The function $y(t)$ satisfies
$$t^{2} y^{\prime \prime}(t)-2 t y^{\prime}(t)+2 y(t)=0,$$
where $y^{\prime}(t)$ and $y^{\prime \prime}(t)$ denote the first and second derivatives of $y(t)$, respectively.
Given $y^{\prime}(0)=1$ and $y^{\prime}(1)=-1$, the maximum value of $y(t)$ over $[0,1]$ is $\_\_\_\_\_\_\_$ (rounded off to two decimal places).
