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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 8 - Techniques of IntegrationProblema 8.4.56
Capitolo 8, Problema 8.4.56

Solve the initial value problems in Exercises 53–56 for y as a function of x.
(x² + 1)² (dy/dx) = √(x² + 1), where y(0) = 1

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Rewrite the given differential equation to isolate \( \frac{dy}{dx} \): \( (x^{2} + 1)^{2} \frac{dy}{dx} = \sqrt{x^{2} + 1} \) becomes \( \frac{dy}{dx} = \frac{\sqrt{x^{2} + 1}}{(x^{2} + 1)^{2}} \).
Simplify the right-hand side by expressing the square root as a power: \( \sqrt{x^{2} + 1} = (x^{2} + 1)^{\frac{1}{2}} \), so \( \frac{dy}{dx} = (x^{2} + 1)^{\frac{1}{2} - 2} = (x^{2} + 1)^{-\frac{3}{2}} \).
Set up the integral to find \( y \) by separating variables: \( dy = (x^{2} + 1)^{-\frac{3}{2}} dx \), then integrate both sides: \( y = \int (x^{2} + 1)^{-\frac{3}{2}} dx + C \).
Use an appropriate substitution or recognize the integral form to evaluate \( \int (x^{2} + 1)^{-\frac{3}{2}} dx \). For example, consider a trigonometric substitution such as \( x = \tan \theta \) or use a standard integral formula.
Apply the initial condition \( y(0) = 1 \) to solve for the constant of integration \( C \) after finding the antiderivative expression for \( y \).

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