Introduction to Systems and Phase Plane Analysis

Let $\mathbf{A}\mathbf{=}\mathbf{D}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{B}\mathbf{=}\mathbf{D}\mathbf{+}\mathbf{2}\mathbf{,}\mathbf{C}\mathbf{=}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}\mathbf{,}$ where $\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}$. For $\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{8}$, compute 

(a) $\mathbf{A}\left[\mathbf{y}\right]$

(b) $\mathbf{B}\left[\mathbf{A}\left[\mathbf{y}\right]\right]$

(c) $\mathbf{B}\left[\mathbf{y}\right]$

(d) $\mathbf{A}\left[\mathbf{B}\left[\mathbf{y}\right]\right]$

(e) $\mathbf{C}\left[\mathbf{y}\right]$

Show that the operator (D-1)(D+2) is the same as the operator ${\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}$.

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,} \\ \mathbf{x}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.
$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{,} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{y}\mathbf{-}\mathbf{4}\mathbf{x} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{x}\mathbf{=}\mathbf{5}\mathbf{,} \\ \mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{1} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{sint}\mathbf{,} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{-}\mathbf{cost} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{u}\right]\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{v}\right]\mathbf{=}{\mathbf{e}}^{\mathbf{t}}\mathbf{,} \\ \left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{u}\right]\mathbf{+}\left(\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{v}\right]\mathbf{=}\mathbf{5} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \left(\mathbf{D}\mathbf{-}\mathbf{3}\right)\left[\mathbf{x}\right]\mathbf{+}\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{t}\mathbf{,} \\ \left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}\left(\mathbf{D}\mathbf{+}\mathbf{4}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{1} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{,} \\ \mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{sint} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{,} \\ \mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{t}} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\right)\left[\mathbf{u}\right]\mathbf{+}\mathbf{5}\mathbf{v}\mathbf{=}{\mathbf{e}}^{\mathbf{t}}\mathbf{,} \\ \mathbf{2}\mathbf{u}\mathbf{+}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\right)\left[\mathbf{v}\right]\mathbf{=}\mathbf{0} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} {\mathbf{D}}^{\mathbf{2}}\left[\mathbf{u}\right]\mathbf{+}\mathbf{D}\left[\mathbf{v}\right]\mathbf{=}\mathbf{2}\mathbf{,} \\ \mathbf{4}\mathbf{u}\mathbf{+}\mathbf{D}\left[\mathbf{v}\right]\mathbf{=}\mathbf{6} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{y} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{,} \\ \mathbf{-}\mathbf{x}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{1} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \frac{\mathbf{dw}}{\mathbf{dt}}\mathbf{=}\mathbf{5}\mathbf{w}\mathbf{+}\mathbf{2}\mathbf{z}\mathbf{+}\mathbf{5}\mathbf{t}\mathbf{,} \\ \frac{\mathbf{dz}}{\mathbf{dt}}\mathbf{=}\mathbf{3}\mathbf{w}\mathbf{+}\mathbf{4}\mathbf{z}\mathbf{+}\mathbf{17}\mathbf{t} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{x}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{,} \\ \mathbf{2}\mathbf{x}\mathbf{+}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{0} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,} \\ \mathbf{-}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0} \end{gathered}$$

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \mathit{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{t}\mathbf{,} \\ \mathit{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

In Problems 19 – 21, solve the given initial value problem.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{;} \mathbf{x}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

In Problems 19 – 21, solve the given initial value problem.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{;} \mathbf{x}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

In Problems 19 – 21, solve the given initial value problem.

$$\begin{gathered} \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{y}\mathbf{;} \mathbf{x}\left(\mathbf{0}\right)\mathbf{=}\mathbf{3}\mathbf{,} \mathbf{x}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \\ \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{x}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

Verify that the solution to the initial value problem

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{-}\mathbf{2}\mathbf{;} \mathbf{x}\left(\mathbf{0}\right)\mathbf{=}\mathbf{2}\mathbf{,} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

