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We wish to write the two sinusoidal terms as one. \begin{align*} 3 \cos 2 t+4 \sin 2 t & =R \cos \left(\omega_{0} t-\delta\right) \\ & =R\left[\cos \omega_{0} t \cos \delta+\sin \omega_{0} t \sin \delta\right] \\ & =(R \cos \delta) \cos \omega_{0} t+(R \sin \delta) \sin \omega_{0} t \end{align*}

Matching the coefficients, we obtain the following system of equations for \(\omega_{0}, R\), and \(\delta\).

\begin{align*} R \cos \delta=3 \quad \quad \quad (1)\\ \omega_{0}=2 \quad \quad \quad (2)\\ R \sin \delta=4 \quad \quad \quad (3) \end{align*}

Square both sides of the first and third equations

\begin{align*} R^{2} \cos ^{2} \delta=9 \\ R^{2} \sin ^{2} \delta=16 \end{align*}

and add their respective sides.

\begin{align*} R^{2} \cos ^{2} \delta+R^{2} \sin ^{2} \delta=9+16 \\ R^{2}\left(\cos ^{2} \delta+\sin ^{2} \delta\right)=25 \\ R^{2}=25 \\ R=5 \end{align*}

Divide the respective sides of equations (1) and (3).

\begin{equation*} \frac{R \sin \delta}{R \cos \delta}=\frac{4}{3} \quad \rightarrow \quad \tan \delta=\frac{4}{3} \quad \rightarrow \quad \delta=\tan ^{-1} \frac{4}{3} \end{equation*}

Therefore,

\begin{equation*} 3 \cos 2 t+4 \sin 2 t=5 \cos \left(2 t-\tan ^{-1} \frac{4}{3}\right) \end{equation*}

\begin{align*} 20 u” + 400 u' + 3920u &= 0\\ 20r^2 + 400r + 3920 &= 0 \end{align*} then \begin{align*} r = -10 \pm 4 \sqrt{6} i \end{align*} We see that \begin{align*} u(t) = & C_1 e^{-10t}\cos(4\sqrt{6}t) + C_2 e^{-10t}\sin(4\sqrt{6}t) \\ u'(t) = & 4\sqrt{6}C_1e^{-10t}\sin(4\sqrt{6}t) - 10 C_1 e^{-10t} \cos(4\sqrt{6}t) \\ & + 4\sqrt{6} C_2 e^{-10t} \cos(4\sqrt{6}t) - 10 C_2 e^{-10t} \sin (4\sqrt{6}t) \end{align*} By knowing that \(u(0)=2\) and \(u'(0)=2\), we find that \(C_1=2\) and \(C_2=\frac{5}{\sqrt{6}}\). Finally, \begin{align*} u(t) = 2 e^{-10t} \cos(4\sqrt{6}t) + \frac{5}{\sqrt{6}} e^{-10t} \sin(4\sqrt{6}t) \end{align*} Therefore \begin{align*} \text{Quasi-frequency} &: 4\sqrt{6}\\ \text{Quasi-period} &: \frac{\pi}{2\sqrt{6}} \end{align*}

Let \(u=x_{1}\).

\begin{align*} x_{1}^{\prime \prime}+0.5 x_{1}^{\prime}+2 x_{1}=0 \end{align*}

Finally, let \(x_{2}=x_{1}^{\prime}\).

\begin{align*} x_{2}^{\prime}+0.5 x_{2}+2 x_{1}=0 \end{align*}

By making these substitutions, the original second-order ODE has become a system of first-order ODEs.

\begin{align*} \left\{\begin{array}{l} x_{1}^{\prime}=x_{2} \\ x_{2}^{\prime}=-2 x_{1}-0.5 x_{2} \end{array}\right. \end{align*}

Let \(u=x_{1}\).

\begin{align*} x_{1}^{\prime \prime}+p(t) x_{1}^{\prime}+q(t) x_{1}=g(t), \quad x_{1}(0)=u_{0}, \quad x_{1}^{\prime}(0)=u_{0}^{\prime} \end{align*}

Finally, let \(x_{2}=x_{1}^{\prime}\).

\begin{align*} x_{2}^{\prime}+p(t) x_{2}+q(t) x_{1}=g(t), \quad x_{1}(0)=u_{0}, \quad x_{2}(0)=u_{0}^{\prime} \end{align*}

By making these substitutions, the original initial value problem has become a system of first-order ODEs,

\begin{align*} \left\{\begin{array}{l} x_{1}^{\prime}=x_{2} \\ x_{2}^{\prime}=-q(t) x_{1}-p(t) x_{2}+g(t) \end{array}\right. \end{align*}

subject to the initial conditions,

\begin{align*} x_{1}(0)=u_{0} \quad \text { and } \quad x_{2}(0)=u_{0}^{\prime} . \end{align*}

\(\mathbf{A}^{T}\) is the transpose of \(\mathbf{A}, \overline{\mathbf{A}}\) is the complex conjugate of \(\mathbf{A}\), and \(\mathbf{A}^{*}=\overline{\mathbf{A}}^{T}\) is the adjoint of \(\mathbf{A}\).

\begin{equation*} \mathbf{A}^{T}=\left(\begin{array}{cc} 3-2 i & 2-i \\ 1+i & -2+3 i \end{array}\right) \end{equation*}

\begin{equation*} \overline{\mathbf{A}}=\left(\begin{array}{cc} 3+2 i & 1-i \\ 2+i & -2-3 i \end{array}\right) \end{equation*}

\begin{equation*} \mathbf{A}^{*}=\left(\begin{array}{cc} 3+2 i & 2+i \\ 1-i & -2-3 i \end{array}\right) \end{equation*}

Start by calculating the determinant. \begin{align*} \operatorname{det}\left(\begin{array}{rr} 1 & 4 \\ -2 & 3 \end{array}\right)=(1)(3)-(4)(-2)=11 \end{align*} Since it’s not zero, an inverse for the given matrix exists. \begin{align*} \left(\begin{array}{rr|rr} 1 & 4 & 1 & 0 \\ -2 & 3 & 0 & 1 \end{array}\right) \end{align*} The aim is to make the left side of the augmented matrix 1’s and 0 ’s as the right side is now. Since the top left entry is 1 already, we move on to the bottom left entry. To make it zero, multiply both sides of the first row by 2 and add it to the second row. \begin{align*} \left(\begin{array}{cc|cc} 1 & 4 & 1 & 0 \\ 0 & 11 & 2 & 1 \end{array}\right) \end{align*} To make the bottom right entry 1 , divide the bottom row by 11 . \begin{align*} \left(\begin{array}{ll|ll} 1 & 4 & 1 & 0 \\ 0 & 1 & \frac{2}{11} & \frac{1}{11} \end{array}\right) \end{align*} To make the top right entry 0 , multiply the bottom row by \(-4\) and add it to the first row.

\begin{align*} \left(\begin{array}{cc|cc} 1 & 0 & \frac{3}{11} & -\frac{4}{11} \\ 0 & 1 & \frac{2}{11} & \frac{1}{11} \end{array}\right) \end{align*}

Therefore, the inverse matrix is

\begin{align*} \left(\begin{array}{cc} \frac{3}{11} & -\frac{4}{11} \\ \frac{2}{11} & \frac{1}{11} \end{array}\right) . \end{align*}