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