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 - 3.1 Homogenous Equations with Constant Coefficients - Problems - Page 144: 13

Answer

$${y}=\frac{5+\sqrt13}{2\sqrt13}\exp^{-\frac{5-\sqrt13}{2}x}+\frac{\sqrt{13}-5}{2\sqrt13}\exp^{-\frac{5+\sqrt13}{2}x}$$

Work Step by Step

$$y''+5y'+3y=0,\quad{y}(0)=1,\quad{y'}(0)=0$$Let $y=e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$. $${\lambda}^2+5{\lambda}+3=0$$ $$\lambda_{1,2}=\frac{-5\pm\sqrt{25-12}}{2}=\frac{-5\pm\sqrt13}{2}$$ The general solution is ${y}=c_{1}\exp^{-(\frac{5-\sqrt13}{2})x}+c_{2}\exp^{-(\frac{5+\sqrt13}{2})x}$. Substituting in the constraints, we obtain $c_{1}+c_{2}=1$ and $-(\frac{5-\sqrt13}{2})c_{1}-(\frac{5+\sqrt13}{2})c_{2}=0\Rightarrow{c}_{1}=\frac{5+\sqrt13}{\sqrt{13}-5}c_{2}$. $$\therefore\frac{2\sqrt13}{\sqrt{13}-5}c_{2}=1\Rightarrow{c}_{2}=\frac{\sqrt{13}-5}{2\sqrt13}\Rightarrow{c}_{1}=\frac{5+\sqrt13}{2\sqrt13}$$ $$\therefore{y}=\frac{5+\sqrt13}{2\sqrt13}\exp^{-\frac{5-\sqrt13}{2}x}+\frac{\sqrt{13}-5}{2\sqrt13}\exp^{-\frac{5+\sqrt13}{2}x}$$
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