linear vector field $Az$ is given by its matrix exponent $e^{tA}z$. In your case
$$
A=\begin{bmatrix}
0&1\\
1&0
\end{bmatrix}
$$
and $z=(x,y)^\top$. Hence you need to calculate
$$
e^{tA}
$$
and see the result.
Addition.
There are different methods to calculate $e^{tA}$, but here I can use simply the defintion
$$
e^{tA}=I+tA+\frac{t^2}{2!}A^2+\ldots
$$
Simply by noting that $A^2=I$ I find
$$
e^{tA}=\begin{bmatrix}
1+\frac{t^2}{2!}+\frac{t^4}{4!}+\ldots& t+\frac{t^3}{3!}+\ldots\\
t+\frac{t^3}{3!}+\ldots&1+\frac{t^2}{2!}+\frac{t^4}{4!}+\ldots
\end{bmatrix}=\begin{bmatrix}
\cosh t&\sinh t\\
\sinh t&\cosh t
\end{bmatrix},
$$
where, as usual,
$$
\cosh t=\frac{e^t+e^{-t}}{2},\quad \sinh t=\frac{e^t-e^{-t}}{2}.
$$
Now you can see you flow by multiplying $e^{tA}$ by the vector of initial conditions.
A: Another way to see this result without needing to use the exponential map is the following:
For $Y = x\frac{\partial}{\partial y} + y\frac{\partial}{ \partial x}$ we obtain the differential equations $x'(t) = y$ and $y'(t) = x$. We note that differentiating either of the equations gives us that $y''(t)=y(t)$ and similarly for $x(t)$. We know that the general solution to each of these equations is $y(t)= d_1e^t +d_2e^{-t}$ and $x(t) = c_1e^t +c_2e^{-t}$. Now our global flow must satisfy $x(0)=y$ and $y(0)=x$, and thus we impose the conditions that $c_1 +c_2 =y$ and $d_1+d_2 =x$. Now given that $x'(t) = y(t)$ we have the following relationship between coefficients: $c_1=d_1$ and $c_2 =-d_2$. Plugging these in to the system of equations above and solving for $c_1$ and $c_2$ we obtain that $c_1 = \frac{x+y}{2}$, and $c_2 = \frac{y-x}{2}$. Thus our flow becomes $\theta_t(x, y) = (\frac{x+y}{2}e^t + \frac{y-x}{2}e^{-t}, \frac{x+y}{2}e^t-\frac{y-x}{2}e^{-t})$.
