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 . 43. y ′ ( t ) = y ( y – 3) ( y + 2)
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 . 43. y ′ ( t ) = y ( y – 3) ( y + 2)
Solution Summary: The author explains that the equilibrium solution of the given differential equation is y(t)=3,
Equilibrium solutionsA 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 = y0is 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.
43. y′ (t) = y (y – 3) (y + 2)
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.
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
[2] Determine the roots of the auxiliary equation y=x2-e2*cosx for a
homogeneous linear equation with real, constant coefficients.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
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