Satisfies $\left|x\left(t\right)\right|\mathbf{+}\left|y\left(t\right)\right|\mathbf{\to }\mathbf{+}\mathbf{\infty }$ as $\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$

In Problems 23 and 24, show that the given linear system is degenerate. In attempting to solve the system, determine whether it has no solutions or infinitely many solutions.

$$\begin{gathered} \left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{-}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{,} \\ \left(\mathbf{D}\mathbf{+}\mathbf{2}\right)\left[\mathbf{x}\right]\mathbf{+}\left(\mathbf{D}\mathbf{+}\mathbf{2}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{t}} \end{gathered}$$

In Problems 23 and 24, show that the given linear system is degenerate. In attempting to solve the system, determine whether it has no solutions or infinitely many solutions.

$$\begin{gathered} \mathbf{D}\left[x\right]\mathbf{+}\left(D+1\right)\left[y\right]\mathbf{=}{\mathbf{e}}^{\mathbf{t}}\mathbf{,} \\ {\mathbf{D}}^{\mathbf{2}}\left[x\right]\mathbf{+}\left(D^2+D\right)\left[y\right]\mathbf{=}\mathbf{0} \end{gathered}$$

In Problems 25 – 28, use the elimination method to find a general solution for the given system of three equations in the three unknown functions x(t), y(t), z(t).

 $$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{z}\mathbf{,} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{z}\mathbf{,} \\ \mathbf{z}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{+}\mathbf{5}\mathbf{z} \end{gathered}$$

In Problems 25 – 28, use the elimination method to find a general solution for the given system of three equations in the three unknown functions x(t), y(t), z(t).

 $$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{z}\mathbf{,} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{z}\mathbf{,} \\ \mathbf{z}\mathbf{\text{'}}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{-}\mathbf{z} \end{gathered}$$

In Problems 25 – 28, use the elimination method to find a general solution for the given system of three equations in the three unknown functions x(t), y(t), z(t)

 $$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{z}\mathbf{,} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{4}\mathbf{y}\mathbf{-}\mathbf{2}\mathbf{z}\mathbf{,} \\ \mathbf{z}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{+}\mathbf{4}\mathbf{z} \end{gathered}$$

In Problems 25 – 28, use the elimination method to find a general solution for the given system of three equations in the three unknown functions x(t), y(t), and z(t).

 $$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{z}\mathbf{,} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{6}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{,} \\ \mathbf{z}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{z} \end{gathered}$$

In Problems 29 and 30, determine the range of values (if any) of the parameter that will ensure all solutions x(t), and y(t) of the given system remain bounded as $\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$.

 $$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{\lambda x}\mathbf{-}\mathbf{y}\mathbf{,} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{y} \end{gathered}$$

In Problems 29 and 30, determine the range of values (if any) of the parameter that will ensure all solutions x(t), and y(t) of the given system remain bounded as $\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$.

 $$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{\lambda y}\mathbf{,} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{y} \end{gathered}$$

Two large tanks, each holding 100 L of liquid, are interconnected by pipes, with the liquid flowing from tank A into tank B at a rate of 3 L/min and from B into A at a rate of 1 L/min (see Figure 5.2). The liquid inside each tank is kept well stirred. A brine solution with a concentration of 0.2 kg/L of salt flows into tank A at a rate of 6 L/min. The (diluted) solution flows out of the system from tank A at 4 L/min and from tank B at 2 L/min. If, initially, tank A contains pure water and tank B contains 20 kg of salt, determine the mass of salt in each tank at a time $\mathbf{t}⩾\mathbf{0}$.

In Problem 31, 3 L/min of liquid flowed from tank A into tank B and 1 L/min from B to A. Determine the mass of salt in each tank at time $\mathbf{t}⩾\mathbf{0}$ if, instead, 5 L/min flows from A into B and 3 L/min flows from B into A, with all other data the same.

In Problem 31, assume that no solution flows out of the system from tank B, only 1 L/min flows from A into B, and only 4 L/min of brine flows into the system at tank A, other data being the same. Determine the mass of salt in each tank at the time $\mathbf{t}⩾\mathbf{0}$.

