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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
All textbooksBriggs 3rd EditionCh. 9 - Differential EquationsProblem 9.1.55c
Chapter 9, Problem 9.1.55c

52-56. In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations.


(Use of Tech) Chemical rate equations The reaction of certain chemical compounds can be modeled using a differential equation of the form y'(t) = -kyⁿ(t), where y(t) is the concentration of the compound, for t ≥ 0, k > 0 is a constant that determines the speed of the reaction, and n is a positive integer called the order of the reaction. Assume the initial concentration of the compound is y(0) = y₀ > 0.


c. Let y₀ = 1 and k = 0.1. Graph the first-order and second-order solutions found in parts (a) and (b). Compare the two reactions. 

Verified step by step guidance
1
Recall the general differential equation for the chemical reaction: \(y'(t) = -k y^{n}(t)\), where \(k > 0\) and \(n\) is the order of the reaction.
For the first-order reaction (\(n=1\)), the differential equation becomes \(y'(t) = -k y(t)\). This is a separable differential equation.
Solve the first-order equation by separating variables: write \(\frac{dy}{dt} = -k y\), then rearrange to \(\frac{dy}{y} = -k dt\). Integrate both sides to find the solution involving an exponential decay.
For the second-order reaction (\(n=2\)), the differential equation is \(y'(t) = -k y^{2}(t)\). Again, separate variables: \(\frac{dy}{dt} = -k y^{2}\) leads to \(\frac{dy}{y^{2}} = -k dt\). Integrate both sides to find the solution, which will be a rational function in \(t\).
With initial condition \(y(0) = y_0 = 1\) and \(k = 0.1\), write explicit expressions for both solutions (first-order and second-order). Then, plot both functions on the same graph for \(t \geq 0\) to compare how the concentration decreases over time for each reaction order.

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Key Concepts

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

Differential Equations in Chemical Kinetics

Chemical rate equations describe how the concentration of a substance changes over time using differential equations. The general form y'(t) = -kyⁿ(t) models the rate of reaction, where y(t) is concentration, k is a positive rate constant, and n is the reaction order. Understanding this setup is essential to analyze how concentration evolves.
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Classifying Differential Equations

Order of Reaction and Its Effect

The order n in the differential equation indicates how the reaction rate depends on concentration. A first-order reaction (n=1) means the rate is proportional to y(t), leading to exponential decay. A second-order reaction (n=2) depends on the square of concentration, resulting in a different decay pattern. Comparing these helps understand reaction speed and behavior.
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Higher Order Derivatives

Solving and Graphing Differential Equations

Solving the differential equation for given initial conditions y(0) = y₀ and constants k allows us to find explicit formulas for y(t). Graphing these solutions for different orders (n=1 and n=2) visually compares how concentration changes over time, highlighting differences in reaction dynamics and decay rates.
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Solving Separable Differential Equations
Related Practice
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.

c. Sketch the solution curve that corresponds to the initial condition y0=1. 


y′(t) = 2y + 4

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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.

c. Sketch the solution curve that corresponds to the initial condition y0=1. 


y′(t) = 6 - 2y

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Textbook Question

Another second-order equation Consider the differential equation y''(t) + k²y(t) = 0, where k is a positive real number.

c. Give the general solution of the equation for arbitrary k > 0 and verify your conjecture.

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Textbook Question

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


c. y′(t) + y = √y

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Textbook Question

Cooling time Suppose an object with an initial temperature of T₀ > 0 is put in surroundings with an ambient temperature of A, where A < T₀/2. Let t₁/₂ be the time required for the object to cool to T₀/2.


c. Why is the condition A < T₀/2 needed?

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Textbook Question

{Use of Tech} Fish harvesting A fish hatchery has 500 fish at t=0, when harvesting begins at a rate of b>0fish/year The fish population is modeled by the initial value problem y′(t)=0.01y−b,y(0)=500 where t is measured in years.


c. Graph the solution in the case that b=60fish/year. Describe the solution.

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