Trouble Solving a system of linear ODEs I am trying to solve:
$x' = 9y$; $x(0) = -2$
$y' = -16x$; $y(0) = -4$.
I am unsure how to even begin. Any help is appreciated.
 A: from the second equation
$$x=-\frac{y'}{16}$$
$$x'=-\frac{y''}{16}$$
substitute in the first equation to get
$$-\frac{y''}{16}=9y$$
$$y''+144y=0$$
this homogeneous D.E so 
the general solution is
$$y=C_1\cos 12t+C_2\sin 12t$$
at $y(0)=-4$
$$-4=C_1(1)$$
$$C_1=-4$$
$$y=-4\cos 12t+C_2\sin 12t$$
now derive it
$$y'=48\sin 12t+12C_2\cos 12t$$
but 
$$x=-\frac{y'}{16}$$
so
$$x=-\frac{1}{16}(48\sin 12t+12C_2\cos 12t)$$
use the $x(0)=-2$ to find the $C_2$
$$-2=-\frac{1}{16}(0+12C_2(1))$$
A: One way would be to solve these functions as a matrix of linear differential equations. However, given the simple nature of this one, we can solve it better by substitution. 
Substituting $y = x'/9$ into the second equation, we have $x''/9 = -16x \rightarrow x'' = -144x$. Do you know how to solve this equation? 
A: $\textbf{hint}$
Taking the derivative of one of the equations we find
$$
x'' = 9y' = 9(-16x) = - (3\cdot 4)^2 x
$$
Solve for $x$ and then plug into the remaining ode and solve for $y$ .
Or you can go with the full matrix method.
