Derive.
$$y'(t)=e^t\cos t-e^t\sin t=y(t)-e^t\sin t,$$ $$y^{''}(t)=y'(t)-e^t\sin t-y(t).$$
Then subtracting the second equation from the first you arrive at $$2y'(t)-y^{''}(t)-2y(t)=0.$$
Derive.
$$y'(t)=e^t\cos t-e^t\sin t=y(t)-e^t\sin t,$$ $$y^{''}(t)=y'(t)-e^t\sin t-y(t).$$
Then subtracting the second equation from the first you arrive at $$2y'(t)-y^{''}(t)-2y(t)=0.$$