Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 17 - Section 17.1 - Second-Order Linear Equations - 17.1 Exercise: 10

Answer

y = $C_{1}e^{\frac{-2+\sqrt {13}}{3}x}$ + $C_{2}e^{\frac{-2-\sqrt {13}}{3}x}$

Work Step by Step

Question: 3y"+4y'-3y=0 We know: y = $e^{rx}$ y' = $re^{rx}$ y" = $r^{2}e^{rx}$ This results in $(3r^{2}+4r-3)e^{rx}$=0 The corresponding characteristic equation is: $3r^{2}+4r-3=0$ Solving the characteristic equation with the abc-formula and complex numbers gives: $\frac{-b\frac{+}{-}\sqrt {b^{2}-4*a*c}}{2*a}$ $\frac{-4\frac{+}{-}\sqrt {(4)^{2}-4*3*-3}}{2*3}$ $r=\frac{-4\frac{+}{-}\sqrt {52}}{6}$ This can also be written as: $r=\frac{-2\frac{+}{-}\sqrt {13}}{3}$ Adding the constants gives the following general solution: y = $C_{1}e^{\frac{-2+\sqrt {13}}{3}x}$ + $C_{2}e^{\frac{-2-\sqrt {13}}{3}x}$
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