I'm trying to find a differential equation involving $Y(s) = \mathcal{L}[y(t)]$ of the Legendre's equation $$ (1-t^{2})y'' -2ty' + \alpha(\alpha+1)y = 0\qquad\qquad ,y(0)=1, \quad y'(0)=1,$$ for some $\alpha \in \mathbb{R}.$
The following is my attempt \begin{align*} \mathcal{L}[(1-t^{2})y'' - 2ty' +\alpha(\alpha+1)y] &= \mathcal{L} [0] \\ \mathcal{L}[y''] - \mathcal{L}[t^{2} y''] - 2 \mathcal{L}[ty'] + \alpha(\alpha+1) \mathcal{L}[y] &= 0 \\ s^{2}Y(s)-1 - \frac{d^2[s^{2}Y(s) -1]}{dt^2} + 2 \frac{d[sY(s)]}{dt} +\alpha(\alpha+1)Y(s) &= 0. \end{align*} Using the product rule on both terms and recollecting everything yields, \begin{align*} s^{2} Y(s) -1 - Y''(s) - Y'(s) 4s - Y(s)2 + 2Y(s) + 2sY'(s) + \alpha(\alpha +1)Y(s) &= 0 \\ Y''(s) +2sY'(s) -(s^{2} + \alpha(\alpha+1))Y(s) &= -1 \end{align*}
The part of which I'm not entirely sure about is taking the derivative with respect to $t$ of $s^2 Y(s)-1$ and $sY(s)$. Here I assumed that $Y(s)$ is implicitly dependent on $t$ and so I used the product rule. If anyone could confirm my reasoning or point where I went wrong that would be appreciated.
$$\mathcal{L}[y''] - \mathcal{L}[t^{2} y''] - 2 \mathcal{L}[ty'] + \alpha(\alpha+1) \mathcal{L}[y] = 0 \\$$ Then you have to apply $$\mathcal{L}(t y)=-\dfrac {d}{ds} \left (\mathcal{L}(y) \right)$$ So that: $$\mathcal{L}(t^2 y'')=\dfrac {d^2}{ds^2} \left (\mathcal{L}(y'') \right)$$ $$\mathcal{L}(y'')=s^2Y(s)-sy(0)-y'(0)$$ $$\mathcal{L}(y'')=s^2Y(s)-s-1$$ Differentiate with respect to $s$: $$ \dfrac {d}{ds} \left (\mathcal{L}(y'') \right)=s^2Y'(s)+2sY(s)-1$$ $$\implies \mathcal{L}(t^2 y'')=\dfrac {d^2}{ds^2} \left (\mathcal{L}(y'') \right)=s^2Y''(s)+4sY'(s)+2Y(s)$$ And: $$ \mathcal{L}(t y')=-\dfrac {d}{ds} \left (\mathcal{L}(y') \right)=-(sY'(s)+Y(s))$$ You differentiate with respect to the variable $s$ not $t$. You should only have in your final answer functions of $s$. not a mix with variable $t$ and variable $s$