Question

Solve (2) of Section 7.6 using the method outlined in the Remarks (page 351) - that is, express (2) of Section 7.6 as a first-order system of four linear equations. Use a CAS or linear algebra software as an aid in finding eigenvalues and eigenvectors of a 4 x 4 matrix. Then apply the initial conditions to your general solution to obtain (4) of Section 7.6.

Short answer

The initial condition value of general solution is

$Y=c_1\left[\left(\begin{array}{c}0\\ -2\\ 0\\ 1\end{array}\right)\cos 2\sqrt{3t}-\left(\begin{array}{c}\frac{\sqrt{3}}{3}\\ 0\\ -\frac{\sqrt{3}}{6}\\ 1\end{array}\right)\sin 2\sqrt{3}t\right]+c_2\left[\left(\begin{array}{c}0\\ -2\\ 0\\ -\frac{\sqrt{3}}{6}\\ 1\end{array}\right)\cos 2\sqrt{3t}+\left(\begin{array}{c}\frac{\sqrt{3}}{3}\\ 0\\ 1\end{array}\right)\sin 2\sqrt{3}t\right]+c_3\left[\left(\begin{array}{c}0\\ \frac{1}{2}\\ 0\\ 1\end{array}\right)\cos \sqrt{2t}-\left(\begin{array}{c}-\frac{\sqrt{2}}{4}\\ 0\\ -\frac{\sqrt{2}}{2}\\ 0\end{array}\right)\sin \sqrt{2t}\right]+c_4\left[\left(\begin{array}{c}-\frac{\sqrt{2}}{4}\\ 0\\ -\frac{\sqrt{2}}{2}\\ 0\end{array}\right)\cos \sqrt{2t}+\left(\begin{array}{c}0\\ \frac{1}{2}\\ 0\\ 1\end{array}\right)\sin \sqrt{2}t\right]$

Step-by-step solution

Determine the linear system

A linear system is a mathematical model of a system that is based on the usage of a linear operator in systems theory.

Linear systems often have considerably simpler characteristics and properties than nonlinear systems.

Determine the matrix form

We want to express

$x_1^{\text{'}\text{'}}+10x_1-4x_2=0-4x_1+x_2^{\text{'}\text{'}}+4x_2=0$

As a first-order system of four linear equations.

So, let $x_1=y_1,x_1^{\text{'}}=y_2,$

x₂= y₂ , and x₂= y₄.

Substituting into (1),

$$\begin{gathered} y_2^{\text{'}}+10y_1-4y_4=0 \\ {y}_2^{\text{'}}=-10y_1+4y_4 \end{gathered}$$

And,

$$\begin{gathered} -4y_1+y_4^{\text{'}}+4y_3=0 \\ {y}_4^{\text{'}}=4y_1-4y_3 \end{gathered}$$

So here we have

$\left\{\begin{array}{l}{y}_1^{\text{'}}=y_2\\ y_2^{\text{'}}=-10y_1+4y_4\\ y_3^{\text{'}}=y_4\\ y_4^{\text{'}\text{'}}=4y_1-4y_3\end{array}\right.$

Which can be written in the matrix form,

$Y\text{'}=\left(\begin{array}{cccc}0& 1& 0& 0\\ -10& 0& 0& 4\\ 0& 0& 0& 1\\ 4& 0& -4& 0\end{array}\right)Y$

Determine the general solution of the system

Using MATLAB we can obtain the eigenvalues and eigenvectors of the system.

The eigenvalues of the system are ${\lambda }_1=\pm 2\sqrt{3}iand{\lambda }_2=\pm \sqrt{2}i$.

As for the eigenvectors, we have

$K_1=\left(\begin{array}{c}\pm \frac{\sqrt{3i}}{3}\\ -2\\ \pm \frac{\sqrt{3i}}{6}\\ 1\end{array}\right)$

$=\left(\begin{array}{c}0\\ -2\\ 0\end{array}\right)+i\left(\begin{array}{c}\pm \frac{\sqrt{3}}{3}\\ 0\\ 1\end{array}\right)$

And,

$$\begin{gathered} K_2=\left(\begin{array}{c}\pm \frac{\sqrt{2i}}{4}\\ \pm \frac{\sqrt{2i}}{2}\\ 1\end{array}\right) \\ =\left(\begin{array}{c}0\\ \frac{1}{2}\\ 0\\ 1\end{array}\right)+i\left(\begin{array}{c}\pm \frac{\sqrt{2}}{4}\\ 0\\ \pm \frac{\sqrt{2}}{2}\\ 0\end{array}\right) \end{gathered}$$

$$\begin{gathered} \gg A=sym\left(\right[0100;-10040;0001;40-40\left]\right) \\ A= \\ [0,1,0,0] \\ [-10,0,4,0] \\ [0,0,0,1] \\ [4,0,-4,0] \\ \gg \left[V,D\right]=eig\left(A\right) \\ V= \\ \left[-\left(12^(1/2\right)*1i)/6,(12^(1/2)*1i)/6,(2^(1/2)*1i)/4,-(2^(1/2)*1i)/4\right] \\ \left[-\left(12^(1/2\right)*1i)/12,-(12^(1/2)*1i)/12,(2^(1/2)*1i)/2,-(2^(1/2)*1i)/2\right] \end{gathered}$$

$$\begin{gathered} [1,1,1,1] \\ D=, \\ \left[-(12^(1/2)*1i),0,0,0\right] \\ [0,(12^(1/2)*1i),0,0] \\ {x}_2\left(t\right)=y_3\left(t\right)=-\frac{\sqrt{3}}{10}\sin 2\sqrt{3}t-\frac{\sqrt{2}}{5}\sin \sqrt{2} \end{gathered}$$

