I was reading in Wikipedia about Legendre's differential equation. I was particularly interested in the simple case of the equation given by $$ \left(1-x^2\right)y'' -2xy' + \lambda(\lambda+1)y = 0 \tag{1} $$ In the article, it is stated that one of the two linearly independent solutions to $(1)$ is given by $$ P_{\lambda}(x) = \,_2F_1 \left(-\lambda, \lambda+1; 1; \frac{1-x}{2}\right) \tag{2} $$ where $\,_2F_1$ represents the hypergeometric function. Since by definition, the hypergeometric function ${}_2F_1(a,b;c;t)$ is a solution to the hypergeometric differential equation given by $$ t(1-t) \ \ddot{y} + \left[c-(a+b+1)t \right] \dot{y} - ab\,y = 0 \tag{3} $$ if $(2)$ is a solution to $(1)$ then I should be able to re-write $(1)$ to look like $(3)$ with the corresponding parameters. With this in mind, taking \begin{align*} a &= - \lambda\\ b &= \lambda +1\\ c &= 1\\ t & = \frac{1-x}{2} \end{align*} From where we also get that $$ \dot{y} = \frac{dy}{dt} = \frac{dy}{dx} \cdot \underbrace{\frac{dx}{dt}}_{\color{blue}{-2}} = -2y' $$ which also tells us that $$ \ddot{y} = -2 y'' $$ Substituting this into $(3)$ gives \begin{align} \frac{1-x}{2}\left(1-\frac{1-x}{2}\right)\ddot{y} + &\left[1-\left(-\lambda+(\lambda +1)+1\right)\frac{1-x}{2} \right] \dot{y} - (-\lambda)(\lambda +1)y = 0\\ \implies \frac{1}{4}\left(1-x^2 \right)\left(-2 y''\right) +& \left[1-2\frac{1-x}{2} \right] \left(-2 y'\right) +\lambda(\lambda +1)y = 0\\ \implies \left(1-x^2 \right)y'' &+ 4x y' -2 \lambda(\lambda +1)y = 0 \tag{4} \end{align} and although $(4)$ looks really similar to $(1)$, they're not the same equation.
Could anyone tell me where did I make a mistake? Or alternatively, does anyone know another simple way to show that the Legendre function can be written in terms of the hypergeometric function? Thank you!
In the proposed derivation, it is correctly establish that $\dot y=-2y'$ but we have also $\ddot y=4y''$ (and not $\ddot y=-2y''$ as stated). With this correction, the ODE are identical.