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Exercises 14.9 Exercises

πŸ’‘ Conceptual Quiz.

Answer the following questions to test your understanding of linear systems.

πŸ“: Abbreviations.

1. True-False.

(a) πŸ“–β“ Which Variable Drives Which?

Consider the system:
\begin{align*} \frac{dx}{dt} \amp = x - y\\ \frac{dy}{dt} \amp = -2y \end{align*}
Select all true statements:
  • \(x\) evolves independently of \(y\text{.}\)
  • \(y\) depends on \(x\text{.}\)
  • This is an uncoupled system.
  • Not quite β€” notice that \(y\) shows up in the \(dx/dt\) equation.
  • You must solve for \(y\) before you can solve for \(x\text{.}\)
  • Yes β€” solve \(y\) first (it is independent), then use it to solve for \(x\text{.}\)

(b) Interdependence.

What makes a system fully coupled? How does that affect how we approach the problem?

(c) πŸ“–β“ Classify the System.

(d) πŸ“–β“ Identifying Features.

Select all statements that are true about partially coupled systems.
  • One equation is independent and can be solved first.
  • Yes, this is a defining feature of partial coupling.
  • Both equations must be solved simultaneously.
  • No, only fully coupled systems require that.
  • The dependent equation uses the solution of the independent one.
  • Exactly. You substitute \(x(t)\) into the second equation to solve for \(y(t)\text{.}\)
  • Partial coupling means both variables evolve independently.
  • That describes an uncoupled system, not a partially coupled one.
  • Only one of the equations involves both variables.
  • Correct. The coupling only appears in one direction.
  • The second variable can be solved without knowing the first.
  • Nope, you need \(x(t)\) to solve for \(y(t)\text{.}\)

(e) πŸ“–β“ Select the True Statements.

Which of the following statements are true for the system:
\begin{align*} \frac{dx}{dt} \amp = -x + 1 \\ \frac{dy}{dt} \amp = -2y \end{align*}
  • The rate of change of \(x\) depends on \(y\text{.}\)
  • The variable \(x\) has no effect on how \(y\) changes.
  • The rate of change of \(y\) depends only on \(y\text{.}\)
  • This is an example of a coupled system.
  • This system has two independent variables.

(f) πŸ“–β“ Independent Solutions.

In an uncoupled system, how do we find the solution for the whole system?
  • Solve each equation separately, then combine the answers into a pair \((x(t), y(t))\text{.}\)
  • Rewrite the system as a second-order equation for just one variable.
  • You could do this in some cases, but for uncoupled systems, it’s simpler to solve each one directly.
  • Use Euler’s Method only β€” they cannot be solved exactly.
  • Uncoupled systems are straightforward to solve exactly.

✍🏻 The Eigenvalue Method.

Practice the eigenvalue method from Solving Linear Systems: find eigenvalues and eigenvectors, build general solutions, and classify the phase portrait using Connecting Eigenvalues to Phase Portraits.

2.

Find the eigenvalues and eigenvectors of \(A = \begin{bmatrix} 4 \amp 2 \\ 1 \amp 3 \end{bmatrix}\text{.}\)
Answer.
\(r_1 = 2\) with \(\vec{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\text{;}\) \(r_2 = 5\) with \(\vec{v}_2 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\)

3.

Find the general solution of the system and classify its phase portrait:
\begin{align*} \frac{dx}{dt} \amp = x + 2y \\ \frac{dy}{dt} \amp = 3x \end{align*}
Answer.
\(\vec{X}(t) = C_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{3t} + C_2 \begin{bmatrix} 2 \\ -3 \end{bmatrix} e^{-2t}\text{;}\) saddle

4.

Solve the initial value problem
\begin{align*} \frac{dx}{dt} \amp = x + y, \amp x(0) \amp = 3, \\ \frac{dy}{dt} \amp = 4x + y, \amp y(0) \amp = 2. \end{align*}
Answer.
\(x(t) = 2e^{3t} + e^{-t},\quad y(t) = 4e^{3t} - 2e^{-t}\)

5.

