Find interval solutions of $y'=2y\cos^2t-\sin t$ I need to find interval solutions of following equation:
$y'=2y\cos^2t-\sin t$
So it looks like odrinary linear equation, so we can solve it first, assuming that $\sin t=0$, and obtain $$y=ce^{\sin t\cos t+t}$$ Then assume it is solution of "full" equation and find $c$, which, I believe, is $$-\int\frac{\sin t}{e^{\sin t\cos t+t}}$$
But what to do next?

You have the equation of the form $y'+p(x)y=q(x)$ It can be solved by a sum of the solution to the homogeneous equation and the particular solution given by integrating factor. The homogeneous solution you have just obtained
$$\int\frac{dy_{h}}{y_{h}}=2\int\cos^{2}(t)dt$$
$$y_{h}(t)=c_{1}e^{(\frac{1}{2}\sin(2t)+t)}$$
The integrating Factor is
$$I=e^{\int{p(t)}dt}=e^{(-\frac{1}{2}\sin(2t)-t)}$$
So, the particular solution is
$$y_{p}(t)=-e^{(\frac{1}{2}\sin(2t)+t)}\int{e^{(-\frac{1}{2}\sin(2t)-t)}}\sin(t)dt$$
So the solution of the equation is
$$y(t)=y_{h}(t)+y_{p}(t)=e^{\frac{1}{2}\sin(2t)+t}\Big[c_{1}-\int{e^{(-\frac{1}{2}\sin(2t)-t)}}\sin(t)dt\Big]$$
