Two coupled second order differential equations I have two coupled equations in the form:

$$f''(x) + g'(x) + f(x) = 0$$
$$g''(x) + f'(x) + g(x) = 0$$

Looking at the form, i can guess a relation of the form $g(x) = \lambda f(x)$.
where $\lambda$ is some constant. I can find the constant by replacing $g(x)$ in the above equations and comparing the coefficients of every derivative. Finally i'm left with a single equation which is easily solvable. 

The question is: is this procedure legal? and is the solution that i get is the most general one?

Thanks
Hint

Make a substitute
$$
f' = p \\
g' = q
$$
then you get a system
\begin{align}
f' &amp;= p \\
g' &amp;= q \\
p' &amp;= -f - q \\
q' &amp;= -g - p
\end{align}
or in matrix form
$$
\left[ \begin{array}{c}
f \\ g \\ p \\ q
\end{array}\right ]' = \left [ \begin{array}{cccc}
0 &amp; 0 &amp; 1 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 1 \\
-1 &amp; 0 &amp; 0 &amp; -1 \\
0 &amp; -1 &amp; -1 &amp; 0
\end{array}\right ] \cdot 
\left[ \begin{array}{c}
f \\ g \\ p \\ q
\end{array}\right ]
$$
which is in form of
$$
\mathbf y' = \mathbf A \cdot \mathbf y
$$
Can you take it from here?
Numerical answer :
Given the IC $f(0)$, $f'(0)$, $g(0)$, $g'(0)$ we get the approximations
$$ 
f_n= 2f_{n-1} -f_{n-2} - dx^2\{ g'_{n-1}+f_{n-1} \} ,~~ n=2,3,4...\\
g_n= 2g_{n-1} -g_{n-2} - dx^2\{f'_{n-1}+g_{n-1} \}   ,~~ n=2,3,4...
$$
where
$$
    f'_{n-1} = ( f_{n-1}-f_{n-2} )/dx\\
    g'_{n-1} = ( g_{n-1}-g_{n-2} )/dx      
$$