Calculus (3rd Edition)

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

Chapter 10 - Introduction to Differential Equations - 10.4 First-Order Linear Equations - Exercises - Page 524: 6


$$y=x \left( x -\ln x +C\right).$$

Work Step by Step

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