Verify that the indicated function is a particular solution

Solution

y\'=sin x+ x cos x+ cos (x) 1/cos(x)* (-sin x)-sin(x) ln(cos x)
=x cos x-sin(x) ln(cos(x))

y\'\'=cos x-x sin x-cos(x) ln(cos(x))-sin(x)*1/cos(x)*(-sin(x))
So
y\'\'+y=cos x+sin^2(x)/cos(x)=(cos^2(x)+sin^2(x))/cos(x)=1/sec(x)
so it verifies the equation
we need to have cos(x)>0 in order for the logarithm to be defined
(in this case automatically cos(x) cannot be 0 , so sec is defined)

cos(x)>0 on (-/2,/2) so we choose I=(-/2,/2)