Solve a differential equation systems$\frac{dx}{dt} = -x-6y $ $\frac{dy}{dt} = 3x+5y$ $\mathbf {Consider \space the \space system}$ $$\frac{dx}{dt} = -x-6y $$
$$\frac{dy}{dt} = 3x+5y$$
$\mathbf {Find\space the\space general\space solution\space of\space the \space system.}$
For this problem, I solved eigenvalues, which are $ 2 \pm 6 \mathbf i $ and got the general solution: 
$$x(t) = k_1e^{2t}[2cos\left(6t\right)]+ k_2e^{2t}[2sin\left(6t\right)]$$
$$y(t) = k_1e^{2t}[-cos(6t)+2sin(6t)]+k_2e^{2t}[-2cos(6t)-sin(6t)]$$
But, I am not sure if this is correct. Please help me to check this answer.
 A: Use Laplace Transform:


*

*Function 1:


$$x'(t)=-x(t)-6y(t)\Longleftrightarrow$$
$$\mathcal{L}_{t}\left[x'(t)\right]_{(s)}=\mathcal{L}_{t}\left[-x(t)-6y(t)\right]_{(s)}\Longleftrightarrow$$
$$sx(s)-x(0)=-x(s)-6y(s)\Longleftrightarrow$$
$$sx(s)+x(s)=x(0)-6y(s)\Longleftrightarrow$$
$$x(s)\left[s+1\right]=x(0)-6y(s)\Longleftrightarrow$$
$$x(s)=\frac{x(0)-6y(s)}{s+1}$$


*

*Function 2:


$$y'(t)=3x(t)+5y(t)\Longleftrightarrow$$
$$\mathcal{L}_{t}\left[y'(t)\right]_{(s)}=\mathcal{L}_{t}\left[3x(t)+5y(t)\right]_{(s)}\Longleftrightarrow$$
$$sy(s)-y(0)=3x(s)+5y(s)\Longleftrightarrow$$
$$sy(s)-5y(s)=3x(s)+y(0)\Longleftrightarrow$$
$$y(s)\left[s-5\right]=3x(s)+y(0)\Longleftrightarrow$$
$$y(s)=\frac{3x(s)+y(0)}{s-5}$$
So, we can find that:


*

*For the function $x(s)$:


$$x(s)=\frac{x(0)-6\cdot\frac{3x(s)+y(0)}{s-5}}{s+1}\Longleftrightarrow x(s)=\frac{x(0)(s-5)-6y(0)}{13+s(s-4)}$$


*

*For the function $y(s)$:


$$y(s)=\frac{3\cdot\frac{x(0)-6y(s)}{s+1}+y(0)}{s-5}\Longleftrightarrow y(s)=\frac{y(0)+3x(0)+y(0)s}{13+s(s-4)}$$
And with the Inverse Laplace Transform, find that:


*

*For the function $x(t)$:


$$\mathcal{L}_{s}^{-1}\left[x(s)\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{x(0)(s-5)-6y(0)}{13+s(s-4)}\right]_{(t)}\Longleftrightarrow$$
$$x(t)=\mathcal{L}_{s}^{-1}\left[\frac{x(0)(s-5)-6y(0)}{13+s(s-4)}\right]_{(t)}\Longleftrightarrow$$
$$x(t)=e^{2t}\left(x(0)\cos(3t)-(x(0)+2y(0))\sin(3t)\right)$$


*

*For the function $y(t)$:


$$\mathcal{L}_{s}^{-1}\left[y(s)\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{y(0)+3x(0)+y(0)s}{13+s(s-4)}\right]_{(t)}\Longleftrightarrow$$
$$y(t)=\mathcal{L}_{s}^{-1}\left[\frac{y(0)+3x(0)+y(0)s}{13+s(s-4)}\right]_{(t)}\Longleftrightarrow$$
$$y(t)=e^{2t}\left(y(0)\cos(3t)+(y(0)+x(0))\sin(3t)\right)$$
So, our solution is:


*

*$$x(t)=e^{2t}\left(x(0)\cos(3t)-(x(0)+2y(0))\sin(3t)\right)$$

*$$y(t)=e^{2t}\left(y(0)\cos(3t)+(y(0)+x(0))\sin(3t)\right)$$

A: $$\frac{dx}{dt} = -x-6y \Rightarrow \frac{d^2x}{dt^2} = -\frac{dx}{dt}-6\frac{dy}{dt}$$
$$\frac{dy}{dt} = 3x+5y \Rightarrow \frac{d^2x}{dt^2} = -\frac{dx}{dt}-18x-30y$$
$$\frac{dx}{dt} = -x-6y \Rightarrow -6y=\frac{dx}{dt} +x \Rightarrow -30y=5\frac{dx}{dt} +5x$$
$$\frac{d^2x}{dt^2} = -\frac{dx}{dt}-18x+5\frac{dx}{dt} +5x$$
$$\frac{d^2x}{dt^2} -4\frac{dx}{dt}+13x=0$$
Auxiliary equation $\lambda^2-4\lambda +13=0$
$\lambda = \frac{4 \pm \sqrt{16-52}}{2}$
$\lambda = 2 \pm 3i$
Slightly different to your values.
