The questions is: Check the stability of the solutions to the following equation:
$$
y^{(11)}-|cos(t)|y^{(10)} +3t^3ArcTan(t)y=1
$$
My take: --- Solved--
I used Abel's formula $$ \dot W=|Cos(t)|W$$ and got that (edit: it's wrong)$$ W(t) = W(0) \cdot e^{|sint|} $$
I'm not sure if it's enough to show that $W \not \xrightarrow[t\to\infty]{} \infty $
in order to ensure stability of the ODE.
Edit: the Wronskian is NOT what I stated: $W(0) \cdot e^{|Sin(t)|}$ because the integral: $$\int_{0}^{t} |Cos(s)| ds \not =|sin(t)| $$ the limit of that integral when t approaches infinity is $\infty$, $$, W\xrightarrow[t\to\infty]{} e^\infty =\infty$$ from that we can infer that one of the columns of the fundemental matrix is not bounded (because the determinant of that matrix (the Wronskian) is approaching infinity). and from that we can say that one solution is not stable, so every other solution is not stable. - (if I made a mistake please correct me)
the Wronskian is NOT what I stated: $W(0) \cdot e^{|Sin(t)|}$ because the integral: $$\int_{0}^{t} |Cos(s)| ds \not =|sin(t)| $$ the term for the wronskian is:$$W= W(0)\cdot e^{\int_{0}^{t} |Cos(s)| ds}$$ the limit of that integral when t approaches infinity is $\infty$\ (always summing positive values), $$W \xrightarrow[t\to\infty]{} e^\infty =\infty$$ from that we can infer that one of the columns of the fundemental matrix is not bounded (because the determinant of that matrix (the Wronskian) is not bounded). and from that we can say that at least one solution approaches infinity and thus not stable, so every other solution is not stable.