Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 9 - Section 9.5 - Linear Equations - 9.5 Exercises - Page 625: 26


$y = x^3-\frac{C_1}{x}+C_2$

Work Step by Step

Let $u = y'$ A first-order linear differential equation is one that can be put into the form: $\frac{dy}{dx} + P(x)~y = Q(x)$ We can consider the given equation: $xy'' + 2y' = 12x^2$ $xu' + 2u = 12x^2$ $u' + \frac{2}{x}u = 12x$ Let $P(x) = \frac{2}{x}$ Let $Q(x) = 12x$ We can find the integrating factor: $I(x) = e^{\int \frac{2}{x} ~dx} = e^{2ln~x} = e^{ln~x^2} = x^2$ We can multiply by the integrating factor: $x^2~u' + 2x~u = 12x^3$ $\frac{d}{dx}(x^2~u) = 12x^3$ We can integrate both sides of the equation: $x^2~u = \int 12x^3~dx$ $x^2~u = 3x^4+C_1$ $u = 3x^2+\frac{C_1}{x^2}$ We can substitute $u = y'$ and integrate both sides of the equation: $u = 3x^2+\frac{C_1}{x^2}$ $y' = 3x^2+\frac{C_1}{x^2}$ $y = x^3-\frac{C_1}{x}+C_2$
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