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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)

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We have the general case of

$$x' = a x + b y \\ y' = c x + d y$$

Taking the derivative of the first equation yields

$$x'' = a x' + b y' = a x' + b(c x + d y)$$

However, from the first equation, we also have that $y = \dfrac{1}{b}(x' - ax)$, for $b \neq 0$, and upon substituting

$$x'' = a x' + b c x + d (x' - a x) = 3x' - 36 x -3(x' - 3 x) = -27 x \implies x'' + 27 x = 0$$

We need two initial conditions and already have $x(0) = 2$, but need $x'(0)$. Using the first equation

$$x'(0) = a x (0) + b(y(0)) = 3(2) - 9(-4) = 42$$

We now have to solve the second order system

$$x'' + 27 x = 0, x(0) = 2, x'(0) = 42$$

We get

$$x(t) = 2 \cos (3 \sqrt{3} t) + \dfrac{14\sqrt{3}}{3} \sin \left(3 \sqrt{3} t\right)$$

Compare that to your current solution. Also, it is now just taking derivatives of this to find $y(t)$.

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  • $\begingroup$ $x(t)=2\cos{(3\sqrt{3}t)}+\frac{14}{\sqrt{3}}\sin{(3\sqrt{3}t)}$ $\endgroup$ – Dhamnekar Winod Nov 7 '18 at 13:42
  • $\begingroup$ Oops, just a transcription error - thanks for catching that, now fixed. $\endgroup$ – Moo Nov 7 '18 at 13:43
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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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  • $\begingroup$ Budria, $x=\frac{\dot{y}+3y}{4}$ $\endgroup$ – Dhamnekar Winod Nov 7 '18 at 16:48
  • $\begingroup$ @DhamnekarWinod a typo in copying, I write my calculations on paper. Corrected. thanks! $\endgroup$ – Rafa Budría Nov 7 '18 at 16:52

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