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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.2.87
Chapter 7, Problem 7.2.87

Solve the initial value problems in Exercises 87 and 88.
87. dy/dx = 1 + 1/x, y(1) = 3

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Identify the given differential equation: \(\frac{dy}{dx} = 1 + \frac{1}{x}\) with the initial condition \(y(1) = 3\).
Rewrite the differential equation to separate variables or recognize it as a direct integration problem since the right side is expressed explicitly in terms of \(x\).
Integrate both sides with respect to \(x\): \(y = \int \left(1 + \frac{1}{x}\right) \, dx\).
Compute the integral by splitting it into two simpler integrals: \(\int 1 \, dx\) and \(\int \frac{1}{x} \, dx\).
Use the initial condition \(y(1) = 3\) to solve for the constant of integration after finding the general solution.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

First-Order Differential Equations

A first-order differential equation involves the first derivative of a function and can often be solved by direct integration or separation of variables. Understanding how to manipulate and integrate such equations is essential for finding the general solution.
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Initial Value Problems (IVP)

An initial value problem specifies the value of the unknown function at a particular point, allowing determination of the unique solution from the family of general solutions. Applying the initial condition helps find the constant of integration.
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Initial Value Problems

Integration of Rational Functions

Solving dy/dx = 1 + 1/x requires integrating terms like 1 and 1/x separately. Knowing the integral of 1/x is ln|x| and the integral of 1 is x is crucial for correctly finding the antiderivative.
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Intro to Rational Functions
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