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
Ch. 9 - Differential Equations
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 9, Problem 9.3.53b
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.
Verified step by step guidance1
Start with the given initial value problem: \(y'(t) = y^2 + 1\) with initial condition \(y(0) = \frac{1}{\sqrt{2}}\).
Rewrite the differential equation in separable form: \(\frac{dy}{dt} = y^2 + 1\) implies \(\frac{dy}{y^2 + 1} = dt\).
Integrate both sides: integrate \(\int \frac{dy}{y^2 + 1}\) on the left and \(\int dt\) on the right. Recall that \(\int \frac{dy}{y^2 + 1} = \arctan(y) + C\).
After integration, write the implicit solution: \(\arctan(y) = t + C\). Use the initial condition \(y(0) = \frac{1}{\sqrt{2}}\) to solve for the constant \(C\) by substituting \(t=0\) and \(y=\frac{1}{\sqrt{2}}\).
Finally, solve for \(y\) explicitly by taking the tangent of both sides: \(y = \tan(t + C)\), which gives the solution to the initial value problem.
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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(y)h(t), allowing variables y and t to be separated on opposite sides of the equation. This enables integration with respect to each variable independently, which is essential for solving the given initial value problem.
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Solving Separable Differential Equations
Initial Value Problem (IVP)
An initial value problem specifies a differential equation along with a condition y(t₀) = y₀. This condition allows determination of the particular solution from the family of general solutions by solving for the integration constant.
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Finite Time Blowup
Finite time blowup occurs when the solution to a differential equation becomes unbounded in a finite time interval. For nonlinear equations like y' = y² + 1, solutions can grow rapidly and approach infinity at a finite time, which is important to analyze after solving the IVP.
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Related Practice
Textbook Question
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
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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. The solution of a stirred tank initial value problem always approaches a constant as t→∞
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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) = 2y + 4
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
27–30. Predator-prey models Consider the following pairs of differential equations that model a predator-prey system with populations x and y. In each case, carry out the following steps.
b. Find the lines along which x'(t) = 0. Find the lines along which y'(t) = 0.
x′(t) = 2x − xy, y′(t) = −y + xy
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