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: 2


$y=C_{0} e^{x^2/2}$

Work Step by Step

The equation $y' =xy$ becomes : $\Sigma^\infty_{n=1}nC_{n} x^{n-1}= x[ \Sigma^\infty_{n=0}C_{n}x^{n}] \implies c_1+\Sigma^\infty_{n=2} \times nc_nx^{n-1}=\Sigma^\infty_{n=0}c_{n} \times x^{n+1}$ $c_1+\Sigma^\infty_{n=0} \times (n+2)c_{n+2} x^{n+1}=\Sigma^\infty_{n=0}c_{n} \times x^{n+1}$ and $c_1+\Sigma^\infty_{n=0}[(n+2)c_{n+2} x^{n+1}-c_n]x^{n+1}=0$ $\implies c_1=0\\c_{n+2}=\dfrac{c_n}{n+2}$ Consider the odd values of $n=3,5...$, then we get $C_{3} =0\\C_{5} = 0$; Consider the even values of $n=2,4...$, then we get $C_{2} = \dfrac{C_0}{2}\\ C_{4} = \dfrac{C_0}{2 \times 4}$ Thus, we have $C_{2n} = (\dfrac{1}{2})^n \dfrac{c_0}{n!}$ Thus, we get $y=\Sigma^\infty_{n=0} (\dfrac{1}{2})^n \dfrac{c_0}{n!}x^{2n}+\Sigma^\infty_{n=0} (0) x^{2n+1}$ Use: $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $ Hence, we have $y=C_{0} e^{x^2/2}$
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