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: 19

Answer

$$ y^{\prime \prime}-2 y^{\prime}+5 y=0, \quad y(\pi / 2)=0, \quad y^{\prime}(\pi / 2)=2 $$ The general solution of the given initial value problem is $$ y(t) =e^{t }( -e^{-\pi / 2 } \sin 2t)= -e^{t-\pi / 2 } \sin 2t $$ For increasing $t$, $ y(t) $ has growing oscillation.

Work Step by Step

$$ y^{\prime \prime}-2 y^{\prime}+5 y=0, \quad y(\pi / 2)=0, \quad y^{\prime}(\pi / 2)=2 \quad (i) $$ We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation $$ r^{2}-2r+5=0,$$ so its roots are $$ \:r_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \:5}}{2\cdot \:1} $$ Thus the possible values of $r$ are $$r_{1}=1+2i ,\quad r_{2}=1-2i .$$ Therefore two solutions of Eq. (1) are $$ y_{1}(t) = e^{(1+2i )t}=e^{t }(\cos 2t + i \sin 2t) $$ and $$ y_{2}(t) = e^{(1-2i )t}=e^{t }(\cos 2t - i \sin 2t) $$ Thus the general solution of the differential equation is $$ y(t) =e^{t }( c_{1} \cos 2t+c_{2} \sin 2t) \quad (ii) $$ where $ c_{1} $and $c_{2}$ are arbitrary constants. To apply the first initial condition, we set $t = \pi / 2$ in Eq. (ii); this gives $ y(\pi / 2) =-e^{\pi / 2 } c_{1} = 0, $ this implies that $ c_{1} = 0 .$ For the second initial condition we must differentiate Eq. (ii) as follows $$ y^{\prime}(t) =e^{t }[( c_{1}+2c_{2}) \sin 2t+(c_{2} -2c_{1}) \cos 2t ] $$ and then set $t =\pi / 2 $. In this way we find that $$ y^{\prime}(\pi / 2) =-e^{ \pi / 2 }(c_{2}-2c_{1}) =2, c_{1} = 0 $$ and we get $ c_{2} = -e^{-\pi / 2 }$. Using these values of $c_{2}=-e^{-\pi / 2 }$ and $c_{1}=0$ in Eq. (ii), we obtain $$ y(t) =e^{t }( -e^{-\pi / 2 } \sin 2t)= -e^{t-\pi / 2 } \sin 2t $$ as the solution of the initial value problem (i). Since $ \sin 2t $ is bounded and $ -e^{t-\pi / 2 }$ goes to $−∞$ as $ t → ∞$ , $ y(t) $ goes to $−∞ $ as $ t → ∞ $. From the graph we can tell that the function $ y(t) $ has growing oscillation.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.