Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 17 - Second-Order Differential Equations - 17.4 Series Solutions - 17.4 Exercises - Page 1220: 1

Answer

$y=C_{0} e^{x}$

Work Step by Step

We have $y=\Sigma^\infty_{n=1}C_{n} x^{n}$ and $y'=\Sigma^\infty_{n=0}C_{n+1} x^n (n+1) \times x^{n}$ The equation $y' - y = 0$ becomes : r, $\Sigma^\infty_{n=0}[C_{n+1}(n+1) - C_{n}]=0$ $C_{1} = C_{0}(n=0)\\C_{2} = \dfrac{C_{0}}{2}=0; (n=1)$; $C_{3} = (\dfrac{1}{3}) (\dfrac{1}{2}) \times C_{0}=0 ; (n=2) $ This gives: $C_{n} = \dfrac{C_0}{n!}$ Use Formula: $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $ Hence, we have $y=C_{0} e^{x}$
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