Question

find a general solution of the linear system below
$x=-8x + 8y, y = 4x-12y$
solve for x(t). choose the correct answer below
a. $x(t)=c_1e^{-4t}+c_2e^{4t}+c_3e^{-16t}+c_4e^{16t}$
b. $x(t)=c_1cos4t + c_2sin4t + c_3cos16t + c_4sin16t$
c. $x(t)=c_1e^{-2t}+c_2e^{2t}+c_3e^{-4t}+c_4e^{4t}$
d. $x(t)=c_1cos2t + c_2sin2t + c_3cos4t + c_4sin4t$
now find y(t) so that y(t) and the solution for x(t) found in the previous step are a general solution to the system of differential equations
y(t) =

Explanation:

Step1: Write the system in matrix form

The system

$$\begin{cases}x''=- 8x + 8y\\y''=4x - 12y\end{cases}$$

can be written as $\mathbf{X}''=

$$\begin{pmatrix}-8&8\\4&-12\end{pmatrix}$$

\mathbf{X}$, where $\mathbf{X}=

$$\begin{pmatrix}x\\y\end{pmatrix}$$

$.

Step2: Find the characteristic equation of the coefficient matrix

Let $A =

$$\begin{pmatrix}-8&8\\4&-12\end{pmatrix}$$

$. The characteristic equation is $\det(A-\lambda I)=0$, where $I$ is the $2\times2$ identity - matrix. So, $

$$\begin{vmatrix}-8-\lambda&8\\4&-12-\lambda\end{vmatrix}$$

=0$. Expanding, we get $(-8 - \lambda)(-12-\lambda)-32 = 0$, which simplifies to $\lambda^{2}+20\lambda + 96 - 32=0$, or $\lambda^{2}+20\lambda + 64 = 0$.

Step3: Solve the characteristic equation

Using the quadratic formula $\lambda=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $\lambda^{2}+20\lambda + 64 = 0$ with $a = 1$, $b = 20$, and $c = 64$. We have $\lambda=\frac{-20\pm\sqrt{400 - 256}}{2}=\frac{-20\pm\sqrt{144}}{2}=\frac{-20\pm12}{2}$. The roots are $\lambda_1=-4$ and $\lambda_2=-16$. The general solution of the homogeneous system of second - order linear differential equations has the form $x(t)=c_1\cos(4t)+c_2\sin(4t)+c_3\cos(16t)+c_4\sin(16t)$ (since for a second - order linear homogeneous differential equation $y''+\omega^{2}y = 0$, the general solution is $y = c_1\cos(\omega t)+c_2\sin(\omega t)$ and our system has two distinct frequencies corresponding to the roots of the characteristic equation). So the answer for $x(t)$ is B.

Step4: Substitute $x(t)$ into the original system to find $y(t)$

We know $x(t)=c_1\cos(4t)+c_2\sin(4t)+c_3\cos(16t)+c_4\sin(16t)$. Then $x''(t)=-16c_1\cos(4t)-16c_2\sin(4t)-256c_3\cos(16t)-256c_4\sin(16t)$.
Substitute $x(t)$ and $x''(t)$ into the first equation $x''=-8x + 8y$.

$$
LATEXBLOCK5
$$

Answer:

B. $x(t)=c_1\cos(4t)+c_2\sin(4t)+c_3\cos(16t)+c_4\sin(16t)$
$y(t)=-c_1\cos(4t)-c_2\sin(4t)-31c_3\cos(16t)-31c_4\sin(16t)$