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 = y 0 is an equilibrium solution of the equation provided f(y 0 ) = 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. b. Sketch the direction field, for t ≥ 0. c. Sketch the solution curve that corresponds to the initial condition y (0) = 1. 38. y ′( t ) = 2 y + 4
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 = y 0 is an equilibrium solution of the equation provided f(y 0 ) = 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. b. Sketch the direction field, for t ≥ 0. c. Sketch the solution curve that corresponds to the initial condition y (0) = 1. 38. y ′( t ) = 2 y + 4
Solution Summary: The author explains that the equilibrium solution of the given differential equation is y(t)=2y+4.
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.
b.Sketch the direction field, for t ≥ 0.
c. Sketch the solution curve that corresponds to the initial condition y (0) = 1.
38. y′(t) = 2y + 4
With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.
y" +p (t)y' +q(t)y = 0 (i),
can be put in more suitable form for finding a solution by making a change of the independent and/or dependent variables. Determine conditions on p and q that enable this equation to be transformed into an equation with constant coefficients by a change of the independent variable. Let r = u (t) be the new independent variable. It is easy to show
(島)
( + ( +p(t) + a()y = 0 (1i).
dy
that
dt
dz dy
d'y
and
2 d'y
d²z dy
dt dz
The differential equation becomes
dt dr
dt
d'y
d'z
de
dz dy
d'y
, 9 and y must be proportional. If g (t) > 0, then choose the constant of proportionality to be 1. Hence, x = u (t) = [g (t)]i dt. With a chosen this way, the coefficient of in equation (ii) is also a constant, provided that the function H =
g'(t) +2p(t)q(t)
is a
dy
In order for equation (ii) to have constant coefficients, the coefficients of
constant. Thus, equation (i) can be transformed into an equation with constant coefficients by a change of the independent variable,…
A force of 4 pounds stretches a spring 1 foot. A mass weighing 3.2 pounds is attached to the spring, and the system is then immersed in a medium that offers a damping force numerically equal to 0.4 times the
instantaneous velocity.
(a) Find the equation of motion if the mass is initially released from rest from a point 1 foot above the equilibrium position.
x(t)
=
ft
(b) Express the equation of motion in the form x(t) = Ae-¹t sin(√w² - 2²t + p), which is given in (23) of Section 3.8. (Round up to two decimal places.)
x(t) =
ft
(c) Find the first time at which the mass passes through the equilibrium position heading upward. (Round your answer to three decimal places.)
S
Assume that N(t) denotes the density of an insect species at time t and P(t) denotes the density of its
predator at time t. The insect species is an agricultural pest, and its predator is used as a biological control
agent. Their dynamics are given below by the system of differential equations. Complete parts (a) through
(c).
dN
= 7N - 5PN
dt
dP
= 4PN - P
dt
.....
(a) Explain why
dN
= 7N describes the dynamics of the insect in the absence of the predator.
dt
If there are no predators present, then P(t) =
for all t. Substitute P =
in the given differential
dN
equations to get
dt
So in the absence of the predators, the above equation describes the
dynamics of the insect population.
dN
Solve the equation,
dt
N(t) =
(Type an expression using t as the variable.)
Describe what happens to the insect population in the absence of the predator.
In the absence of the predator, the insect population
Chapter D1 Solutions
Student Solutions Manual, Single Variable for Calculus: Early Transcendentals
A Problem Solving Approach To Mathematics For Elementary School Teachers (13th Edition)
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