I want to calculate the local flow of the following smooth vector field on $\mathbb{R}^2$, $X:\mathbb{R}^2\to T\mathbb{R}^2$, defined by $X=(x^2-x^1)\frac{\partial}{\partial x^1}-(x^1+x^2)\frac{\partial}{\partial x^2}$.
To do this, I have to resolve the following sistem of differential equations:
\begin{cases} \frac{d}{dt}\theta^p(t)^1=\theta^p(t)^2-\theta^p(t)^1 \\ \frac{d}{dt}\theta^p(t)^2=-(\theta^p(t)^1+\theta^p(t)^2)\\ \theta^p(0)=p \end{cases}
for all $p\in\mathbb{R}^2$. (Where $\theta^p(t)^i $ is the $i$-component function of $\theta^p(t)$ )
I know the solution should be $\theta^p(t)=(p^1e^{-t}\cos t+p^2e^{-t}\sin t,-p^1e^{-t}\sin t+p^2e^{-t}\cos t)$ defined for all $t\in \mathbb{R}$ and for all $p\in\mathbb{R}^2$.
Any hint/ suggestion/ strategy/ observation on how to resolve this?
Hint. Your system can be written as: $$\begin{pmatrix}\dot{\theta_p^1}\\\dot{\theta_p^2}\end{pmatrix}=\begin{pmatrix}-1& 1\\-1&-1\end{pmatrix}\begin{pmatrix}\theta_p^1\\\theta_p^2\end{pmatrix},$$ which is a first order linear equation in $\mathbb{R}^2$ and notice that one has: $$\exp\left(t\begin{pmatrix}-1& 1\\-1&-1\end{pmatrix}\right)=\begin{pmatrix}e^{-t}\cos(t)&e^{-t}\sin(t)\\-e^{-t}\sin(t)&e^{-t}\cos(t)\end{pmatrix},$$ whence the result.
Remark. I must explain how I compute this matrix exponential, I diagonalize it over $\mathbb{C}$, finding its eigenvalues are $(-1-i)t$ and $(-1+i)t$ and the change of basis matrix is given by $\begin{pmatrix}i&-i\\1&1\end{pmatrix}$.
Reminder. Let $A$ be a square matrix, then one has: $$\dot{x}=Ax\iff\frac{\mathrm{d}}{\mathrm{d}t}e^{tA}x=0\iff x=x(0)e^{-tA},$$ the point being that the derivative of $t\mapsto e^{tA}$ is $t\mapsto Ae^{tA}$ and that $e^{-tA}e^{tA}=I$.