Answer
See below
Work Step by Step
Given $y_1(x)=e^{x}\cos 2x\\
y_2(x)=e^{x}\sin 2x\\
y_3(x)=e^{-4x}\\
\rightarrow y_1'''(x)=e^{x}\cos 2x-2e^x\sin 2x=e^x(\cos 2x -2\sin 2x)\\
y_1''(x)=e^{x}\cos 2x-2e^x\sin 2x-2e^x\sin 2x-4e^x\cos 2x=-e^x(4\sin 2x +3\cos 2x)\\
y_1'(x)=e^{x}\cos 2x-2e^x\sin 2x=e^x(\cos 2x -2\sin 2x)$
Do the same for $y_2,y_3$, we get:
$\rightarrow y_2'''(x)=4e^{x}\cos 2x-8e^x\sin 2x-3e^x\sin 2x-6e^x\cos 2x=-2e^x\cos 2x-11e^x\sin 2x\\
y_2''(x)=e^{x}\sin 2x+2e^x\cos 2x+2e^x\cos 2x-4e^x\sin 2x=4e^x\cos 2x-3e^x\sin 2x\\
y_2'(x)=e^{x}\sin 2x+2e^x\cos 2x\\
\rightarrow y_3'''(x)=27e^{3x}\\
y_3''(x)=9e^{3x}\\
y_3'(x)=3e^{3x}$
Obtain $det=\begin{vmatrix}
y_1 & y_2 & y_3\\
y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3''
\end{vmatrix}\\=\begin{vmatrix}
e^{x}\cos 2x & e^{x}\sin 2x & e^{3x} \\e^{x}(\cos 2x -2\sin 2x) & e^{x}(\sin 2x+2e^x\cos 2x & 3e^{3x} \\-e^{-x}(4\sin 2x+3\cos 2x) & 4e^{x}\cos 2x-3e^x\sin 2x) & 9e^{3x}
\end{vmatrix}\\
=15e^{5x}\cos ^2 2x+16e^{5x}\sin^2 2x\ne 0$
