solve x′=(2−1 42)x+(1sin2t 2cos2t)e2tx'=(\begin{matrix} 2&-1\ 4&2\end{matrix} )x+(\begin{matrix} 1sin2t\ 2cos2t\end{matrix} )e^{2t}

asked by guest
on Apr 07, 2025 at 12:21 pm



Mathbot Says...

I wasn't able to parse your question, but the HE.NET team is hard at work making me smarter.


+ Submit Question