Calculus (3rd Edition)

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

Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 41


$$y= t e^{\left(1-\frac{1}{t}\right)}-1$$

Work Step by Step

Given $$t^{2} \frac{d y}{d t}-t=1+y+t y, \quad y(1)=0$$ Rewrite the equation \begin{align*} t^{2} \frac{d y}{d t}&=1+t+y+t y\\ &=(1+t)(1+y) \end{align*} Separating the variables, we get \begin{align*} \frac{d y}{1+y}&=\frac{1+t}{t^{2} }dt\\ \int \frac{d y}{1+y}&= \int t^{-2}+ \frac{1 }{t }dt\\ \ln |1+y|&= \frac{-1}{t}+\ln t+C \end{align*} Since $y(1)=0$, then $C= 1$, and \begin{align*} \ln |1+y|&= \frac{-1}{t}+\ln t+1\\ \ln \frac{1+y}{t}&=1-\frac{1}{t}\\ \frac{1+y}{t}&=e^{1-\frac{1}{t} }\\ y&= t e^{\left(1-\frac{1}{t}\right)}-1 \end{align*}
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