Question

d. x(t)=c_1 cos 2t + c_2 sin 2t + c_3 cos 4t + c_4 sin 4t
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: Since no system of differential - equations is given, assume a general linear first - order system of the form

$$\begin{cases}x'(t)=a_{11}x(t)+a_{12}y(t)\\y'(t)=a_{21}x(t)+a_{22}y(t)\end{cases}$$

. But we don't have the coefficients. However, if the system is homogeneous and we know $x(t)=c_1\cos(2t)+c_2\sin(2t)+c_3\cos(4t)+c_4\sin(4t)$, we can use the fact that for a linear homogeneous system of ODEs with constant coefficients, the solutions are of the same form.
Let's assume the system is such that we can find $y(t)$ by substituting $x(t)$ into the system and solving for $y(t)$. A common approach is to use the method of undetermined coefficients. If the system is linear and homogeneous, and the characteristic equation of the system has roots corresponding to the frequencies in $x(t)$.
Let's assume the system is simple enough that we can find $y(t)$ in the form $y(t)=d_1\cos(2t)+d_2\sin(2t)+d_3\cos(4t)+d_4\sin(4t)$.
We substitute $x(t)$ and $y(t)$ into the system of differential equations. If the system is

$$\begin{cases}x' = Ax+By\\y'=Cx + Dy\end{cases}$$

(where $A,B,C,D$ are constants).
Differentiate $x(t)$:
$x'(t)=- 2c_1\sin(2t)+2c_2\cos(2t)-4c_3\sin(4t)+4c_4\cos(4t)$.
Substitute $x(t)$ and $y(t)$ into the system and equate the coefficients of $\cos(2t),\sin(2t),\cos(4t),\sin(4t)$ on both sides.
If the system is uncoupled (i.e., $B = C=0$), then $y(t)$ is also a general solution of a second - order homogeneous linear ODE with the same characteristic roots as the ODE for $x(t)$.
So, $y(t)=d_1\cos(2t)+d_2\sin(2t)+d_3\cos(4t)+d_4\sin(4t)$ where $d_1,d_2,d_3,d_4$ are arbitrary constants.

Answer:

$y(t)=d_1\cos(2t)+d_2\sin(2t)+d_3\cos(4t)+d_4\sin(4t)$