I get:

\(\displaystyle y=\cos(t)\ln(\cos(t))+t\sin(t)\)

\(\displaystyle y'=-\sin(t)-\sin(t)\ln(\cos(t))+t\cos(t)+\sin(t)=-\sin(t)\ln(\cos(t))+t\cos(t)\)

\(\displaystyle y''=\sin^2(t)\sec(t)-\cos(t)\ln(\cos(t))-t\sin(t)+\cos(t)=\cos(t)+\sin^2(t)\sec(t)-y=(\cos^2(t)+\sin^2(t))\sec(t)-y=\sec(t)-y\)

And thus:

\(\displaystyle y''+y=\sec(t)\)