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: 17

Answer

$$ y^{\prime \prime}+4 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1 $$ The general solution of the given initial value problem is $$ y(t) =\frac{1}{2} \sin 2t $$ For increasing $t$, $ y $ will be steadily oscillating
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Work Step by Step

$$ y^{\prime \prime}+4 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1 \quad (i) $$ We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation $$ r^{2}+4=0,$$ so its roots are $$ \quad \:r_{1,\:2}=\frac{\pm \sqrt{-4\cdot \:1\cdot \:4}}{2\cdot \:1} $$ Thus the possible values of $r$ are $$r_{1}=2i ,\quad r_{2}=-2i .$$ Therefore two solutions of Eq. (1) are $$ y_{1}(t) = e^{2i}=(\cos 2t + i \sin 2t) $$ and $$ y_{2}(t) = e^{-2i}=(\cos 2t - i \sin 2t) $$ Thus the general solution of the differential equation is $$ y(t) =( c_{1} \cos 2t+c_{2} \sin 2t) \quad (ii) $$ where $ c_{1} $ and $c_{2}$ are arbitrary constants. To apply the first initial condition, we set $t = 0$ in Eq. (ii); this gives $ y(0) = c_{1} = 0. $ For the second initial condition we must differentiate Eq. (ii) as follows $$ y^{\prime}(t) =( -2c_{1} \sin 2t+2c_{2} \cos 2t) $$ and then set $t = 0 $. In this way we find that $$ y^{\prime}(0) =2c_{2} =1 $$ from which $ c_{2} = \frac{1}{2}$. Using these values of $c_{1}=0$ and $c_{2}=\frac{1}{2}$ in Eq. (ii), we obtain $$ y(t) =\frac{1}{2} \sin 2t $$ as the solution of the initial value problem (i). For increasing $t$, $ y $ will be steadily oscillating
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