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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 9, Problem 9.1.42

33–42. Solving initial value problems Solve the following initial value problems.
p'(x) = 2/(x² + x), p(1) = 0

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1
Identify the given differential equation and initial condition: \(p'(x) = \frac{2}{x^{2} + x}\) with \(p(1) = 0\).
Rewrite the derivative notation as \(\frac{dp}{dx} = \frac{2}{x^{2} + x}\) to recognize it as a separable differential equation.
Simplify the denominator by factoring: \(x^{2} + x = x(x + 1)\), so the equation becomes \(\frac{dp}{dx} = \frac{2}{x(x + 1)}\).
Integrate both sides with respect to \(x\): \(p(x) = \int \frac{2}{x(x + 1)} \, dx + C\), where \(C\) is the constant of integration.
Use partial fraction decomposition to express \(\frac{2}{x(x + 1)}\) as \(\frac{A}{x} + \frac{B}{x + 1}\), find \(A\) and \(B\), then integrate each term separately. Finally, apply the initial condition \(p(1) = 0\) to solve for \(C\).

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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 technique simplifies solving by integrating each side independently.
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Integration of Rational Functions

Integrating rational functions often involves techniques like partial fraction decomposition to rewrite the integrand into simpler fractions. This method is essential for integrating expressions like 2/(x² + x) effectively.
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Initial Value Problems (IVP)

An initial value problem specifies a differential equation along with a condition at a particular point, such as p(1) = 0. Solving an IVP involves finding the general solution and then using the initial condition to determine the specific constant.
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Initial Value Problems