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Dec 21, 2016 at 11:49 comment added Robert Bryant @LuigiScorzato: Yes, 'NO'. But it's worse than that: Because your equation is not $\mathbb{C}$-linear, multiplying a given solution $y(t)$ by a non-real complex number will not yield another solution, so making an $n$-by-$n$ complex matrix $\Phi(t)$ whose columns are $n$ particular solutions will yield a matrix whose ($\mathbb{C}$-valued) determinant depends strongly on the $\mathbb{R}$-span of those particular solutions in the $\mathbb{R}$-vector space of all solutions (which has real dimension $2n$). Any formula for how the determinant changes with $t$ would have to take this into account.
Dec 21, 2016 at 10:16 comment added Luigi Scorzato Thanks a lot! Very insightful. I understand, however, that this is not the determinant I was looking for, and therefore your answer to my original question is NO. Correct?
Dec 21, 2016 at 10:14 vote accept Luigi Scorzato
Dec 15, 2016 at 12:48 history edited Robert Bryant CC BY-SA 3.0
Rearranged the answer to make it more coherent and readable.
Dec 15, 2016 at 1:15 history answered Robert Bryant CC BY-SA 3.0