Answer
See below
Work Step by Step
Given $y_1(x)=\cos 2x\\
y_2(x)=\sin 2x\\
y_3(x)=e^x\\
\rightarrow y_1'''(x)=8\sin 2x\\
y_1''(x)=-4\cos 2x\\
y_1'(x)=-2\sin 2x$
Substituting into the given equation:
$y_1'''-y''_1+4y_1'-4y_1\\
=8\sin 2x+4\cos 2x-8\sin 2x-4\cos 2x\\
=0$
Do the same for $y_2,y_3$, we get:
$y_2'''-y''_2+4y_2'-4y_2\\
=-8\cos 2x+4\sin 2x+8\cos 2x-4\sin 2x\\
=0$
$y_3'''-y''_3+4y_3'-4y_3\\
=e^x-e^x+4e^x-4e^x\\
=0$
Obtain $det=\begin{vmatrix}
y_1 & y_2 & y_3\\
y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3''
\end{vmatrix}=\begin{vmatrix}
\cos 2x&\sin 2x& e^x \\ -2\sin 2x &-2\cos 2x& e^x\\-4\cos 2x& -4\sin 2x & e^x
\end{vmatrix}\\
=\cos 2x(-2\cos 2x)e^x+\sin 2x.(-4\cos 2x).e^x+e^x.(-2\sin 2x).(-4\sin 2x)-(-4\cos 2x).(-2\cos 2x).e^x-(-4\sin 2x).e^x.\cos 2x-e^x.(-2\sin 2x).\sin 2x\\
=2e^x\cos^x-4e^x\sin 2x \cos 2x+8e^x\sin^2 2x+8e^x\cos^2 2x+4e^x\sin 2x \cos 2x+2e^x\sin^2 2x\\
=10e^x\sin^2 2x+10e^x\cos ^2 x\\
=10e^x (\sin^2 2x+\cos ^2 x)\\
=10e^x \ne 0$
