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I am asked to find the stability of the following ODE: \begin{equation*} \dot{y} = y^{2} + 2\cos(t)\sin(t) - \sin^{4}(t) \end{equation*} by linearizing around a particular solution $\eta = \sin^{2}(t)$, and investigating the solution.

My understanding is that I should look at the first variational equation \begin{equation*} \dot{y} = A(t)y \end{equation*} where \begin{equation*} A(t) = df_{y}(t,\eta), \quad A(t+T) = A(t), \quad f(t,y) = \dot{y}(t). \end{equation*}

Thus, I get \begin{alignat*}{2} f(t,y) &= y^{2} + 2\cos(t)\sin(t) - \sin^{4}(t) &&\Rightarrow \\ f_{y}(t,y) &= 2y &&\Rightarrow \\ A(t) &= 2\sin^{2}(t). \end{alignat*}

Now $A(t)$ is periodic with period $T=2\pi$, and my next understanding is that stability of the solution depends on the sign of the eigenvalue for this one-dimensional system which are the Floquet multipliers (or characteristic exponents). This is given by the formula (using notion from Teschl's ODE text): \begin{alignat*}{2} \bar{a} &= \frac{1}{T}\int_{0}^{T}A(s)ds \\ &= 1 \end{alignat*} which seems to imply that the solution is not stable. Is this understanding correct? Either way, I think part of my confusion comes from not understanding the connection between the first variation and the principal matrix solution of the original equation. Any help towards that would be greatly appreciated.

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Yes, looks fine. Actually, you can solve the linearization of the perturbation equation $$ y'(t)=2\sin^2(t)\,y(t)=(1-\cos(2t))\,y(t) $$ directly to get $$ y(t)=\exp(t-\tfrac12\sin(2t))\,y(0) $$ which is clear evidence of the unstable behavior.

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