Textbook Question
Explain how the growth rate function determines the solution of a population model.
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 9.3.30
17–32. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem.
y'(t) = y³sin t, y(0) = 1
Verified step by step guidance1
Identify whether the differential equation is separable. The given equation is \(y'(t) = y^{3} \sin t\). Since the right side can be expressed as a product of a function of \(y\) and a function of \(t\), it is separable.
Rewrite the differential equation using Leibniz notation: \(\frac{dy}{dt} = y^{3} \sin t\). Then separate variables by dividing both sides by \(y^{3}\) and multiplying both sides by \(dt\): \(\frac{1}{y^{3}} dy = \sin t \, dt\).
Integrate both sides: \(\int \frac{1}{y^{3}} dy = \int \sin t \, dt\). This will give you two antiderivatives, one in terms of \(y\) and one in terms of \(t\), plus a constant of integration.
Solve the integrals: Recall that \(\int y^{-3} dy = \int y^{-3} dy\) and \(\int \sin t \, dt = -\cos t + C\). Write the integrated form explicitly, including the constant of integration on one side.
Apply the initial condition \(y(0) = 1\) to find the constant of integration. Substitute \(t=0\) and \(y=1\) into the integrated equation and solve for the constant. Then express \(y\) explicitly as a function of \(t\) if possible.
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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 differential equation is separable if it can be written as a product of a function of y and a function of t, allowing the variables to be separated on opposite sides of the equation. This form enables integration with respect to each variable independently, simplifying the solution process.
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Solving Separable Differential Equations
Initial Value Problems (IVP)
An initial value problem specifies the value of the unknown function at a particular point, providing a unique solution to a differential equation. Solving an IVP involves finding the general solution and then applying the initial condition to determine the constant of integration.
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Integration Techniques for Solving ODEs
Once variables are separated, integration is used to solve the resulting expressions. This often involves integrating standard functions like powers of y and trigonometric functions of t. Proper integration and algebraic manipulation yield the explicit or implicit solution to the differential equation.
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Related Practice
Textbook Question
The general solution of a first-order linear differential equation is y(t) = Ce⁻¹⁰ᵗ − 13. What solution satisfies the initial condition y(0) = 4?
Textbook Question
21–24. Logistic equations Consider the following logistic equations. In each case, sketch the direction field, draw the solution curve for each initial condition, and find the equilibrium solutions. A detailed direction field is not needed. Assume t ≥ 0 and tP ≥ 0.
P′(t) = 0.05P − 0.001P²; P(0) = 10, P(0) = 40, P(0) = 80
Textbook Question
5–10. First-order linear equations Find the general solution of the following equations.
y'(x) = −y + 2
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Textbook Question
21–24. Logistic equations Consider the following logistic equations. In each case, sketch the direction field, draw the solution curve for each initial condition, and find the equilibrium solutions. A detailed direction field is not needed. Assume t ≥ 0 and tP ≥ 0.
P′(t) = 0.05P(1−P/800); P(0) = 100, P(0) = 400, P(0) = 700
Textbook Question
12–16. Sketching direction fields Use the window [-2, 2] x [-2, 2] to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. A detailed direction field is not needed.
y'(t) = 4−y, y(0) = −1