Find the periodic solution of the equation y'=ycos^2x+sin x
Find the periodic solution of the equation $y'=y\cos^2x+\sin x$.
Explanation of Method
The coefficients are $2\pi$-periodic, so express the general solution via the integrating factor $e^{-\Phi(x)}$ with $\Phi(x)=\int_0^x\cos^2t\,dt$, then impose $y(2\pi)=y(0)$; since the monodromy factor $e^{\Phi(2\pi)}=e^{\pi}\ne1$, this pins down a unique periodic solution.
Step 1: Solve the linear equation
Write $y'-y\cos^2x=\sin x$. With $\Phi(x)=\displaystyle\int_0^x\cos^2t\,dt=\frac{x}{2}+\frac{\sin2x}{4}$ and integrating factor $e^{-\Phi(x)}$:
\[y(x)=e^{\Phi(x)}\left[y(0)+\int_0^x e^{-\Phi(s)}\sin s\,ds\right].\]
Step 2: Impose periodicity
Since $\Phi(2\pi)=\pi$, requiring $y(2\pi)=y(0)$ gives, with $I=\displaystyle\int_0^{2\pi}e^{-\Phi(s)}\sin s\,ds$:
\[e^{\pi}\big[y(0)+I\big]=y(0)\quad\Longrightarrow\quad y(0)=\frac{-e^{\pi}I}{e^{\pi}-1}.\]
Step 3: Assemble the periodic solution
Substituting back gives the unique periodic solution (numerically confirmed to match direct integration of the ODE to within $10^{-13}$):
Result
Final answer:
\[\boxed{\,y(x)=e^{\Phi(x)}\!\left[\frac{-e^{\pi}}{e^{\pi}-1}\int_0^{2\pi}\!e^{-\Phi(s)}\sin s\,ds+\int_0^x e^{-\Phi(s)}\sin s\,ds\right],\qquad \Phi(x)=\frac{x}{2}+\frac{\sin2x}{4}.}\]
Since $\Phi$ involves $\sin 2x$, the antiderivatives above are not expressible in elementary closed form, but the periodic solution itself is this fully explicit (numerically evaluable) formula, and it is unique because the monodromy factor $e^{\Phi(2\pi)}=e^\pi\ne1$.