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 165: 37

Answer

$$ t^{2} y^{\prime \prime}+3 t y^{\prime}+1.25 y=0 , \quad t > 0 $$ the general solution of the given differential equation is $$ y(t) =t ^{-1} [ c_{1} \cos ( \frac{1}{2}\ln t)+ c_{2}\sin ( \frac{1}{2} \ln t) ] $$ where $ c_{1} $and $c_{2}$ are arbitrary constants.

Work Step by Step

$$ t^{2} y^{\prime \prime}+3 t y^{\prime}+1.25 y=0 , \quad t > 0 \quad (1). $$ Let $ x = \ln t$ . Then we have $$ \frac{d y}{d t}=\frac{d y}{d x} \frac{d x}{d t}=\frac{1}{t} \frac{d y}{d x} $$ $$ \frac{d^{2} y}{d t^{2}}=\frac{d}{d t}\left(\frac{1}{t} \frac{d y}{d x}\right)=\frac{-1}{t^{2}} \frac{d y}{d x}+\frac{1}{t} \frac{d}{d t}\left(\frac{d y}{d x}\right)=\frac{1}{t^{2}}\left(-\frac{d y}{d x}+\frac{d^{2} y}{d x^{2}}\right) $$ So if we use $x$ instead of $t$ as the variable, the equation (with unknown $y $ and variable $ x$) reads $$ \frac{d^{2} y}{d x^{2}}+ 2\frac{d y}{d x}+ 1.25 y=0, \quad (2) $$ We assume that $ y = e^{rx}$, and it then follows that $r$ must be a root of the characteristic equation $$ r^{2}+2r+1.25=100r^2+200r+125=0, $$ so its roots are $$ \:r_{1,\:2}=\frac{-200\pm \sqrt{200^2-4\cdot \:100\cdot \:125}}{2\cdot \:100}\quad $$ 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 $$ y_{1}(x) = e^{(-1+i\frac{1}{2} )x}= e^{(-x)} (\cos \frac{1}{2}x + i \sin \frac{1}{2} x ) $$ and $$ y_{2}(x) = e^{(-1-i\frac{1}{2} )x}= e^{(-x)} (\cos \frac{1}{2}x - i \sin \frac{1}{2} x ) $$ Thus the general solution of the differential equation is $$ y(x) = e^{(-x)} ( c_{1} \cos \frac{1}{2}x+ c_{2}\sin \frac{1}{2} x ) \quad\quad\quad (2) $$ where $ c_{1} $ and $c_{2}$ are arbitrary constants. Back to t: $$ \begin{split} y(t) & = e^{(-\ln t)} ( c_{1} \cos \frac{1}{2}\ln t+ c_{2}\sin \frac{1}{2} \ln t ) \\ & = e^{(\ln t ^{-1})} ( c_{1} \cos \frac{1}{2}\ln t+ c_{2}\sin \frac{1}{2} \ln t ) \\ & =t ^{-1} [ c_{1} \cos ( \frac{1}{2}\ln t)+ c_{2}\sin ( \frac{1}{2} \ln t) ] \end{split} $$ where $ c_{1} $ and $c_{2}$ are arbitrary constants.
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