Answer
\[y''\left( t \right)+2y'\left( t \right)+5y\left( t \right)=0\]
Work Step by Step
\[\begin{align}
& y''\left( t \right)+2y'\left( t \right)+5y\left( t \right)=0 \\
& \text{Where }y\left( t \right)={{e}^{-t}}\left( \sin 2t-2\cos 2t \right) \\
& \\
& \text{Calculate }y'\left( t \right) \\
& y'\left( t \right)=\frac{d}{dt}\left[ {{e}^{-t}}\left( \sin 2t-2\cos 2t \right) \right] \\
& \text{By the product rule} \\
& y'\left( t \right)={{e}^{-t}}\left( 2\cos 2t+4\sin 2t \right)-{{e}^{-t}}\left( \sin 2t-2\cos 2t \right) \\
& \text{Simplifying} \\
& y'\left( t \right)=2{{e}^{-t}}\cos 2t+4{{e}^{-t}}\sin 2t{}-{{e}^{-t}}\sin 2t+2{{e}^{-t}}\cos 2t \\
& y'\left( t \right)=4{{e}^{-t}}\cos 2t+3{{e}^{-t}}\sin 2t{{e}^{-t}} \\
& y'\left( t \right)={{e}^{-t}}\left( 4\cos 2t+3\sin 2t \right) \\
& \text{Calculate }y''\left( t \right) \\
& y''\left( t \right)=\frac{d}{dt}\left[ {{e}^{-t}}\left( 4\cos 2t+3\sin 2t \right) \right] \\
& \text{By the product rule} \\
& y''\left( t \right)={{e}^{-t}}\left( -8\sin 2t+6\cos 2t \right)-{{e}^{-t}}\left( 4\cos 2t+3\sin 2t \right) \\
& y''\left( t \right)={{e}^{-t}}\left( -8\sin 2t+6\cos 2t-4\cos 2t-3\sin 2t \right) \\
& y''\left( t \right)={{e}^{-t}}\left( 2\cos 2t-11\sin 2t \right) \\
& \text{Substitute }y(t), y'\left( t \right),\text{ and }y''\left( t \right)\text{ into } \\
& y''\left( t \right)+2y'\left( t \right)+5y\left( t \right)=0 \\
& {{e}^{-t}}\left( 2\cos 2t-11\sin 2t \right)+2{{e}^{-t}}\left( 4\cos 2t+3\sin 2t \right) \\
& +5{{e}^{-t}}\left( \sin 2t-2\cos 2t \right)=0 \\
& \text{Simplify} \\
& {{e}^{-t}}\left( 2\cos 2t-11\sin 2t \right)+{{e}^{-t}}\left( 8\cos 2t+6\sin 2t \right) \\
& +{{e}^{-t}}\left( 5\sin 2t-10\cos 2t \right)=0 \\
& \text{Combine like terms} \\
& {{e}^{-t}}\left( 2\cos 2t-11\sin 2t+8\cos 2t+6\sin 2t+5\sin 2t-10\cos 2t \right) \\
& =0 \\
& {{e}^{-t}}\left( 0 \right)=0 \\
\end{align}\]