The general solution of the system is

$Y=c_1\left[\left(\begin{array}{c}0\\ -2\\ 0\\ 1\end{array}\right)\cos 2\sqrt{3t}-\left(\begin{array}{c}\frac{\sqrt{3}}{3}\\ 0\\ -\frac{\sqrt{3}}{6}\\ 1\end{array}\right)\sin 2\sqrt{3}t\right]+c_2\left[\left(\begin{array}{c}0\\ -2\\ 0\\ -\frac{\sqrt{3}}{6}\\ 1\end{array}\right)\cos 2\sqrt{3t}+\left(\begin{array}{c}\frac{\sqrt{3}}{3}\\ 0\\ 1\end{array}\right)\sin 2\sqrt{3}t\right]+c_3\left[\left(\begin{array}{c}0\\ \frac{1}{2}\\ 0\\ 1\end{array}\right)\cos \sqrt{2t}-\left(\begin{array}{c}-\frac{\sqrt{2}}{4}\\ 0\\ -\frac{\sqrt{2}}{2}\\ 0\end{array}\right)\sin \sqrt{2t}\right]+c_4\left[\left(\begin{array}{c}-\frac{\sqrt{2}}{4}\\ 0\\ -\frac{\sqrt{2}}{2}\\ 0\end{array}\right)\cos \sqrt{2t}+\left(\begin{array}{c}0\\ \frac{1}{2}\\ 0\\ 1\end{array}\right)\sin \sqrt{2}t\right]$

Determine the initial value condition

We were given the initial value condition

$Y\left(0\right)=\left(\begin{array}{c}0\\ 1\\ 0\\ -1\end{array}\right)$,

So,

$\left(\begin{array}{c}0\\ 1\\ 0\\ -1\end{array}\right)=c_1\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)+c_2\left(\begin{array}{c}\frac{\sqrt{3}}{3}\\ 0\\ -\frac{\sqrt{3}}{6}\\ 1\end{array}\right)+c_3\left(\begin{array}{c}0\\ \frac{1}{2}\\ 0\\ 1\end{array}\right)+c_4\left(\begin{array}{c}-\frac{\sqrt{2}}{4}\\ 0\\ -\frac{\sqrt{2}}{2}\\ 0\end{array}\right)$

Here we have

$$\begin{gathered} \frac{\sqrt{3}}{3}{c}_2-\frac{\sqrt{2}}{4}{c}_4=0.......\left(1\right) \\ -2c_1+\frac{1}{2}{c}_3=1.......\left(2\right) \\ -\frac{\sqrt{3}}{6}{c}_2-\frac{\sqrt{2}}{2}{c}_4=0.......\left(3\right) \\ {c}_1+c_2+c_3=-1........\left(4\right) \end{gathered}$$

Multiplying (3) by 1/2 then subtracting from (1) we will get

$c_2=0And{c}_4=0$

Since c₂=0, equation (4) becomes

$c_1+c_3=-1$

We can multiply it by 2 then add it to (2), we will get

$\frac{5}{2}{c}_3=-1,\to c_3=-\frac{2}{5}$

Finally, to solve for c₁ we substitute the values we obtained into (4), where

$c_1-\frac{2}{5}=-1\to c_1=-\frac{3}{5}$

So, the solution that satisfies the given initial condition is

$Y\left(t\right)=-\frac{3}{5}\left[\left(\begin{array}{c}0\\ -2\\ 0\\ 1\end{array}\right)\cos 2\sqrt{3t}-\left(\begin{array}{c}\frac{\sqrt{3}}{3}\\ 0\\ -\frac{\sqrt{3}}{6}\\ 1\end{array}\right)\sin 2\sqrt{3}t\right]-\frac{2}{5}\left[\begin{array}{c}-\frac{\sqrt{2}}{4}\\ \frac{1}{2}\\ 0\\ \cos \sqrt{2t}\end{array}\right]$

Where,

$$\begin{gathered} x_1\left(t\right)=y_1\left(t\right)=\frac{\sqrt{3}}{5}\sin 2\sqrt{3}t-\frac{\sqrt{2}}{10}\sin \sqrt{2}t \\ {x}_2\left(t\right)=y_3\left(t\right)=-\frac{\sqrt{3}}{10}\sin 2\sqrt{3}t-\frac{\sqrt{2}}{5}\sin \sqrt{2} \end{gathered}$$

The initial condition value of general solution is

$Y=c_1\left[\left(\begin{array}{c}0\\ -2\\ 0\\ 1\end{array}\right)\cos 2\sqrt{3t}-\left(\begin{array}{c}\frac{\sqrt{3}}{3}\\ 0\\ -\frac{\sqrt{3}}{6}\\ 1\end{array}\right)\sin 2\sqrt{3}t\right]+c_2\left[\left(\begin{array}{c}0\\ -2\\ 0\\ -\frac{\sqrt{3}}{6}\\ 1\end{array}\right)\cos 2\sqrt{3t}+\left(\begin{array}{c}\frac{\sqrt{3}}{3}\\ 0\\ 1\end{array}\right)\sin 2\sqrt{3}t\right]+c_3\left[\left(\begin{array}{c}0\\ \frac{1}{2}\\ 0\\ 1\end{array}\right)\cos \sqrt{2t}-\left(\begin{array}{c}-\frac{\sqrt{2}}{4}\\ 0\\ -\frac{\sqrt{2}}{2}\\ 0\end{array}\right)\sin \sqrt{2t}\right]+c_4\left[\left(\begin{array}{c}-\frac{\sqrt{2}}{4}\\ 0\\ -\frac{\sqrt{2}}{2}\\ 0\end{array}\right)\cos \sqrt{2t}+\left(\begin{array}{c}0\\ \frac{1}{2}\\ 0\\ 1\end{array}\right)\sin \sqrt{2}t\right]$