Ch. 9 - Differential Equations

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 9 - Differential Equations

Problem 9.3.42b

Chapter 9, Problem 9.3.42b

42–43. Implicit solutions for separable equations For the following separable equations, carry out the indicated analysis.
b. Find the value of the arbitrary constant associated with each initial condition. (Each initial condition requires a different constant.)

y'(t) = t²/(y² + 1); y(−1) = 1, y(0) = 0, y(−1) = −1

Verified step by step guidance

Start with the given differential equation: \(y'(t) = \frac{t^{2}}{y^{2} + 1}\). Recognize that this is a separable differential equation, meaning you can rewrite it to separate variables \(y\) and \(t\) on opposite sides.

Rewrite the equation by expressing \(y'(t)\) as \(\frac{dy}{dt}\) and separate variables: multiply both sides by \((y^{2} + 1) dt\) to get \((y^{2} + 1) dy = t^{2} dt\).

Integrate both sides: \(\int (y^{2} + 1) dy = \int t^{2} dt\). This will give you an implicit solution involving \(y\) and \(t\) plus an arbitrary constant \(C\).

After integrating, you will have an equation of the form \(\frac{y^{3}}{3} + y = \frac{t^{3}}{3} + C\). Use each initial condition to find the corresponding constant \(C\) by substituting the given \(t\) and \(y\) values into this equation.

Solve for \(C\) for each initial condition: for \(y(-1) = 1\), \(y(0) = 0\), and \(y(-1) = -1\), plug in the values and isolate \(C\). This gives you the specific constant for each initial condition.

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Key Concepts

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

Separable Differential Equations

A separable differential equation can be written as dy/dt = g(t)h(y), allowing the variables y and t to be separated on opposite sides of the equation. This enables integration of each side independently to find an implicit or explicit solution. Recognizing separability is key to solving the given equation.

Initial Conditions and Arbitrary Constants

When solving differential equations, integration introduces an arbitrary constant representing a family of solutions. Applying initial conditions, such as y(t₀) = y₀, allows determination of the specific constant that fits each condition, yielding a unique solution curve for each initial value.

Implicit Solutions

Sometimes, after integrating, the solution cannot be explicitly solved for y in terms of t. Instead, the solution is given implicitly as a relation involving both variables and the constant of integration. Understanding how to work with implicit solutions is essential for interpreting and applying the results.

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Related Practice

Textbook Question

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

b. Using the exact solution given, compute the errors in the Euler approximations at t=0.2 and t=0.4.

y′(t) = −y, y(0) = 1; y(t) = e⁻ᵗ

Textbook Question

Blowup in finite time Consider the initial value problem y'(t) = yⁿ + 1, y(0) = y₀, where n is a positive integer.

b. Solve the initial value problem with n = 2 and y₀ = 1/√2.

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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.

b. Sketch the direction field, for t≥0.

y′(t) = 6 - 2y

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Textbook Question

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} Drug infusion The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation m'(t) + km(t) = I, where m(t) is the mass of the drug in the blood at time t ≥ 0, k is a constant that describes the rate at which the drug is absorbed, and I is the infusion rate.

b. Graph the solution for I = 10 mg/hr and k = 0.05 hr⁻¹.

Textbook Question

{Use of Tech} Chemical rate equations Let y(t) be t he concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation dy/dt=-ky^n for t≥0, where k>0 is a rate constant and the positive integer n is the order of the reaction.

b. Solve the initial value problem for a second-order reaction (n=2) assuming y(0)=y0.

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Textbook Question

{Use of Tech} Torricelli’s law An open cylindrical tank initially filled with water drains through a hole in the bottom of the tank according to Torricelli’s law (see figure). If h(t) is the depth of water in the tank for t≥0 s, then Torricelli’s law implies h′(t)=−k√h, where k is a constant that includes g=9.8m/s², the radius of the tank, and the radius of the drain. Assume the initial depth of the water is h(0)=Hm.

b. Find the solution in k=0.1the case that and H=0.5m.

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