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: 36


$$ t^{2} y^{\prime \prime}+4t y^{\prime}+2y=0 , \quad t > 0 $$ the general solution of the given differential equation is $$ y(t) =c_{1} t^{-1} +c_{2} t^{-2} $$

Work Step by Step

$$ t^{2} y^{\prime \prime}+4t y^{\prime}+2y=0 , \quad t > 0 \quad (1). $$ 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 $$ \frac{d^{2} y}{d x^{2}}+ 3\frac{d y}{d x}+ 2 y=0, \quad (2) $$ We assume that $ y = e^{rx}$, and it then follows that $r$ must be a root of the characteristic equation $$ r^{2}+3r+2=0, $$ so its roots are $$r_{1}=-1 ,\quad r_{2}= -2 .$$ Therefore the general solution of the differential equation (2) is $$ y(x) = c_{1} e^{-x}+c_{2} e^{-2x} $$ where $ c_{1} $and $c_{2}$ are arbitrary constants. Back to t: $$ \begin{split} y(t) & = c_{1} e^{-( \ln t )}+c_{2} e^{-2 ( \ln t)} \\ & = c_{1} e^{( \ln t^{-1} )}+c_{2} e^{( \ln t^{-2})} \\ & =c_{1} t^{-1} +c_{2} t^{-2} \end{split} $$ where $ c_{1} $and $c_{2}$ are arbitrary constants.
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