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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 9 - Differential EquationsProblema 9.3.15
Capitolo 9, Problema 9.3.15

5–16. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable.
u'(x) = e²ˣ⁻ᵘ

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Rewrite the given differential equation \(u'(x) = e^{2x - u}\) in a form that separates the variables \(u\) and \(x\). This means expressing it as \(\frac{du}{dx} = e^{2x} \cdot e^{-u}\), which can be rearranged to isolate \(u\) terms on one side and \(x\) terms on the other.
Separate the variables by multiplying both sides by \(e^{u}\) and \(dx\), giving $e^{u} du = e^{2x} dx$. This sets up the equation so that all \(u\) terms are on the left and all \(x\) terms are on the right.
Integrate both sides: compute \(\int e^{u} du\) on the left and \(\int e^{2x} dx\) on the right. Remember to include the constant of integration after integrating.
After integration, you will have an implicit equation involving \(u\) and \(x\). Solve this equation algebraically to express \(u\) explicitly as a function of \(x\).
Finally, write the general solution \(u(x)\) including the constant of integration, which represents the family of solutions to the differential equation.

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Separable Differential Equations

A separable differential equation can be written as a product of a function of the dependent variable and a function of the independent variable. This allows the variables to be separated on opposite sides of the equation, enabling integration with respect to each variable independently.
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Solving Separable Differential Equations

Integration of Exponential Functions

Integrating exponential functions involves recognizing the form e^(ax + b) and applying the rule ∫e^(ax + b) dx = (1/a)e^(ax + b) + C. This is essential when solving differential equations where the right-hand side contains exponential expressions.
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Integrals of General Exponential Functions

Expressing the Solution Explicitly

After integrating, the solution often involves an implicit relation between variables. Expressing the solution explicitly means solving for the dependent variable as a function of the independent variable, which may require algebraic manipulation or applying inverse functions.
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Simplifying Trig Expressions
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27–30. Newton’s Law of Cooling Solve the differential equation for Newton’s Law of Cooling to find the temperature function in the following cases. Then answer any additional questions.


A pot of boiling soup (100°C) is put in a cellar with a temperature of 10°C. After 30 minutes, the soup has cooled to 80°C. When will the temperature of the soup reach 30°C 

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Explain how a stirred tank reaction works.

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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)

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23–26. Loan problems The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for t≥0 graph the solution, and determine the first month in which the loan balance is zero.


B′(t) = 0.005B − 500, B(0) = 50,000

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15–16. {Use of Tech} Solving logistic equations Write a logistic equation with the following parameter values. Then solve the initial value problem and graph the solution. Let r be the natural growth rate, K the carrying capacity, and P₀ the initial population.


r=0.2, K=300, P₀=50

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33–42. Solving initial value problems Solve the following initial value problems.

y'(x) = 4 sec² 2x, y(0) = 8

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