Feedback System with Pooling Delay. Many physical and biological systems involve time delays. A pure time delay has its output the same as its input but shifted in time. A more common type of delay is pooling delay. An example of such a feedback system is shown in Figure 5.3 on page 251. Here the level of fluid in tank B determines the rate at which fluid enters tank A. Suppose this rate is given by ${\mathbf{R}}_{\mathbf{1}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{\alpha }\left[\mathbf{V}\mathbf{-}{\mathbf{V}}_{\mathbf{2}}\left(\mathbf{t}\right)\right]$ where $\mathbf{\alpha }$ and V are positive constants and ${\mathbf{V}}_{\mathbf{2}}\left(\mathbf{t}\right)$ is the volume of fluid in tank B at time t.

           $$\begin{gathered} \frac{{\mathbf{dV}}_{\mathbf{1}}}{\mathbf{dt}}\mathbf{=}\mathbf{\alpha }\left(\mathbf{V}\mathbf{-}{\mathbf{V}}_{\mathbf{2}}\left(\mathbf{t}\right)\right)\mathbf{-}{\mathbf{KV}}_{\mathbf{1}}\left(\mathbf{t}\right)\mathbf{,} \\ \frac{{\mathbf{dV}}_{\mathbf{2}}}{\mathbf{dt}}\mathbf{=}{\mathbf{KV}}_{\mathbf{1}}\left(\mathbf{t}\right)\mathbf{-}{\mathbf{R}}_{\mathbf{3}} \end{gathered}$$

A house, for cooling purposes, consists of two zones: the attic area zone A and the living area zone B (see Figure 5.4). The living area is cooled by a 2 – ton air conditioning unit that removes 24,000 Btu/hr. The heat capacity of zone B is $\frac{\mathbf{1}}{\mathbf{2}}{}^{\mathbf{\circ }}\mathbf{F}$ per thousand Btu. The time constant for heat transfer between zone A and the outside is 2 hr, between zone B and the outside is 4 hr, and between the two zones is 4 hr. If the outside temperature stays at $\mathbf{100}\mathbf{°}\mathbf{F}$, how warm does it eventually get in the attic zone A? (Heating and cooling buildings was treated in Section 3.3 on page 102.)

A building consists of two zones A and B (see Figure 5.5). Only zone A is heated by a furnace, which generates 80,000 Btu/hr. The heat capacity of zone A is per thousand Btu. The time constant for heat transfer between zone A and the outside is 4 hr, between the unheated zone B and the outside is 5 hr, and between the two zones is 2 hr. If the outside temperature stays at , how cold does it eventually get in the unheated zone B?

In Problem 36, if a small furnace that generates 1000 Btu/hr is placed in zone B, determine the coldest it would eventually get in zone B has a heat capacity of $\mathbf{2}\mathbf{°}\mathbf{F}$ per thousand Btu.

Arms Race. A simplified mathematical model for an arms race between two countries whose expenditures for defense are expressed by the variables x(t) and y(t) is given by the linear system

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{a}\mathbf{;} \mathbf{x}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{+}\mathbf{b}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{4}\mathbf{,} \end{gathered}$$ 

 Where a and b are constants that measure the trust (or distrust) each country has for the other. Determine whether there is going to be disarmament (x and y approach 0 as t increases), a stabilized arms race (x and y approach a constant as $\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$ ), or a runaway arms race (x and y approach $\mathbf{+}\mathbf{\infty }$ as $\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$).

Let A, B and C represent three linear differential operators with constant coefficients; for example,

$$\begin{gathered} \mathbf{A}\mathbf{:}\mathbf{=}{\mathbf{a}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\mathbf{,}\mathbf{B}\mathbf{:}\mathbf{=}{\mathbf{b}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{b}}_{\mathbf{0}}\mathbf{,} \\ \mathbf{C}\mathbf{:}\mathbf{=}{\mathbf{c}}_{\mathbf{2}}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{1}}\mathbf{D}\mathbf{+}{\mathbf{c}}_{\mathbf{0}}\mathbf{,} \end{gathered}$$

