How can I solve $y''+y'\tan x-y\cos^2x=2\cos^4x$?

Question

The original form is: $$(d^2y/dx^2)\cos x+(dy/dx)\sin x-y\cos^3x=2\cos^5x$$ And after some simplifying I got $$y''+y'\tan x-y\cos^2x=2\cos^4x$$ But seems like I can't solve it with all the textbooks I searched for some clue.

Answer 1

The equation is singular at $\frac\pi2+k\pi$, so around $x=0$ one can parametrize with $y(x)=z(\sin x)$. Then \begin{align} y'(x)&=\cos x\,z'(\sin x)\\ y''(x)&=\cos^2x\,z''(\sin(x))-\sin x\,z'(x)\\ \hline \cos x\,y''(x)+\sin x\,y'(x)&=\cos^3x\,z''(\sin(x))\\ &=2\cos^5x+\cos^3z(\sin(x))\\ \,z''(\sin(x))-z(\sin(x))&=2\cos^2x=2(1-\sin^2x)\\ \hline \implies z''(s)-z(s)&=2(1-s^2) \end{align} This now is a standard linear DE with constant coefficients.

Answer 2

After change $t=\sin x$ we get ode $$y''-y=2(1-t^2)$$ with solution $$y=c_1e^t+c_2e^{-t}+2t^2+2.$$ Then $$y=c_1e^{\sin x}+c_2e^{-\sin x}+2\sin^2 x+2.$$