Answer
$y= c_1\cos x+c_2 \sin x+(\sin x) \ln (\sec x+\tan x) -1$
Work Step by Step
Given: $y''+y=\sec^2 x$
The Auxiliary equation is: $r^2+1=0 \implies r=\pm i$
The roots are imaginary so the complimentary solution is:
$y_c= c_1\cos x+c_2 \sin x$
The particular solution is:
$y_p=u_1\cos x+u_2 \sin x \\ y'_p=-u_1 \sin x+u_2 \cos x\\ y_p=(-u'_1\sin x+u'_2 \cos x)+(-u_1\cos x-u_2 \sin x)$
Thus, $u_1=-\sec x; u_2=\ln (\sec x+\tan x) $
Hence, $y=y_c+y_p= c_1\cos x+c_2 \sin x+(\sin x) \ln (\sec x+\tan x) -1$
