# 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?

• I've tried this myself using Variation of Parameters and obtained exactly the same integral you did. The integral you obtained cannot be expressed in terms of standard mathematical functions as far as I know, and as far as Wolfram|Alpha knows. Commented Mar 25, 2017 at 12:51

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 $$y(t)=y_{h}(t)+y_{p}(t)=e^{\frac{1}{2}\sin(2t)+t}\Big[c_{1}-\int{e^{(-\frac{1}{2}\sin(2t)-t)}}\sin(t)dt\Big]$$