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