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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 9, Problema 9.1.42

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

Guida verificata passo dopo passo
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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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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Solving Separable Differential Equations

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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Intro to Rational Functions

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