 Where the a’s, b’s, and c’s are constants. Verify the following properties:

      $$\begin{gathered} \mathit{A}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{B}\mathbf{ }\mathbf{=}\mathbf{ }\mathit{B}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{A}\mathbf{,}\mathbf{ } \\ \mathit{A}\mathit{B}\mathbf{ }\mathbf{=}\mathbf{ }\mathit{B}\mathit{A} \end{gathered}$$

 (b) Associative laws:

       $$\begin{gathered} \left(\mathbf{A}\mathbf{+}\mathbf{B}\right)\mathbf{+}\mathbf{C}\mathbf{=}\mathbf{A}\mathbf{+}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{,} \\ \left(\mathbf{AB}\right)\mathbf{C}\mathbf{=}\mathbf{A}\left(\mathbf{BC}\right)\mathbf{.} \end{gathered}$$

 (c) Distributive law: $\mathbf{A}\left(\mathbf{B}\mathbf{+}\mathbf{C}\right)\mathbf{=}\mathbf{AB}\mathbf{+}\mathbf{AC}$

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{6}$

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{cos}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{y}\mathbf{)}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

${\mathbf{y}}^{\mathbf{4}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}{\mathbf{y}}^{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{cost}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathbf{y}}^{\mathbf{3}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}$

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

${\mathbf{y}}^{\mathbf{6}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\left[\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right]}^{\mathbf{3}}\mathbf{-}\mathbf{sin}\mathbf{\left(}\mathbf{y}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right)}\mathbf{+}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}......\mathbf{=}{\mathbf{y}}^{\mathbf{\left(}\mathbf{5}\mathbf{\right)}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{-}\mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{t}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{,}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

[hint] ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{x}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{,}{\mathbf{x}}_{\mathbf{3}}\mathbf{=}\mathbf{y}\mathbf{,}{\mathbf{x}}_{\mathbf{4}}\mathbf{=}\mathbf{y}\mathbf{\text{'}}$

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

$$\begin{gathered} \mathbf{3}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \\ \mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{6}\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2} \end{gathered}$$

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

$$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{t}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4} \\ \mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1} \end{gathered}$$

Sturm–Liouville Form. A second-order equation is said to be in Sturm–Liouville form if it is expressed as $\left[p\left(t\right)y\text{'}\left(t\right)\right]\mathbf{\text{'}}\mathbf{+}\mathbf{q}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$. Show that the substitutions ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{y}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{py}\mathbf{\text{'}}$ result in the normal form $\mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{p}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathbf{qx}}_{\mathbf{1}}$. If $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{a}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{b}$ are the initial values for the Sturm–Liouville problem, what are ${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{and}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$?

In Section 3.6, we discussed the improved Euler’s method for approximating the solution to a first-order equation. Extend this method to normal systems and give the recursive formulas for solving the initial value problem.

In Problems 10–13, use the vectorized Euler method with = 0.25 to find an approximation for the solution to the given initial value problem on the specified interval.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{]}$

Predator-Prey Model. The Volterra–Lotka predator-prey model predicts some rather interesting behavior that is evident in certain biological systems. For example, suppose you fix the initial population of prey but increase the initial population of predators. Then the population cycle for the prey becomes more severe in the sense that there is a long period of time with a reduced population of prey followed by a short period when the population of prey is very large. To demonstrate this behavior, use the vectorized Runge–

Kutta algorithm for systems with $\mathbf{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$ to approximate the populations of prey x and of predators y over the period [0, 5] that satisfy the Volterra–Lotka system $\mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{(}\mathbf{3}\mathbf{-}\mathbf{y}\mathbf{)}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{)}$

 under each of the following initial conditions:

$$\begin{gathered} \left(\mathbf{a}\mathbf{\left)}\mathbf{x}\mathbf{\right(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{4} \\ \right(\mathbf{b}\mathbf{\left)}\mathbf{x}\mathbf{\right(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{5} \\ (\mathbf{c}\mathbf{\left)}\mathbf{x}\mathbf{\right(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{7} \end{gathered}$$