Back17–20. Increasing and decreasing solutions Consider the following differential equations. A detailed direction field is not needed.
d. Sketch the direction field and verify that it is consistent with parts (a)–(c).
y'(t) = (y−2)(y+1)
21–24. Logistic equations Consider the following logistic equations. In each case, sketch the direction field, draw the solution curve for each initial condition, and find the equilibrium solutions. A detailed direction field is not needed. Assume t ≥ 0 and tP ≥ 0.
P′(t) = 0.05P(1−P/800); P(0) = 100, P(0) = 400, P(0) = 700
21–24. Logistic equations Consider the following logistic equations. In each case, sketch the direction field, draw the solution curve for each initial condition, and find the equilibrium solutions. A detailed direction field is not needed. Assume t ≥ 0 and tP ≥ 0.
P′(t) = 0.05P − 0.001P²; P(0) = 10, P(0) = 40, P(0) = 80
Direction fields Consider the direction field for the equation y′=y(2−y) shown in the figure and initial conditions of the form y(0)=A.
a. Sketch a solution on the direction field with the initial condition y(0)=1.
Direction fields The direction field for the equation y′(t)=t−y, for |t|≤4 and |y|≤4, is shown in the figure.
b. Use the direction field to sketch the solution curve that passes through the point (0,−1/2).
Direction fields Consider the direction field for the equation y′=y(2−y) shown in the figure and initial conditions of the form y(0)=A.
d. For what values of A are the corresponding solutions decreasing, for t≥0
Direction field analysis Consider the first-order initial value problem y'(t)=ay+b,y(0)=A for t≥0 where a, b, and A are real numbers.
c. Draw a representative direction field in the case that a<0. Show that if A>−b/a, then the solution decreases for t≥0, and that if A<−b/a, then the solution increases for t≥0.
46–48. Analyzing models The following models were discussed in Section 9.1 and reappear in later sections of this chapter. In each case, carry out the indicated analysis using direction fields.
Drug infusion The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation m′(t)+km(t)=I, where m(t) is the mass of the drug in the blood at time t≥0, K is a constant that describes the rate at which the drug is absorbed, and I is the infusion rate. Let I=10mg/hr and k=0.05 hr^−1.
a. Draw the direction field, for 0≤t≤100, 0≤y≤600.
38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.
a. Find the equilibrium solutions.
y′(t) = 6 - 2y
38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.
a. Find the equilibrium solutions.
y′(t) = y(2 - y)
38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.
b. Sketch the direction field, for t≥0.
y′(t) = 6 - 2y