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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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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주요 개념

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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.

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.

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.

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