50-55. 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 = yo is an equilibrium solution of the equation provided F(yo) = 0 (because then y'(t) = 0, and the solu- tion remains constant for all t). Note that equilibrium solutions cor- respond to horizontal line segments in the direction field. Note also that for autonomous equations, the direction field is independent of t. Consider the following equations. a. Find all equilibrium solutions. b. Sketch the direction field on either side of the equilibrium solutions for t z 0. c. Sketch the solution curve that corresponds to the initial condition y(0) = 1. 50. y'(t) = 2y + 4 51. у'() — у? 52. y' () — у(2 — у) 53. y'() — У(у -3)

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50-55. 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 = yo is an equilibrium solution of the
equation provided F(yo) = 0 (because then y'(t) = 0, and the solu-
tion remains constant for all t). Note that equilibrium solutions cor-
respond to horizontal line segments in the direction field. Note also
that for autonomous equations, the direction field is independent of t.
Consider the following equations.
a. Find all equilibrium solutions.
b. Sketch the direction field on either side of the equilibrium solutions
for t z 0.
c. Sketch the solution curve that corresponds to the initial condition
y(0) = 1.
50. y'(t) = 2y + 4
51. у'() — у?
52. y' () — у(2 — у)
53. y'() — У(у -3)
Transcribed Image Text:50-55. 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 = yo is an equilibrium solution of the equation provided F(yo) = 0 (because then y'(t) = 0, and the solu- tion remains constant for all t). Note that equilibrium solutions cor- respond to horizontal line segments in the direction field. Note also that for autonomous equations, the direction field is independent of t. Consider the following equations. a. Find all equilibrium solutions. b. Sketch the direction field on either side of the equilibrium solutions for t z 0. c. Sketch the solution curve that corresponds to the initial condition y(0) = 1. 50. y'(t) = 2y + 4 51. у'() — у? 52. y' () — у(2 — у) 53. y'() — У(у -3)
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