Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 10 - Introduction to Differential Equations - Chapter Review Exercises - Page 526: 24


$$y= -x+\frac{1}{2} +\frac{40}{9}x^2 .$$

Work Step by Step

Rewrite the equation in the form $y'-\frac{2}{x}y=1-\frac{1}{x}$. This is a linear equation and has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{ \int \frac{-2}{x}dx}=e^{-2\ln x}=\frac{1}{ x^2}$$ Now the general solution is \begin{align} y& =\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ & = x^2 \left( \int ( \frac{1}{x^2}-\frac{1}{x^3})dx+C\right)\\ & =x^2 \left(- \frac{1}{x}+\frac{1}{2x^2}+C\right)\\ & = -x+\frac{1}{2} +Cx^2\end{align} Since, $y(3/2)=9$, then $C=\frac{40}{9}$. Thus, the general solution is: $$y= -x+\frac{1}{2} +\frac{40}{9}x^2 .$$
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