Answer
see solution.
Work Step by Step
$\;\;\;\;\;\;y_{1}=5\;\;\;\;\;\;y_{2}=sin^2(t)\;\;\;\;\;y_{3}=cos(2t)$
We assume the contrary, or that $y_{1},y_{2},y_{3}$ do form a fundamental set of solutions on $I$. The theorem then implies that $y_{1},y_{2},y_{3}$ are linearly independent on $I$.
$y_{3}=\;cos(2t)\;= cos^2(t)-sin^2(t)=1-2sin^2(t)\;=\;\frac{1}{5}.5\;-2.sin^2(t)\;=\;\frac{1}{5}.y_{1}-2.y_{2}$
By the definition of the linear dependence, we see that $y_{1},y_{2},y_{3}$ are linearly dependent on $I$.
So, the wronskian $W(y_{1},y_{2},y_{3})$ is equal to zero.
