Textbook Question
First-Order Linear Equations
Solve the differential equations in Exercises 1–14.
tan θ dr/dθ + r = sin²θ, 0 < θ < π/2
Ch. 9 - First-Order Differential Equations
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 9, Problem 9.1.12
Integral Equations
In Exercises 7–12, write an equivalent first-order differential equation
and initial condition for y.
y = ln x + ∫ₓᵉ √ (t² + (y(t))²) dt
Verified step by step guidance1
Start with the given integral equation: \(y = \ln x + \int_{x}^{e} \sqrt{t^{2} + (y(t))^{2}} \, dt\).
Recognize that the integral has a variable lower limit \(x\) and a constant upper limit \(e\). To differentiate both sides with respect to \(x\), apply the Leibniz rule for differentiation under the integral sign.
Differentiate the right side with respect to \(x\): the derivative of \(\ln x\) is \(\frac{1}{x}\), and by the Leibniz rule, the derivative of the integral \(\int_{x}^{e} f(t) \, dt\) with respect to \(x\) is \(-f(x)\) because the lower limit is \(x\).
Set up the differential equation by differentiating both sides: \(\frac{dy}{dx} = \frac{1}{x} - \sqrt{x^{2} + (y)^{2}}\).
Use the original equation to find the initial condition by substituting \(x = e\): \(y(e) = \ln e + \int_{e}^{e} \sqrt{t^{2} + (y(t))^{2}} \, dt\), which simplifies to \(y(e) = 1\) since the integral from \(e\) to \(e\) is zero.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Integral Equations and Their Relation to Differential Equations
An integral equation expresses a function in terms of an integral involving the function itself. Converting such an equation into a differential equation involves differentiating both sides to eliminate the integral, allowing the use of differential equation techniques to find solutions.
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Fundamental Theorem of Calculus
This theorem connects differentiation and integration, stating that differentiation can undo integration. When differentiating an integral with a variable limit, the derivative equals the integrand evaluated at that limit, which is essential for transforming the given integral equation into a differential equation.
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Initial Conditions in Differential Equations
Initial conditions specify the value of the unknown function at a particular point, ensuring a unique solution to a differential equation. In this problem, identifying the initial condition from the integral equation or given values is crucial for solving the resulting first-order differential equation.
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Related Practice
Textbook Question
In Exercises 39–42, use Euler’s method with the specified step size to estimate the value of the solution at the given point x*. Find the value of the exact solution at x*.
y′ = √x/y, y > 0, y(0) = 1, dx = 0.1, x* = 1
Textbook Question
Carbon monoxide pollution An executive conference room of a corporation contains 4500 ft³ of air initially free of carbon monoxide. Starting at time t = 0, cigarette smoke containing 4% carbon monoxide is blown into the room at the rate of 0.3 ft³/min. A ceiling fan keeps the air in the room well circulated and the air leaves the room at the same rate of 0.3 ft³/min. Find the time when the concentration of carbon monoxide in the room reaches 0.01%.
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Textbook Question
Write the formula for a logistic function that has values between y = 0 and y = 1, crosses the line y = 1/2 at x = 0, and has slope 5 at this point.
Textbook Question
Use Euler’s method with dx = 0.2 to estimate y(1) if y′ = y and y(0) = 1. What is the exact value of y(1)?
Textbook Question
In Exercises 39–42, use Euler’s method with the specified step size to estimate the value of the solution at the given point x*. Find the value of the exact solution at x*.
y' = 2y²(x-1), y(2) = -1/2, dx = 0.1, x* = 3