## Calculus 8th Edition

Published by Cengage

# Chapter 17 - Second-Order Differential Equations - 17.2 Nonhomogeneous Linear Equations - 17.2 Exercises - Page 1208: 23

#### Answer

$y=c_1\cos x+c_2 \sin x+(\sin x) \ln (\sec x+\tan x) -1$

#### Work Step by Step

We are given that $y''+y=\sec^2 x$ The Auxiliary equation is: $r^2+1=0 \implies r=\pm i$ When the roots are imaginary so the complimentary solution is: $y_c= c_1\cos x+c_2 \sin x$ The particular solution is: $y_p=u_1\cos x+u_2 \sin x; y'_p=-u_1 \sin x+u_2 \cos x; y_p=(-u'_1\sin x+u'_2 \cos x)+(-u_1\cos x-u_2 \sin x)$ So, we have $u_1=-\sec x; u_2=\ln (\sec x+\tan x)$ Hence, $y=y_c+y_p$ or, $y= c_1\cos x+c_2 \sin x+(\sin x) \ln (\sec x+\tan x) -1$

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