In Exercises 26 through 32, solve the system with the given initial value

Solve the initial value problems posed in Exercises 1 through 5. Graph the solution. = 5x with jc(0) = 7 dt

Solve the initial value problems posed in Exercises 1 through 5. Graph the solution.dx = 0.71 jc with jc(0) = -e dt

Solve the initial value problems posed in Exercises 1 through 5. Graph the solution.dP = 0.03 P with P(0) = 7 dt

Solve the initial value problems posed in Exercises 1 through 5. Graph the solution.dy = 0.81 with v(0) = 0.8

Solve the initial value problems posed in Exercises 1 through 5. Graph the solution.^ = 0.8y with v(0) = -0.8 dt

Solve the nonlinear differential equations in Exercises 6 through 11 using the method of separation of variables: Write the differential equation dx/dt = / ( jc) as dx/f{x) = dt and integrate both sides. = ! ., = I dt x

Solve the nonlinear differential equations in Exercises 6 through 11 using the method of separation of variables: Write the differential equation dx/dt = / ( jc) as dx/f{x) = dt and integrate both sides.dx t = jc , jc(0) = 1. Describe the behavior of your solution as t increases.

Solve the nonlinear differential equations in Exercises 6 through 11 using the method of separation of variables: Write the differential equation dx/dt = / ( jc) as dx/f{x) = dt and integrate both sides.^ = yfx, jc(0) = 4 dt

Solve the nonlinear differential equations in Exercises 6 through 11 using the method of separation of variables: Write the differential equation dx/dt = / ( jc) as dx/f{x) = dt and integrate both sides.^ = xk (with k ^ 1), jc(0) = 1 dt

Solve the nonlinear differential equations in Exercises 6 through 11 using the method of separation of variables: Write the differential equation dx/dt = / ( jc) as dx/f{x) = dt and integrate both sides.dx 1 - = 7 T *() = dt cos(jc)

Solve the nonlinear differential equations in Exercises 6 through 11 using the method of separation of variables: Write the differential equation dx/dt = / ( jc) as dx/f{x) = dt and integrate both sides.^ = 1 4- x2, jc(0) = 0

Find a differential equation of the form dx/dt = kx for which jc(r) = 3' is a solution.

In 1778, a wealthy Pennsylvanian merchant named Jacob DeHaven lent $450,000 to the Continental Congress to support the troops at Valley Forge. The loan was never repaid. Mr. DeHavens descendants have taken the U.S. government to court to collect what they believe they are owed. The going interest rate at the time was 6%. How much were the DeHavens owed in 1990 a. if interest is compounded yearly? b. if interest is compounded continuously? (Source: Adapted from The New York Times, May 27, 1990.)

The carbon in living matter contains a minute proportion of the radioactive isotope C-14. This radiocarbon arises from cosmic-ray bombardment in the upper atmosphere and enters living systems by exchange processes. After the death of an organism, exchange stops, and the carbon decays. Therefore, carbon dating enables us to calculate the time at which an organism died. Let jc (/) be the proportion of the original C-14 still present t years after death. By definition, jc(0) = 1 = 100%. We are told that x(t) satisfies the differential equation dx ~di 1 8270 x. a. Find a formula for x(t). Determine the half-life of C-14 (that is, the time it takes for half of the C-14 to decay). b. The Iceman. In 1991, the body of a man was found in melting snow in the Alps of Northern Italy. A well- known historian in Innsbruck, Austria, determined that the man had lived in the Bronze Age, which started about 2000 B.C. in that region. Examination of tissue samples performed independently at Zurich and Oxford revealed that 47% of the C-14 present in the body at the time of his death had decayed. When did this man die? Is the result of the carbon dating compatible with the estimate of the Austrian historian?

Justify the Rule of 69: If a quantity grows at a constant instantaneous rate of k%, then its doubling time is about 69/k. Example: In 2008 the population of Madagascar was about 20 million, growing at an annual rate of about 3%, with a doubling time of about 69/3 = 23 years. Consider the system dx ~di Xx 0 0 X2

For the values ofk\ and X2 given in Exercises 16 through 19, sketch the trajectories for all nine initial values shown in the following figure. For each of the points, trace out both the future and the past of the system.

For the values ofk\ and X2 given in Exercises 16 through 19, sketch the trajectories for all nine initial values shown in the following figure. For each of the points, trace out both the future and the past of the system.

For the values ofk\ and X2 given in Exercises 16 through 19, sketch the trajectories for all nine initial values shown in the following figure. For each of the points, trace out both the future and the past of the system.

For the values ofk\ and X2 given in Exercises 16 through 19, sketch the trajectories for all nine initial values shown in the following figure. For each of the points, trace out both the future and the past of the system.

Consider the system dx/dt = Ax with A = ^ ^ Sketch a direction field for Ax. Based on your sketch, de scribe the trajectories geometrically. From your sketch, can you guess a formula for the solution with 5o = ^ Verify your guess by substituting into the equations.

Consider the system dx/dt = Ax with A = 0 0 Sketch a direction field of Ax. Based on your sketc describe the trajectories geometrically. Can you find the solutions analytically?

Consider a linear system dx/dt = Ax of arbitrary size. Suppose x\(t) and *2(0 are solutions of the system. Is the sum x(t) = x\ (t) + x2(t) a solution as well? How do you know?

Consider a linear system dx/dt = Ax of arbitrary size. Suppose x\(t) is a solution of the system and k is an arbitrary constant. Is 5(r) = kx 1 (t) a solution as well? How do you know?

Let A be an nxn matrix and k a scalar. Consider the following two systems: dt dc dt = (A + kln)c. (I) (U) Show that if 5(f) is a solution of system (I), then c(t) = ektx(t) is a solution of system (II).