Find the general solution of the system and classify its phase portrait:
\begin{align*} \frac{dx}{dt} \amp = -x - 2y \\ \frac{dy}{dt} \amp = 2x - y \end{align*}
Answer.
\(x(t) = e^{-t}(C_1\cos 2t + C_2\sin 2t),\quad y(t) = e^{-t}(C_1\sin 2t - C_2\cos 2t)\text{;}\) spiral sink

6.

Without solving, classify the phase portrait of a linear system whose coefficient matrix has the given eigenvalues:
  1. \(\displaystyle r = -3,\ -1\)
  2. \(\displaystyle r = -2,\ 5\)
  3. \(\displaystyle r = \pm 3i\)
  4. \(\displaystyle r = 1 \pm 2i\)
Answer.
(a) node (sink); (b) saddle; (c) center; (d) spiral source

Exercise Group.

Solve the following systems of differential equations. The worked solutions below take a different road than the eigenvalue method: they apply the Laplace transform from the previous chapters to each equation, turning the system into simultaneous algebraic equations. Both methods produce the same solutions β€” as you work through these, notice how the exponential rates that appear are exactly the eigenvalues of each system’s coefficient matrix.

7.

\(\left\{ \begin{array}{rll} \dfrac{dx}{dt} \amp = -x+y,\amp x(0) = 0,\\ \dfrac{dy}{dt} \amp = 2x, \amp y(0) = 1. \end{array} \right.\)
Answer.
\(x(t) = \frac{1}{3}e^t - \frac{1}{3}e^{-2t},\quad y(t) = \frac{1}{3}e^{-2t} + \frac{2}{3}e^t \)

8.

\(\left\{ \begin{array}{rll} \dfrac{dx}{dt} \amp = x - 2y,\amp x(0) = -1,\\ \dfrac{dy}{dt} \amp = 5x - y,\amp y(0) = 2. \end{array} \right.\)
Answer.
\(x(t) = -\cos(3t) - \frac{5}{3} \sin(3t),\quad y(t) = 2\cos(3t) - \frac{7}{3}\sin(3t) \)

9.

\(\left\{ \begin{array}{rll} y' - 2x = 1, \amp x(0) = 0,\\ x' + y' - 3x - 3y = 2, \amp y(0) = 0. \end{array} \right.\)
Answer.
\(x(t) = -\frac{1}{2} + \frac{2}{5}e^{3t} + \frac{1}{10}e^{-2t},\quad y(t) = -\frac{1}{6}+ \frac{4}{15}e^{3t} - \frac{1}{10} e^{-2t} \)

10.

Suppose a mixture containing 0.3 kg of sugar per liter runs into a tank initially filled with 400 L of water containing 2 kg of sugar. The liquid enters at 10 L/min,the mixture is kept uniform by stirring, and the mixture flows out at the same rate.
Answer the following:
  1. Find the mass of sugar in the tank after \(t \) minutes.
  2. Find the concentration of sugar in the tank after 10 minutes.
  3. Use your answer to question to determine the concentration of sugar in the tank after a long time. Does your answer make sense in terms of the physical scenario? Explain.
Answer.
(a) \(S(t) = 120 - 118e^{-t/40}\) kg; (b) \(\approx 0.070\) kg/L; (c) \(0.3\) kg/L, the inflow concentration.

11.

A brine solution of salt flows at a constant rate of 4 L/min into a large tank that initially holds 100 L of pure water. The solution inside the tank is kept well-stirred and flows out of the tank at a rate of 3 L/min. Suppose the concentration of salt in the brine entering the tank is \(0.2\) kg/L. Then
  1. Determine the mass of the salt in the tank after \(t \) minutes.
  2. What is the concentration of salt after one hour?
Answer.
(a) \(\ds S(t) = 0.2(100+t) - \frac{2\times 10^7}{(100+t)^3}\) kg; (b) \(\approx 0.17\) kg/L.

12. From a Second-Order Equation to a System.

Rewrite the second-order equation
\begin{equation*} y'' - y = 0 \end{equation*}
as a system of first-order equations.
Answer.
\(u' = v, \quad v' = u\)

Exercise Group.

