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. $x(t)=c_1cos2t + c_2sin2t + c_3cos4t + c_4sin4t$

Explanation:

Step1: Assume solutions of the form

Assume $x = e^{rt}$ and $y = e^{rt}$. Then $x''=r^{2}e^{rt}$ and $y'' = r^{2}e^{rt}$. Substituting into the system $x''=-8x + 8y$ and $y''=4x-12y$, we get the system of equations:

$$
LATEXBLOCK0
$$

Dividing by $e^{rt}$ (since $e^{rt}
eq0$), we have the homogeneous - linear system \(

$$\begin{cases}r^{2}+8x - 8y = 0\\-4x+r^{2}+12y = 0\end{cases}$$

\).
The coefficient matrix \(A=

$$\begin{bmatrix}r^{2}+8&-8\\-4&r^{2}+12\end{bmatrix}$$

\).

Step2: Find the determinant of the coefficient matrix

The determinant \(\text{det}(A)=(r^{2}+8)(r^{2}+12)-(- 8)\times(-4)=r^{4}+20r^{2}+96 - 32=r^{4}+20r^{2}+64\).
Let \(u = r^{2}\), then the equation becomes \(u^{2}+20u + 64=0\).
Using the quadratic formula \(u=\frac{-20\pm\sqrt{20^{2}-4\times64}}{2}=\frac{-20\pm\sqrt{400 - 256}}{2}=\frac{-20\pm\sqrt{144}}{2}=\frac{-20\pm12}{2}\).
We get \(u_1=-4\) and \(u_2=-16\).
Since \(u = r^{2}\), then \(r^{2}=-4\) gives \(r=\pm2i\) and \(r^{2}=-16\) gives \(r=\pm4i\).

Step3: Write the general solution for \(x(t)\)

The general solution for a second - order linear homogeneous differential equation with complex roots \(r = a\pm bi\) is \(x(t)=C_1e^{at}\cos(bt)+C_2e^{at}\sin(bt)\). When \(a = 0\), \(r=\pm2i\) and \(r=\pm4i\), the general solution for \(x(t)\) is \(x(t)=C_1\cos(2t)+C_2\sin(2t)+C_3\cos(4t)+C_4\sin(4t)\).

Answer:

D. \(x(t)=C_1\cos(2t)+C_2\sin(2t)+C_3\cos(4t)+C_4\sin(4t)\)