find a general solution of the linear system below.\n$x=-8x + 8y$, $y=4x - 12y$\nsolve for $x(t)$. choose…

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: [ \begin{cases} r^{2}e^{rt}=-8e^{rt}+8ye^{rt}\ r^{2}e^{rt}=4xe^{rt}-12e^{rt} \end{cases} ] Dividing by $e^{rt}$ (since $e^{rt}\neq0$), 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))