Solving a first order coupled system ODE with 2 second order ODE I was trying to solve a second order PDE and inside of the Lagrange-Charpit system, a first order coupled system popped out.
You can find the problem here: http://www.math.ucla.edu/~yanovsky/handbooks/PDEs.pdf on page 86, 87 if you want to get an idea of the whole problem. Here is the part I am interested in:
$$
\left\{ 
\begin{array}{c}
x'=p_1 \\ 
p_1'=-x
\end{array}
\right. 
$$
This is the coupled system, and they say it resolves to these two second-order equations:
$$x''
+
x
=0
,x
(
s,
0) =
s,
x
'
(
s,
0) =
p
(
s,
0)  =  1
,
\\p
''
+
p
=0
,p
(
s,
0) = 1
,p
'
(
s,
0) =
−
x
(
s,
0) =
−
s
$$
And also they state the solutions to be:
$$x
(
s, t
)=
s
·
cos
t
+sin
t,\\
p
(
s, t
)=cos
t
−
s
·
sin
t.$$
The method I know for solving such system in based on eigenvalues & eigenvectors to build a decoupling matrix. In the book, they use another method, as it is stated - they break the system into 2 second order equations and then solve them. 
Can someone explain in more detail, how the method is applied?
EDIT:
I Tried to solve the equation with the eigenvalue & eigenvector approach, but I got to the result:
$$
\left\{ 
\begin{array}{c}
x(s,t)=C_1cos(t) + iC_2sin(t) \\ 
p_1(s,t)=iC_2cos(t) - C_1sin(t)
\end{array}
\right. 
$$
Which can give me the result, which is stated there exactly if the integration constants had the values: 
$$
\left\{ 
\begin{array}{c}
C_1= s \\ 
C_2= -i
\end{array}
\right. 
$$
Is there anything that suggests these values?
 A: You have a coupled system of first order ODEs
\begin{align}
x' &= p \quad (1) \\
p' &= -x \quad (2)
\end{align}
This can be turned into two second order ODEs by differentiating both $x'$ and $p'$ again i.e
\begin{align}
x'' &= p' \quad \text{from (1)} \\
&= -x \quad \text{from (2)} \\\\
p'' &= -x' \quad \text{from (2)} \\
&= -p \quad \text{from (1)}
\end{align}
The boundary conditions can also be derived using your ODE system and the initial conditions for the PDE problem
\begin{align}
x(s,0) &= s, \quad x'(s,0) = 1 \\
p(s,0) &= 1, \quad p'(s,0) = -s
\end{align}
They are the same ODE, so we will just solve the one in $x$ here. There are two ways to solve this apart from using the matrix formulation. First, the ODE is linear so making an ansatz $x = e^{\lambda t}$ gives us $\lambda^{2} = -1 \implies \lambda = \pm i$, hence
\begin{align}
x &= Ae^{it} + Be^{-it} \\
&= A(\cos(t) + i\sin(t)) + B(\cos(t) - i \sin(t)) \\
&= C_{1} \cos(t) + C_{2} \sin(t)
\end{align}
where $C_{1} = A + B, C_{2} = i(A - B)$. The second way is to multiply the ODE by $x'$ and integrate i.e
\begin{align}
x' x'' &= -x x' \\
\implies \frac{1}{2} x'^{2} &= - \frac{1}{2} x^{2} + k_{1} \\
\implies x' &= \pm \sqrt{k_{1} - x^{2}} \\
\implies \int \frac{dx}{\sqrt{k_{1} - x^{2}}} &= \pm \int dt \\
\implies \arcsin \left( \frac{x}{k_{1}} \right) &= \pm t + k_{2} \\
\implies x &= k_{1} \sin(k_{2} \pm t) \\
&= c_{1} \cos(t) + c_{2} \sin(t)
\end{align}
where $c_{1} = k_{1} \sin(\pm k_{2}), c_{2} = k_{1} \cos(\pm k_{2})$. Now, applying your conditions we find
\begin{align}
x(0) &= s \\
&= C_{1} \cos(0) + C_{2} \sin(0) \\
&= C_{1} \\
x'(0) &= 1 \\
&= -s \sin(0) + C_{2} \cos(0) \\
&= C_{2}
\end{align}
Hence the particular solution is given by $x(s,t) = s \cos(t) + \sin(t)$. The same approach is used to solve the ODE in $p$.
A: As to the added question:
You can determine the constants in the eigenvalue solution simply by evaluating the solution formula at $t=0$
$$
s=x(s,0)=C_1\cos(0)+iC_2\sin(0)=C_1\\
1=p(s,0)=iC_2\cos(0)-C_1\sin(0)=iC_2
$$
giving exactly the same result as the cited solution.
