<s>[INST] Solution to differential equation system and solution to its conversion into 2nd order differential equation Following is the differential equation system with IVP
$\vec{x}'=\small\begin{pmatrix}3&-9\\4&-3\end{pmatrix}\vec{x},   
\vec{x}(0)=\small\begin{pmatrix}2\\-4\end{pmatrix}$ The particular solution to this differential equation system is given below.
$\vec{x}(t)=\frac23\small\begin{pmatrix}3\cos{(3\sqrt{3}t)}\\\cos{(3\sqrt{3}t)}+\sqrt{3}\sin{(3\sqrt{3}t)}\end{pmatrix}+\frac{14}{3\sqrt{3}}\small\begin{pmatrix}3\sin{(3\sqrt{3}t)}\\\sin{(3\sqrt{3}t)}-\sqrt{3}\cos{(3\sqrt{3}t)}\end{pmatrix}...(1)$
When i converted this differential equation system into 2nd order differential equation, I got $y"+13y'-7y=0, y(0)=2,y'(0)=-4$
Now, the particular solution to this 2nd order equation is $-\frac{2\sqrt{6}}{3}\sin{(\sqrt{6}t)}+2\cos{(\sqrt{6}t)}...(2)$
Now why there is a difference between these two solutions namely (1) and (2)
 [/INST] That is not the second order equation I get. The system is:
$$\left\{
    \begin{matrix}
    \dot x=3x-9y \\
    \dot y=4x-3y \\
    \end{matrix}
\right.$$
From the second equation we get
$$\ddot y=4\dot x-3\dot y\tag 1$$
and
$$4x=\dot y+3y\tag 2$$
Substituting the value for $\dot x$ from the first equation into $(1)$
$$\ddot y=12x-36y-3\dot y\tag 3$$
And substituting $x$ from $(2)$ into $(3)$
$$\ddot y=3\dot y+9y-36y-3\dot y$$
$$\ddot y+27y=0$$
It has as solution:
$$y=c_1\sin\left(3\sqrt{3}t\right)+c_2\cos\left(3\sqrt{3}t\right)$$
Further, I get $\dot y(0)=20$ as $\dot y(t)=4x(t)-3y(t)$ and hence $\dot y(0)=4·2-3(-4)$
The equation is different and, waiting for the initial values to apply, the argument for the sine and cosine is the same as for the solution for the system.
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