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Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 9 - First-Order Differential EquationsProblem 9.1.12
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 guidance
1
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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Classifying Differential Equations

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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Fundamental Theorem of Calculus Part 1

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

Using Euler’s Method

In Exercises 15–20, use Euler’s method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places.


y' = x(1-y), y(1) = 0, dx = 0.2

Textbook Question

Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.


y' = y/x + cos ((y-x)/x)

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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

Use Euler’s method with dx = 1/3 to estimate y(2) if y′ = x sin y and y(0) = 1. What is the exact value of y(2)?

Textbook Question

Solve the homogeneous equations in Exercises 5–10. First put the equation in the form of a homogeneous equation.

(x.exp(y/x) + y)dx - x dy = 0

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