Ch. 9 - Differential Equations

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 9 - Differential Equations

Problem 9.2.43a

Chapter 9, Problem 9.2.43a

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(y - 3)(y + 2)

Verified step by step guidance

Identify the given autonomous differential equation: $y'(t) = y(y - 3)(y + 2)$.

Recall that equilibrium solutions occur where the derivative $y'(t)$ is zero for all $t$, meaning $f(y) = 0$.

Set the right-hand side equal to zero: $y(y - 3)(y + 2) = 0$.

Solve the equation $y(y - 3)(y + 2) = 0$ by finding the values of $y$ that make each factor zero.

The solutions to this equation give the equilibrium solutions $y = 0$, $y = 3$, and $y = -2$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Autonomous Differential Equations

An autonomous differential equation is one where the derivative y' depends only on the variable y, not explicitly on the independent variable t. This means the rate of change of y depends solely on y itself, simplifying analysis and allowing the use of phase line methods.

Equilibrium Solutions

Equilibrium solutions occur when y' = 0, meaning the function y(t) remains constant over time. For autonomous equations, these are found by solving f(y) = 0. Graphically, equilibrium solutions correspond to horizontal lines in the direction field where the slope is zero.

Factoring and Solving Polynomial Equations

To find equilibrium points in equations like y' = y(y - 3)(y + 2), factor the right-hand side and set each factor equal to zero. Solving these polynomial equations yields the constant values of y where the system is in equilibrium.

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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample

b. If k>0 and b>0 then y(t)=0 is never a solution of y'(t)=ky−b.

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Textbook Question

29–32. {Use of Tech} Errors in Euler’s method Consider the following initial value problems.

a. Find the approximations to y(0.2) and y(0.4) using Euler’s method with time steps of Δt = 0.2, 0.1, 0.05, and 0.025.

y′(t) = y/2, y(0) = 2; y(t) = 2eᵗᐟ²

Textbook Question

33–36. {Use of Tech} Computing Euler approximations Use a calculator or computer program to carry out the following steps.

a. Approximate the value of y(T) using Euler’s method with the given time step on the interval [0,T].

y′(t) = t/y, y(0) = 4; Δt = 0.1, T = 2; y(t) = √(t² + 16)

Textbook Question

23–26. Stirred tank reactions For each of the following stirred tank reactions, carry out the following analysis.

a. Write an initial value problem for the mass of the substance.

A one-million-liter pond is contaminated by a chemical pollutant with a concentration of 20 g/L. The source of the pollutant is removed, and pure water is allowed to flow into the pond at a rate of 1200 L/hr. Assuming the pond is thoroughly mixed and drained at a rate of 1200 L/hr, how long does it take to reduce the concentration of the solution in the pond to 10% of the initial value?

Textbook Question

A bad loan Consider a loan repayment plan described by the initial value problem

B'(t)=0.03B−600,B(0)=40,000,

where the amount borrowed is B(0)=\$40,000, the monthly payments are \$600, and B(t) is the unpaid balance in the loan.

b. What is the most that you can borrow under the terms of this loan without going further into debt each month?

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Textbook Question

Convergence of Euler's method Suppose Euler's method is applied to the initial value problem y′(t) = ay, y(0) = 1, which has the exact solution y(t) = eᵃᵗ. For this exercise, let h denote the time step (rather than Δt). The grid points are then given by tₖ = kh. We let uₖ be the Euler approximation to the exact solution y(tₖ), for k = 0, 1, 2, ...

b. Show by substitution that uₖ = (1 + ah)ᵏ is a solution of the equations in part (a), for k = 0, 1, 2, ...