Two large tanks, each holding 100 L of liquid, are interconnected by pipes, with the liquid flowing from tank 1 into tank 2 at 3 L/min and from tank 2 into tank 1 at 1 L/min. The liquid in each tank is kept well stirred. A brine solution with a salt concentration of 0.01 kg/L flows into tank 1 at a rate of 4 L/min. The (diluted) solution flows out of the system from tank 1 at 2 L/min and from tank 2 at 2 L/min.Initially, tank 1 contains pure water and tank 2 contains 6 kg of salt.

13.

Write a system of differential equations for\(x_1(t) \) (the amount of salt, in kg, in tank 1 at time \(t \) )and \(x_2(t) \) (the amount of salt, in kg, in tank 2 at time \(t \) ).
Answer.
\(x'_1 =- \frac{1}{20}x_1+ \frac{1}{100}x_2 + 0.4,x'_2 = \frac{3}{100}x_1 - \frac{3}{100}x_2 \)

14.

How many initial conditions will you need to find a particular solution? Write the initial conditions.
Answer.
two \(x_1(0) = 0 \) and \(x_2(0) = 6 \)

15.

Why can’t we solve each of the DEs you found individually using techniques we already know?
Answer.
\(x_1(t) \) and \(x_2(t) \) are both unknown functions, and each DE contains BOTH of the unknowns

16.

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{align*} \frac{dx}{dt} \amp = 2y - x + a \amp x(0) = 1\\ \frac{dy}{dt} \amp = 4x - 3y + b \amp y(0) = 4 \end{align*}
where \(a \) and \(b \) are constants that measure the trust (or distrust!) each country has for the other.
  1. What might be appropriate units for \(x(t) \) and \(y(t)\text{?}\)
  2. How can we interpret the fact that the coefficient of \(y \) is positive in the equation for \(\frac{dx}{dt}\text{?}\) Does it make sense? Would the DE make sense if the coefficient were negative instead?
  3. How can we interpret the fact that the coefficient of \(y \) is negative in the equation for \(\frac{dy}{dt}\text{?}\) Does it make sense? Would the DE make sense if the coefficient were positive instead?
  4. Suppose country X trusts country Y, but country Y doesn’t trust country X. What can we say about \(a\) and \(b\text{?}\) Should they be positive, negative, or zero?
  5. Rewrite the system of differential equations in matrix form by using the vector \(\vec{X} = \begin{bmatrix} x\\y \end{bmatrix}.\) Notice that for this system, the right hand side should be of the form \(A\vec{X} + \vec{B}.\)
  6. Write the initial conditions as a vector as well. That is, complete the following. \(\vec{X}(0) = \begin{bmatrix} ?\\? \end{bmatrix}\)
Answer.
(e) \(\ds \frac{d\vec{X}}{dt} = \begin{bmatrix} -1 \amp 2 \\ 4 \amp -3 \end{bmatrix}\vec{X} + \begin{bmatrix} a \\ b \end{bmatrix}\text{;}\) (f) \(\ds \vec{X}(0) = \begin{bmatrix} 1 \\ 4 \end{bmatrix}\)

Exercise Group.

Work through the following problems.

17.

Suppose compartments A and B are filled with fluids and are separated by a permeable membrane. The figure is a compartmental representation of the exterior and interior of a cell. Suppose, too, that a nutrient necessary for cell growth passes through the membrane.
Two-compartment diagram of a cell interior and exterior separated by a permeable membrane, with a nutrient passing through.
Figure 315. A two-compartment model of a cell: compartments \(A\) and \(B\) separated by a permeable membrane through which the nutrient passes.
A model for the amounts \(x(t)\) and \(y(t)\) of the nutrient in compartments A and B, respectively, at time \(t\) is given by the linear system of differential equations:
\begin{align*} \frac{dx}{dt} \amp = \kappa\left(\frac{y}{V_B}-\frac{x}{V_A}\right), \\ \frac{dy}{dt} \amp = \kappa\left(\frac{x}{V_A}-\frac{y}{V_B}\right). \end{align*}
where \(V_A\) and \(V_B\) are the volumes of the compartments (so \(\sfrac{x}{V_A}\) and \(\sfrac{y}{V_B}\) are the concentrations), and \(\kappa > 0 \) is a permeability factor. Let \(x(0) = x_0\) and \(y(0) = y_0\text{.}\)
  1. In the first equation, explain why it makes sense for \(y\) to have a positive coefficient and \(x \) to have a negative coefficient.
  2. Notice that if we add these two equations, we get \(\frac{dx}{dt} + \frac{dy}{dt} = 0, \) or, equivalently, \(\frac{d}{dt}[x+y] = 0.\) Explain what this means and why it makes sense.
  3. Rewrite the system of differential equations in matrix form by using the vector \(\vec{X} = \begin{bmatrix} x\\y \end{bmatrix}.\)
  4. Write the initial conditions as a vector as well. That is, complete the following. \(\vec{X}(0) = \begin{bmatrix} ?\\? \end{bmatrix}\)
Answer.
(c) \(\ds \frac{d\vec{X}}{dt} = \kappa\begin{bmatrix} -\sfrac{1}{V_A} \amp \sfrac{1}{V_B} \\ \sfrac{1}{V_A} \amp -\sfrac{1}{V_B} \end{bmatrix}\vec{X}\text{;}\) (d) \(\ds \vec{X}(0) = \begin{bmatrix} x_0 \\ y_0 \end{bmatrix}\)

