Need help with simple system of differential equations thanks to your help I advanced in computing differential equations, but now I encountered another problem I need help with - this time it is a system of differential equations:
$$x_1'=-x_2$$ 
  $$x_2'=x_1$$
I know that the answer should contain trigonometric functions, (sine and cosine) but I have no idea how to start. I tried to divide first equation/second equation and I got something like:
$$\frac{x_1'}{x_2'}=-\frac{x2}{x1}$$
Then I rewrited x1' as $$\frac{dx1}{dt}$$ and did the same with x2. I got rid of dt this way and got a:
$$x_1dx_1=-x_2dx_2$$
Which lead me to result:
$$x_1=\sqrt(const-x_1^2)$$
After inserting x1 to the $$x_2'=x_1$$ equation I got some results, but neither of them contains sine or cosine. Could you point me what am I doing wrong?
 A: One approach (other than just guessing) is to note that
\begin{align*}
x_1'' &= -x_2' \\
&= -x_1,
\end{align*}
so $x_1$ satisfies the ODE
\begin{equation}
x_1'' + x_1 = 0.
\end{equation}
This can be solved using standard methods for linear second order ODEs with constant coefficients.
Another approach, using linear algebra, is to work directly with the first order system
\begin{equation}
x'(t) = 
\underbrace{\begin{bmatrix} 0 & -1 \\1 & 0 \end{bmatrix}}_{A}
x(t). \quad (\spadesuit)
\end{equation}
The eigenvalues of the coefficient matrix $A$ are $\lambda_1 = i, \lambda_2 = -i$.
Finding corresponding eigenvectors $v_1$ and $v_2$ will yield the solutions
\begin{align}
u_1(t) &= e^{\lambda_1 t} v_1, \\
u_2(t) &= e^{\lambda_2 t} v_2.
\end{align}
Every solution to ($\spadesuit$) is a linear combination of $u_1$ and $u_2$:
\begin{equation}
x(t) = c_1 u_1(t) + c_2 u_2(t).
\end{equation}
A: I got this,
y' = -x and x' = y
y = x' so y' = x''
x'' = -x
x = cos(t) ; is one solution, x=Acos(t+k) represents solution set 
but ACos(t+k) = A(ei(t+k) + e-i(t+k))/2
which = 0.5*Aekeit+0.5*Ae-ke-it 
which is the same form as littleO's solution 
