Elementary Differential Equations and Boundary Value Problems 9th Edition

Published by Wiley
ISBN 10: 0-47038-334-8
ISBN 13: 978-0-47038-334-6

Chapter 2 - First Order Differential Equations - 2.1 Linear Equations; Method of Integrating Factors - Problems - Page 41: 41

Answer

The solution is $$y(t)=\frac{1}{t^{2}}[-t . cos(t)+sin(t)+ c]=\frac{sin(t)}{t^{2}}-\frac{-cos(t)}{t}+\frac{c}{t^{2}}. $$ where $c$ is a constant.

Work Step by Step

$$ty^{'}+2y=sin(t), t\gt0$$ dividing by $t$ we have : $$y^{'}+\frac{2}{t}y=\frac{1}{t}sin(t), t\gt0 , (* )$$ we know that : $$p(t)=\frac{2}{t}, g(t)=\frac{1}{t}sin(t)$$ The homogeneous solution is $$y(t)=Ae^{\int{-p(t)dt}}=Ae^{\int{-\frac{2}{t}dt}}=Ae^{-2ln{t}}=A t^{-2} $$ where $A$ is a constant. By using the method of variation of parameters, we assume that the solution of Eq. (*) is of the form $$ y(t)=A(t)e^{\int{-p(t)dt}}=A(t). t^{-2}, (**) $$ where A is now a function of t. By substituting for y in the given differential equation, show that A(t) must satisfy the condition $$A^{'}(t)=g(t)e^{\int{p(t)dt}}=\frac{1}{t}sin(t). t^{2}=t. sin(t) $$ by integration we find $$A(t)=\int {t. sin(t)dt } $$ integration by parts shows that : $$A(t)=-t. cos(t)+sin(t)+c $$ where $c$ is a constant. Then substitute for A(t) in Eq. (**) and determine y. $$y(t)=\frac{1}{t^{2}}[-t . cos(t)+sin(t)+ c]=\frac{sin(t)}{t^{2}}-\frac{-cos(t)}{t}+\frac{c}{t^{2}}. $$
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