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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
Chapter 9, Problem 9.PE.8

In Exercises 1–22, solve the differential equation.


y' = (y²-1)x⁻¹

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1
Rewrite the given differential equation in a more explicit form: \(y' = \frac{dy}{dx} = \frac{y^{2} - 1}{x}\).
Recognize that this is a separable differential equation because the right side can be expressed as a product of a function of \(y\) and a function of \(x\).
Separate the variables by rewriting the equation as \(\frac{dy}{y^{2} - 1} = \frac{dx}{x}\).
Integrate both sides: integrate \(\int \frac{1}{y^{2} - 1} \, dy\) on the left and \(\int \frac{1}{x} \, dx\) on the right.
After integration, solve the resulting equation for \(y\) implicitly or explicitly, and include the constant of integration.

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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 y and a function of x, allowing variables to be separated on opposite sides of the equation. This form enables integration with respect to each variable independently to find the solution.
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Integration Techniques

Solving separable equations requires integrating both sides after separation. Familiarity with integrating rational functions, such as partial fraction decomposition, is essential when dealing with expressions like (y² - 1) in the denominator or numerator.
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Initial Conditions and General Solutions

After integrating, the solution typically includes an arbitrary constant representing a family of solutions. Applying initial conditions, if given, helps determine the particular solution. Understanding the difference between general and particular solutions is key.
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