## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 10 - Introduction to Differential Equations - 10.4 First-Order Linear Equations - Exercises - Page 524: 11

#### Answer

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

#### Work Step by Step

Rewrite the equation as $$y'-x^{-1}y=-1$$ 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 -\frac{1}{x}dx +C\right)\\ &= x\left( -\ln x+C\right) \end{align} Thus, the general solution is $$y= x\left( -\ln x+C\right) .$$

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