Domanda del libro di testo
A family of exponential functions
b. Verify that the arc length of the curve y=f(x) on the interval [0, ln 2] is A(2^a-1) - 1/4a²A (2^-a - 1).
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Ch. 9 - Differential Equations
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- 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: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 9, Problema 9.4.12
11–16. Initial value problems Solve the following initial value problems.
y'(x) = −y + 2, y(0) = −2
Guida verificata passo dopo passo1
Identify the type of differential equation given. Here, the equation is a first-order linear ordinary differential equation of the form \(y'(x) + y = 2\).
Rewrite the differential equation in standard linear form: \(y' + p(x)y = q(x)\), where \(p(x) = 1\) and \(q(x) = 2\).
Find the integrating factor \(\mu(x)\) using the formula \(\mu(x) = e^{\int p(x) \, dx}\). Since \(p(x) = 1\), calculate \(\mu(x) = e^{\int 1 \, dx} = e^{x}\).
Multiply both sides of the differential equation by the integrating factor \(e^{x}\) to get \(e^{x} y' + e^{x} y = 2 e^{x}\). Recognize that the left side is the derivative of \(e^{x} y\) with respect to \(x\).
Integrate both sides with respect to \(x\): \(\int \frac{d}{dx} (e^{x} y) \, dx = \int 2 e^{x} \, dx\). Then solve for \(y(x)\) and apply the initial condition \(y(0) = -2\) to find the constant of integration.
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First-Order Differential Equations
A first-order differential equation involves the first derivative of a function and the function itself. Solving such equations means finding a function y(x) that satisfies the given relationship between y' and y. In this problem, y'(x) = -y + 2 is a linear first-order differential equation.
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07:39
Classifying Differential Equations
Initial Value Problems (IVP)
An initial value problem specifies the value of the unknown function at a particular point, here y(0) = -2. This condition allows us to find a unique solution to the differential equation by determining the constant of integration after solving the general solution.
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Percorso guidato
05:03Initial Value Problems
Solving Linear Differential Equations Using Integrating Factors
Linear first-order differential equations can be solved using an integrating factor, which is a function that simplifies the equation into an exact derivative. Multiplying both sides by this factor allows integration and leads to the general solution, which can then be adjusted using the initial condition.
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06:06Solving Separable Differential Equations
Pratica correlata
Domanda del libro di testo
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.
(t² + 1)y′(t) + 2ty = 3t², y(2) = 8
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Domanda del libro di testo
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) = 0.004B − 800, B(0) = 40,000
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Domanda del libro di testo
Does the function y(t) = 2t satisfy the differential equation y'''(t) + y'(t) = 2?
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Domanda del libro di testo
17–32. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem.
y'(t) = yeᵗ, y(0) = −1
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