Let A be an n x n matrix and k a scalar. Consider the following two systems: f = A5- dt dc = kAc. dt (I) (II) Show that if 5(f) is a solution of system (I), then c(t) = x(kt) is a solution of system (II). Compare the vector fields of the two systems.

In Exercises 26 through 32, solve the system with the given initial value.

In Exercises 26 through 32, solve the system with the given initial value.

In Exercises 26 through 32, solve the system with the given initial value.

In Exercises 26 through 32, solve the system with the given initial value.

In Exercises 26 through 32, solve the system with the given initial value.

In Exercises 26 through 32, solve the system with the given initial value.

In Exercises 26 through 32, solve the system with the given initial value. Sketch rough phase portraits for the dynamical systems given in Exercises 32 through 39.

Sketch rough phase portraits for the dynamical systems given in Exercises 32 through 39.

Sketch rough phase portraits for the dynamical systems given in Exercises 32 through 39.

Sketch rough phase portraits for the dynamical systems given in Exercises 32 through 39.

Sketch rough phase portraits for the dynamical systems given in Exercises 32 through 39.

Sketch rough phase portraits for the dynamical systems given in Exercises 32 through 39.

Sketch rough phase portraits for the dynamical systems given in Exercises 32 through 39.

Sketch rough phase portraits for the dynamical systems given in Exercises 32 through 39.

Find a 2 x 2 matrix A such that the system dx/dt = Ax has 5(f) = as one of its solutions.

Consider a noninvertible 2x2 matrix A with two distinct eigenvalues. (Note that one of the eigenvalues must be 0.) Choose two eigenvectors ui and V 2 with eigenvalues k\ = 0 and A .2 as shown in the accompanying figure. Suppose A-2 is negative. Sketch a phase portrait for the system dx/dt = Ax> clearly indicating the shape and long-term behavior of the trajectories.

Consider the interaction of two species of animals in a habitat. We are told that the change of the populations j t ( f ) and y(t) can be modeled by the equations = 1.4* 1.2 y dt dy dt = 0.8* - \Ay where time f is measured in years. a. What kind of interaction do we observe (symbiosis, competition, or predator-prey)? b. Sketch a phase portrait for this system. From the nature of the problem, we are interested only in the first quadrant. c. What will happen in the long term? Does the outcome depend on the initial populations? If so, how?

Answer the questions posed in Exercise 42 for the following system: dx c = - 2 , + 4 ,

Answer the questions posed in Exercise 42 for the following system: dx 3 7 = x + i y dy . d i = l x - r

Two herds of vicious animals are fighting each other to the death. During the fight, the populations J t ( f ) and y(t) of the two species can be modeled by the following system:.2 = - 4 y dx ~dt dy_ dt a. What is the significance of the constants 4 and 1 in these equations? Which species has the more vicious (or more efficient) fighters? b. Sketch a phase portrait for this system. c. Who wins the fight (in the sense that some individuals of that species are left while the other herd is eradicated)? How does your answer depend on the initial populations?

Repeat Exercise 45 for the system dx -Ji= ~ py dy dI = ~qX where p and q are two positive constants.3

The interaction of two populations of animals is modeled by the differential equations dx - x -t- ky dt dy_ dt = kx Ay for some positive constant k. a. What kind of interaction do we observe? What is the practical significance of the constant k? b. Find the eigenvalues of the coefficient matrix of the system. What can you say about the signs of these eigenvalues? How does your answer depend on the value of the constant k ? c. For each case you discussed in part (b), sketch a rough phase portrait. What does each phase portrait tell you about the future of the two populations?

Repeat Exercise 47 for the system dx = x + ky dt - r = x 4y where k is a positive constant.

Here is a continuous model of a persons glucose regulatory system. (Compare this with Exercise 7.1.52.) Let g(t) and h(t) be the excess glucose and insulin concentrations in a persons blood. We are told that g = -g-0.2h dh = 0.6g - 0.2/z dt where time t is measured in hours. After a heavy holiday dinner, we measure g(0) = 30 and h(0) = 0. Find closed formulas for g(t) and h(t). Sketch the trajectory.

Consider a linear system dx/dt = Ax, where A isa2x2 matrix that is diagonalizable over R. When is the zero state a stable equilibrium solution? Give your answer in terms of the determinant and the trace of A.

Let x (t) be a differentiable curve in ! matrix. Show that d dx j ^ sx) = s ^

Find all solutions of the system and S an n x n dx ~dt Xy where k is an arbitrary constant. Hint: Exercises 21 and 24 are helpful. Sketch a phase portrait. For which choices of X is the zero state a stable equilibrium solution?

Solve the initial value problem dx __ p q dt ~ q p x with xq = Sketch the trajectory for the cases when p is positive, negative, or 0. In which cases does the trajectory approach the origin? (Hint: Exercises 20, 24, and 25 are helpful.)

Consider a door that opens to only one side (as most doors do). A spring mechanism closes the door automatically. The state of the door at a given time t (measured in seconds) is determined by the angular displacement 0(t) (measured in radians) and the angular velocity (o(t) = dO/dt. Note that 0 is always positive or zero (since the door opens to only one side), but co can be positive or negative (depending on whether the door is opening or closing). When the door is moving freely (nobody is pushing or pulling), its movement is subject to the following differential equations: d0 = a) (the definition of co) dt da> = - 20 - 3co dt (20 reflects the force of the spring, and 3co models friction). a. Sketch a phase portrait for this system. b. Discuss the movement of the door represented by the qualitatively different trajectories. For which initial states does the door slam (i.e., reach 0 = 0 with velocity co < 0)?

Answer the questions posed in Exercise 54 for the system dO _ dt ~ O) d(L> Tt = - Pe-qa>