Answer
$W(1,t,cos(t),sin(t))(t)=1$.
Work Step by Step
we verify the given functions are the solutions of the differential equation by plugging them into it:
$y=1\;\;\;\rightarrow \;\;\;{1}^{(4)}+{1}''=\;0+0=0\\\\$
$y=t\;\;\;\rightarrow \;\;\;{t}^{(4)}+{t}''=\;0+0=0\\\\$
$y=cos(t)\;\;\;\rightarrow \;\;\;{cos(t)}^{(4)}+{cos(t)}''=\;cos(t)-cos(t)=0\\\\$
$y=sin(t)\;\;\;\rightarrow \;\;\;{sin(t)}^{(4)}+{sin(t)}''=\;sin(t)-sin(t)=0\\\\$
$W(1,t,cos(t),sin(t))(t)=\begin{vmatrix}
1 &t & cos(t) & sin(t)\\
{1}' &{t}' & {cos(t)}' & {sin(t)}'\\
{1}'' &{t}''&{cos(t)}'' & {sin(t)}''\\
{1}'''&{t}'''&{cos(t)}'''&{sin(t)}'''
\end{vmatrix}\;\;=\;\;\begin{vmatrix}
1 & t& cos(t) & sin(t)\\
0& 1&-sin(t) & cos(t)\\
0 & 0 &-cos(t) & -sin(t)\\
0 &0& sin(t) & -cos(t)
\end{vmatrix}\;\;=\;\;1.\begin{vmatrix}
1 &-sin(t) &cos(t) \\
0 &-cos(t) & -sin(t)\\
0 &sin(t) &-cos(t)
\end{vmatrix}\;=\;1.\begin{vmatrix}
-cos(t) &-sin(t) \\
sin(t) &-cos(t)
\end{vmatrix}\;\;=\;1.[cos^2(t)\;+\;sin^2(t)]\;=\;1.[1]=1\\\\$
$W(1,t,cos(t),sin(t))(t)=1$
