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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 9, Problem 9.4.44a

Solving Bernoulli equations Use the method outlined in Exercise 43 to solve the following Bernoulli equations.


a. y′(t) + y = 2y²

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Identify the Bernoulli equation in the form \(y'(t) + P(t)y = Q(t)y^n\). Here, the equation is \(y'(t) + y = 2y^2\), so \(P(t) = 1\), \(Q(t) = 2\), and \(n = 2\).
Divide the entire equation by \(y^n = y^2\) (assuming \(y \neq 0\)) to rewrite it as \(y'(t) y^{-2} + y y^{-2} = 2\), which simplifies to \(y'(t) y^{-2} + y^{-1} = 2\).
Make the substitution \(z = y^{1-n} = y^{1-2} = y^{-1}\). Then, compute \(z'(t)\) in terms of \(y'(t)\): since \(z = y^{-1}\), we have \(z' = -y^{-2} y'\).
Rewrite the original equation in terms of \(z\) and \(z'\). Using \(z' = -y^{-2} y'\), rearrange to express \(y'(t) y^{-2} = -z'\). Substitute into the equation from step 2 to get \(-z' + z = 2\).
Rearrange the equation to the linear form \(z' - z = -2\). Solve this first-order linear differential equation for \(z(t)\) using an integrating factor, then substitute back \(y = z^{-1}\) to find the solution for \(y(t)\).

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Key Concepts

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

Bernoulli Differential Equation

A Bernoulli differential equation is a first-order nonlinear ODE of the form y' + P(x)y = Q(x)y^n, where n ≠ 0 or 1. It can be transformed into a linear differential equation by an appropriate substitution, making it easier to solve.
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Substitution Method for Bernoulli Equations

To solve a Bernoulli equation, use the substitution v = y^(1-n), which converts the nonlinear equation into a linear one in terms of v. This allows the use of standard methods for linear ODEs to find the solution.
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Euler's Method

Solving Linear First-Order Differential Equations

Once transformed, the equation becomes linear and can be solved using an integrating factor. The integrating factor is typically e^(∫P(x)dx), which simplifies the equation to an exact derivative, enabling integration and solution.
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Related Practice
Textbook Question

Growth rate functions

a. Show that the logistic growth rate function f(P)=rP(1−P/K) has a maximum value of rK/4 at the point P=K/2.

Textbook Question

{Use of Tech} Logistic equation for an epidemic When an infected person is introduced into a closed and otherwise healthy community, the number of people who contract the disease (in the absence of any intervention) may be modeled by the logistic equation

dP/dt=kP(1−P/A),P0=P_0,

where K is a positive infection rate, A is the number of people in the community, and P0 is the number of infected people at t=0. The model also assumes no recovery.

a. Find the solution of the initial value problem, for t≥0, in terms of K, A, and P0.

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Textbook Question

U.S. population projections According to the U.S. Census Bureau, the nation’s population (to the nearest million) was 296 million in 2005 and 321 million in 2015. The Bureau also projects a 2050 population of 398 million. To construct a logistic model, both the growth rate and the carrying capacity must be estimated. There are several ways to estimate these parameters. Here is one approach:


a. Assume t = 0 corresponds to 2005 and that the population growth is exponential for the first ten years; that is, between 2005 and 2015, the population is given by P(t) = P(0)exp(rt). Estimate the growth rate r using this assumption.

Textbook Question

A second-order equation Consider the differential equation y''(t) - k²y(t) = 0 where k > 0 is a real number.


a. Verify by substitution that when k = 1, a solution of the equation is y(t) = C₁eᵗ + C₂e⁻ᵗ. You may assume this function is the general solution.

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 500-L tank is initially filled with pure water. A copper sulfate solution with a concentration of 20 g/L flows into the tank at a rate of 4 L/min. The thoroughly mixed solution is drained from the tank at a rate of 4 L/min.

Textbook Question

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

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