Textbook Question
5–16. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable.
y'(t) = eʸᐟ²sin 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 9.5.21
20–22. {Use of Tech} Solving the Gompertz equation Solve the Gompertz equation in Exercise 19 with the given values of r, K, and M₀. Then graph the solution to be sure that M(0) and lim(t→∞) M(t) are correct.
r = 0.05, K = 1200, M₀ = 90
Verified step by step guidance1
Recall the Gompertz differential equation, which is commonly written as \(\frac{dM}{dt} = r M \ln\left(\frac{K}{M}\right)\), where \(M(t)\) is the population at time \(t\), \(r\) is the growth rate, and \(K\) is the carrying capacity.
To solve this equation, start by separating variables or using an integrating factor approach. The general solution to the Gompertz equation can be expressed as \(M(t) = K \exp\left(-C e^{-r t}\right)\), where \(C\) is a constant determined by the initial condition.
Apply the initial condition \(M(0) = M_0\) to find the constant \(C\). Substitute \(t=0\) and \(M(0) = M_0\) into the solution formula to get \(M_0 = K \exp(-C)\), which implies \(C = -\ln\left(\frac{M_0}{K}\right)\).
Substitute the value of \(C\) back into the solution to get the explicit formula for \(M(t)\): \(M(t) = K \exp\left(-\left(-\ln\left(\frac{M_0}{K}\right)\right) e^{-r t}\right)\), which simplifies to \(M(t) = K \exp\left(\ln\left(\frac{M_0}{K}\right) e^{-r t}\right)\).
Finally, use the given values \(r=0.05\), \(K=1200\), and \(M_0=90\) to write the specific solution. Then, graph \(M(t)\) over a suitable range of \(t\) to verify that \(M(0) = 90\) and that \(\lim_{t \to \infty} M(t) = K = 1200\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Gompertz Equation
The Gompertz equation is a type of differential equation used to model growth processes, often in biology or demography. It describes how a quantity grows rapidly at first and then slows down as it approaches a limiting value, called the carrying capacity. Understanding its form and parameters is essential to solving and interpreting the solution.
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08:02
Parameterizing Equations
Initial Conditions and Parameters
Initial conditions like M₀ specify the starting value of the function at time zero, while parameters r and K control the growth rate and the upper limit (carrying capacity) respectively. Correctly applying these values is crucial for finding the particular solution that fits the given scenario.
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05:03Initial Value Problems
Limit Behavior and Graphing Solutions
Analyzing the limit of M(t) as t approaches infinity helps verify the long-term behavior of the solution, ensuring it approaches the carrying capacity K. Graphing the solution visually confirms the initial value and asymptotic behavior, providing insight into the model’s accuracy and dynamics.
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Related Practice
Textbook Question
33–38. {Use of Tech} Solutions in implicit form Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one function, be sure to indicate which function corresponds to the solution of the initial value problem.
yy'(x) = 2x/(2 + y)², y(1) = −1
Textbook Question
Solution of the logistic equation Use separation of variables to show that the solution of the initial value problem
P'(t) = rP (1-P/K), P(0) = P₀
is P(t) = K/((K/P₀ − 1)e⁻ʳᵗ + 1)
Textbook Question
21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.
u''(x) = 55x⁹ + 36x⁷ - 21x⁵ + 10x⁻³
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Textbook Question
Consider the differential equation y'(t) = t² - 3y² and the solution curve that passes through the point (3, 1). What is the slope of the curve at (3, 1)?
Textbook Question
39–42. Special equations A special class of first-order linear equations have the form a(t)y'(t)+a'(t)y(t)=f(t), where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form
a(t)y'(t) + a'(t)y(t) = d/dt (a(t)y(t)) = f(t).
Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following initial value problems.
t³y′(t) + 3t²y = (1 + t)/t, y(1) = 6
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