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 3 - Second Order Linear Equations - 3.3 Complex Roots of the Characteristic Equation - Problems - Page 163: 13


$$ y^{\prime \prime}+2 y^{\prime}+1.25 y=0 $$ The general solution of that equation is given by $$ y(t) =e^{-t}( c_{1} \cos\frac{t}{2} +c_{2} \sin\frac{t}{2} ) $$ where $ c_{1} $and $c_{2}$ are arbitrary constants.

Work Step by Step

$$ y^{\prime \prime}+2 y^{\prime}+1.25 y=0 \quad\quad\quad \quad (1) $$ We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation $$ r^{2}+2 r+1.25=100 r^{2}+200 r+125=0, $$ so its roots are $$ \:r_{1,\:2}=\frac{-200\pm \sqrt{200^2-4\cdot \:100\cdot \:125}}{2\cdot \:100} $$ Thus the possible values of $r$ are $$r_{1}=-1+i\frac{1}{2} ,\quad r_{2}=-1-i\frac{1}{2} .$$ Therefore two solutions of Eq. (1) are $$ y_{1}(t) = e^{(-1+i\frac{1}{2})t} =e^{-t}(\cos \frac{t}{2} + i \sin \frac{t}{2}) $$ and $$ y_{1}(t) = e^{(-1-i\frac{1}{2})t} =e^{-t}(\cos \frac{t}{2} - i \sin \frac{t}{2}) $$ Thus the general solution of the differential equation is $$ y(t) =e^{-t}( c_{1} \cos\frac{t}{2} +c_{2} \sin\frac{t}{2} ) $$ where $ c_{1} $and $c_{2}$ are arbitrary constants.
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