System of two ODEs with reverse sign Consider the ordinary differential equations $$\dfrac{d}{dt}x_1(t)=x_2(t)$$ $$\dfrac{d}{dt}x_2(t)=-x_1(t)$$
for $t\in \mathbb{R}$. What are the solutions?
We have $\dfrac{d^2}{dt^2}x_1(t)=\dfrac{d}{dt} x_2(t)=-x_1(t)$ and also $\dfrac{d^2}{dt^2}x_2(t)=-\dfrac{d}{dt} x_1(t)=-x_2(t)$. So $x_1(t)$ and $x_2(t)$ satisfy the equation $\dfrac{d^2}{dt^2}u(t)+u(t)=0$. They could be cosine and sine. What else could they be?

Hint:
What if you wrote the system as a $2x2$ matrix and found the eigenvalues?
The system would be:
$$x' = Ax = \begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix}$$
Spoiler - Do Not Peek

 $x_1(t) = c_1 \cos t + c_2 \sin t$, $x_2(t) = -c_1 \sin t + c_2 \cos t$

Of course, we could have also solved the second order system you wrote:
$$u'' + u = 0 \rightarrow m^2 + 1 = 0 \rightarrow m_{1,2} = \pm ~i$$
This gives a solutions with sine and cosine terms.
We could also plot a phase portrait of the system as:

