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: 14

Answer

$$ 9 y^{\prime \prime}+9 y^{\prime}-4 y=0 $$ The general solution of that equation is given by $$ y=c_{1} e^{ \frac{1}{3} t} +c_{2} e^{ -\frac{4}{3}t} $$ where $ c_{1} $and $c_{2}$ are arbitrary constants.

Work Step by Step

$$ 9 y^{\prime \prime}+9 y^{\prime}-4 y=0 \quad \quad (1) $$ We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation $$ 9r^{2}+9 r-4=0, $$ so its roots are $$ r_{1,\:2}=\frac{-9\pm \sqrt{9^2-4\cdot \:9\left(-4\right)}}{2\cdot \:9} $$ Thus the possible values of $r$ are $$ r_{1}= \frac{1}{3} ,\quad r_{2}=-\frac{4}{3} .$$ the general solution of Eq. (1) is $$ y=c_{1} e^{ \frac{1}{3} t} +c_{2} e^{ -\frac{4}{3}t} $$ where $ c_{1} $and $c_{2}$ are arbitrary constants.
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