Answer
$y=y_c+y_p= c_1\cos x+c_2 \sin x+(\sin x) \ln (\sec x+\tan x) -1$
Work Step by Step
Write the Auxiliary equation for the given differential equation.
$r^2+1=0$ This gives: $r= i,-i$
Since, the roots are imaginary, thus the complimentary solution will be:
$y_c= c_1\cos x+c_2 \sin x$
Now, the particular solution is:
$y_p=u_1\cos x+u_2 \sin x \implies y'_p=-u_1 \sin x+u_2 \cos x$
and $ y_p=(-u'_1\sin x+u'_2 \cos x)+(-u_1\cos x-u_2 \sin x)$
After simplifications , we get $u_1=-\sec x$ and $ u_2=\ln (\sec x+\tan x) $
Hence, we get $y=y_c+y_p= c_1\cos x+c_2 \sin x+(\sin x) \ln (\sec x+\tan x) -1$