18.

A tank contains 100 liters of water. Salt is being added at a rate of 5 grams per minute. If the mixture is kept uniform by stirring and salt is being removed from the tank at a rate of 2 liters per minute, set up the differential equation that models the amount of salt \(y(t)\) in the tank at any time \(t\).
Answer.
\begin{equation*} \frac{dy}{dt} \amp = 5 - 2 \frac{y(t)}{100 - 2t} \end{equation*}

19. Euler’s Method for Systems.

Given the initial-valued system
\begin{align*} y' \amp = 3z,\quad \amp y(2) \amp = 3, \\ z' \amp = y+z^2, \amp z(2) \amp = -4 \end{align*}
approximate \(y(2.2)\) and \(z(2.2)\) using two iterations of Euler’s method with a step size \(h = 0.1\text{.}\)
Answer.
\(y(2.2) \approx 1.17\text{,}\) \(z(2.2) \approx -1.479\)

20. Euler’s Method for Systems.

Consider the initial-valued system
\begin{align*} x' \amp = 2x - y + t, \amp x(0) = 6, \\ y' \amp = x, \amp y(0) = 2. \end{align*}
Approximate \(x(0.2) \) and \(y(0.2) \) using \(h = 0.1\text{.}\)
Answer.
\(x(0.2) \approx 8.15 \) \(\hspace{0.5cm}y(0.2) \approx 3.3. \)

✍🏻 Reduction of Order.

Reduce each of the following to a system of first-order differential equations.

22.

\(x^2 y''' = 2y' \)
Answer.
\(u' = v,\quad v' = w,\quad w' = \dfrac{2v}{x^2}\) where \(u = y\text{,}\) \(v = y'\text{,}\) and \(w = y''\)

23.

\(y'' + x = 12,\quad x' + 2yt = \cos t \)
Answer.
\(u' = v,v' = 12 - w,w' = \cos t - 2ut \) where \(u = y,v = y', \) and \(w = x \)

24. Euler’s Method for Higher Order Systems.

Use Euler’s method to approximate \(y(0.2)\) for the solution to the IVP
\begin{equation*} y'' + 2y' + y^2 = 0,\quad y(0) = 1,\quad y'(0) = 2 \end{equation*}
using a step size of \(h = 0.1\text{.}\)
Answer.
\(y(0.2) \approx 1.35 \)

25. Euler’s Method.

Approximate the solution to the IVP
\begin{equation*} y'' - 2y = e^{t-3}\cos t,\quad y(3) = -1,\quad y'(3) = 0 \end{equation*}
at \(t = 3.2\) using one iteration of Euler’s Method with \(h = 0.2\text{.}\)
Answer.
\(y(3.2) \approx -1\)

26. Euler’s Method.

Approximate \(y(3.2)\) if \(y\) is the solution to IVP:
\begin{equation*} y'' - 2y = e^{t-3} \cos t,\quad y(3) = -1,\quad y'(3) = 0. \end{equation*}
Answer.
\(y(3.2) \approx -1.0299 \)
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