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: 28


$y= \frac{1}{3} +\frac{1}{5}t^2+\frac{1}{2}t+Ct^{-3} $

Work Step by Step

Rewrite the equation in the form $y'+\frac{3}{t}y=\frac{1}{t}+t+2$ This is a linear equation and has the integrating factor as follows $$\alpha(t)= e^{\int P(t)dt}=e^{ \int \frac{3}{t} dt}=e^{3\ln t}=t^3.$$ Now the general solution is \begin{align} y& =\alpha^{-1}(t)\left( \int\alpha(t) Q(t)dt +C\right)\\ & =t^{-3}\left( \int (t^2+t^4+2t^3)dt+C\right)\\ & =t^{-3} \left( \frac{1}{3}t^3+\frac{1}{5}t^5+\frac{2}{4}t^4+C\right)\\ & = \frac{1}{3} +\frac{1}{5}t^2+\frac{1}{2}t+Ct^{-3} .\end{align}
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