Anyone can help me to convert the following maple pdsolve expressed by the hypergeom function to the $LegendreP(n,b,x)$ or $Q$ function?
\begin{equation} dsolve\Big( (1-x^2)\cdot \frac{d^2 y(x)}{dx^2} + n(n+1) \cdot y(x) = 0 \Big); \\ y(x) = c_1 (-1+x^2) \cdot hypergeom \Big (\Big[\frac{3}{4} - \frac{1}{4} \cdot \sqrt{4 \cdot n(n+1)+1}, \frac{3}{4} + \frac{1}{4} \cdot \sqrt{4 \cdot n(n+1)+1} \Big], \Big[\frac{1}{2}\Big],x^2 \Big) + c_2 (x^3-x) \cdot hypergeom \Big( \Big[\frac{5}{4} + \frac{1}{4} \cdot \sqrt{4 \cdot n(n+1)+1}, \frac{5}{4} - \frac{1}{4} \cdot \sqrt{4 \cdot n(n+1)+1} \Big], \Big[\frac{1}{2} \Big],x^2 \Big) \end{equation}
I just want $y(x)$ to be expressed in the form of $LegendreP(n,b,x)$.
This is not an answer to the question.
Using two different CAS, the solution given for $$(1-x^2)y''+n(n+1)y=0$$ is $$y=c_1 \times _2F_1\left(-\frac{n+1}{2},\frac{n}{2};\frac{1}{2};x^2\right)+c_2\times x \times _2F_1\left(\frac{n+1}{2},-\frac{n}{2};\frac{3}{2};x^2\right)$$ For $n=1$, this would give $$y=c_1 \left(x^2-1\right)+c_2 \left(\left(x^2-1\right) \tanh ^{-1}(x)-x\right)$$ For $n=2$, this would give $$y=c_1 x \left(x^2-1\right)+c_2 \left(-3 x^2+3 \left(x^2-1\right) x \tanh ^{-1}(x)+2\right)$$ For $n=3$, this would give $$y=c_1 \left(x^4-\frac{6 x^2}{5}+\frac{1}{5}\right)+ c_2 \left(-15 x^3+13x+3 \left(5 x^4-6 x^2+1\right) \tanh ^{-1}(x)\right)$$ For $n=4$, this would give $$y=c_1 \left(x^5-\frac{10 x^3}{7}+\frac{3 x}{7}\right)+c_2 \left(-105 x^4+115 x^2-16+15 \left(7 x^4-10 x^2+3\right) x \tanh ^{-1}(x)\right)$$
You can try this and simplify the result going from $\log(.)$ to $\tanh ^{-1}(.)$.
Update
Concerning, as asked later in comments, about polynomial forms of the solutions of $$(1-x^2)y''+ny=0$$ the general solution of which being $$y=c_1 \times \, _2F_1\left(-\frac{\sqrt{4 n+1}+1}{4} ,\frac{\sqrt{4 n+1}-1}4 ;\frac{1}{2};x^2\right)+$$ $$\qquad c_2 \times x \times \, _2F_1\left(-\frac{\sqrt{4 n+1}-1}{4} ,\frac{\sqrt{4 n+1}+1}4;\frac{3}{2};x^2\right)$$ only those for which the first and second arguments of the hypergeometric functions are multiples of $\frac 12$ will show this kind of forms.
For example,
$$\left( \begin{array}{ccc} n & \, _2F_1\left(-\frac{\sqrt{4 n+1}+1}{4} ,\frac{\sqrt{4 n+1}-1}4 ;\frac{1}{2};x^2\right) \\ 2 & 1-x^2 \\ 6 & \frac{1}{2} \left(2-3 x^2\right)+\frac{3}{2} x \left(x^2-1\right) \tanh ^{-1}(x) \\ 12 & \left(1-5 x^2\right) \left(1-x^2\right) \\ 20 & \frac{1}{16} \left(105 x^4-115 x^2+16\right)-\frac{15}{16} \left(7 x^5-10 x^3+3 x\right) \tanh ^{-1}(x) \\ 30 & \left(1-x^2\right) \left(21 x^4-14 x^2+1\right) \end{array} \right)$$ $$\left( \begin{array}{ccc} n & \, _2F_1\left(-\frac{\sqrt{4 n+1}-1}{4} ,\frac{\sqrt{4 n+1}+1}4;\frac{3}{2};x^2\right) \\ 2 & \frac{\left(1-x^2\right) \tanh ^{-1}(x)}{2 x}+\frac{1}{2} \\ 6 & 1-x^2 \\ 12 & \frac{1}{16} \left(13-15 x^2\right)+\frac{3 \left(5 x^4-6 x^2+1\right) \tanh ^{-1}(x)}{16 x} \\ 20 & \frac{1}{3} \left(3-7 x^2\right) \left(1-x^2\right) \\ 30 & \frac{1}{128} \left(315 x^4-420 x^2+113\right)-\frac{15 \left(21 x^6-35 x^4+15 x^2-1\right) \tanh ^{-1}(x)}{128 x} \end{array} \right)$$