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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.4.54

Solve the initial value problems in Exercises 53–56 for y as a function of x.
√(x² - 9) (dy/dx) = 1, where x > 3, y(5) = ln 3

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1
Rewrite the given differential equation in the form \( \frac{dy}{dx} = \frac{1}{\sqrt{x^2 - 9}} \) to isolate \( \frac{dy}{dx} \).
Express the differential equation as \( dy = \frac{1}{\sqrt{x^2 - 9}} dx \) to prepare for integration on both sides.
Integrate both sides: \( \int dy = \int \frac{1}{\sqrt{x^2 - 9}} dx \). The left side integrates to \( y \), and the right side requires recognizing the integral form.
Recall that \( \int \frac{1}{\sqrt{x^2 - a^2}} dx = \ln|x + \sqrt{x^2 - a^2}| + C \). Use this formula with \( a = 3 \) to write the general solution for \( y \).
Apply the initial condition \( y(5) = \ln 3 \) by substituting \( x = 5 \) and \( y = \ln 3 \) into the general solution to solve for the constant \( C \).

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Key Concepts

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

Separable Differential Equations

A separable differential equation can be written as a product of a function of x and a function of y, allowing variables to be separated on opposite sides of the equation. This enables integration with respect to each variable independently to find the general solution.
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Initial Value Problems (IVP)

An initial value problem specifies a differential equation along with a condition that the solution must satisfy at a particular point. This condition allows determination of the unique solution curve passing through the given point.
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Integration Techniques and Inverse Functions

Solving the differential equation involves integrating expressions that may include radicals like √(x² - 9). Recognizing standard integral forms and applying inverse hyperbolic or logarithmic functions is essential to express the solution explicitly.
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