I've read a piece in MathWorld where a distribution function $P(A)$ satisfies the ODE $$A^{3}(1-2A)P^{(4)}(A)+A^{2}P^{(3)}(A)-4A^{2}P''(A)+8AP'(A)-8P(A)-96(2A-1)=0.\label{a}\tag{1}$$
It is then inferred that the raw moments of $P$ are $$\mu_{n}'=\frac{3\cdot2^{3-n}\left[\left(n+2\right)H_{n+1}+1\right]}{(n+1)(n+2)^{3}(n+3)^{2}}\label{b}\tag{2}$$
Where $\mu_n'=\mathbb{E}[X^n]=\int_0^1x^ndP(x)$ are the raw moments and $H_n = \sum_{k=1}^n\frac{1}{k}$ is the harmonic number.
My question is: how is (2) derived from (1)? My guess is this has to do with the Laplace transform but this is my first encounter with this type of ODEs so I don't know how they may be solved.
This is an interesting problem, but it seems like it requires a lot of work (at least for me). I can show you how one can solve the equation and find the probability distribution $P(x)$ "explicitly" (at least up to possibly repeated integrals of explicit functions). Let me replace the independent variable $A$ by $x$, i.e. $x=A$. $$x^3(1-2x)P^{(4)}(x) + x^2P^{(3)}(x) - 4x^2P^{''}(x) + 8xP^{'}(x) - 8P(x) - 96(2x-1) = 0$$ which is the same as $$(x^3 - 2x^4)P^{(4)}(x) + x^2P^{(3)}(x) - 4x^2P^{''}(x) + 8xP^{'}(x) - 8P(x) = 192x - 96.$$ First, differentiate the equation with respect to $x$ once and obtain \begin{align*} &(x^3-2x^4)P^{(5)}(x) + (3x^2-8x^3)P^{(4)}(x) \\ & + x^2P^{(4)}(x) + 2xP^{(3)}(x) \\ & - 4x^2P^{(3)}(x) - 8xP^{''}(x) \\ & + 8x P^{''}(x) + 8P^{'}(x) \\ & - 8P^{'}(x) = 192. \end{align*} and after cancelling out the repeating terms with opposite signs we obtain the equation $$(x^3-2x^4)P^{(5)}(x) + (3x^2-8x^3)P^{(4)}(x) + x^2P^{(4)}(x) + 2xP^{(3)}(x)- 4x^2P^{(3)}(x) = 192.$$ Grouping together the terms with the same derivatives turns the equation into $$(x^3-2x^4)P^{(5)}(x) + (4x^2-8x^3)P^{(4)}(x) + (2x - 4x^2)P^{(3)}(x) = 192$$ which is the same as $$x^3(1-2x)P^{(5)}(x) + 4x^2(1 - 2x)P^{(4)}(x) + 2x(1 - 2x)P^{(3)}(x) = 192.$$ Divide both sides of the equation by $(1-2x)$ and obtain $$x^3 P^{(5)}(x) + 4 x^2 P^{(4)}(x) + 2 x P^{(3)}(x) = \frac{192}{1-2x}.$$ Let us set $f(x) = P^{''}(x)$. Then the latter differential equation becomes $$x^3 f^{'''}(x) + 4 x^2 f^{''}(x) + 2 x f^{'}(x) = \frac{192}{1-2x}.$$ Next, perform a change of the independent variable $x=e^u$ and $u = \log{x}$ for $x>0$ or $x=-e^u$ and $u = \log{(- x)}$ for $x<0$. We work with the $x>0$ case, the case $x<0$ is analogous.
Set $g(u) = f(e^u)$. Differentiate three times the latter substitution with respect to $u$ to obtain: \begin{align*} g{'}(u) &= f^{'}(e^u) e^u;\\ g{''}(u) &= f^{''}(e^u) e^{2u} + f^{'}(e^u) e^u; \\ g{'''}(u) &= f^{'''}(e^u) e^{3u} + 2 f^{''}(e^u) e^{2u} + f^{''}(e^u) e^{2u} + f^{'}(e^u) e^{u};\\ &= f^{'''}(e^u) e^{3u} + 3 f^{''}(e^u) e^{2u} + f^{'}(e^u) e^{u}. \end{align*} Remembering that $e^u = x$, we obtain \begin{align*} g{'}(u) &= x f^{'}(x);\\ g{''}(u) &= x^2 f^{''}(x) + x f^{'}(x); \\ g{'''}(u) &= x ^3 f^{'''}(x) + 3 x^2 f^{''}(x) + x f^{'}(x). \end{align*} Reexpressing the derivatives of $f$ in terms of the derivatives of $g$, we get \begin{align*} x f^{'}(x) &= g'(u);\\ x^2 f^{''}(x) & = g{''}(u) - x f^{'}(x) \\ &= g{''}(u) - g^{'}(u);\\ x ^3 f^{'''}(x) &= g{'''}(u) - 3 x^2 f^{''}(x) - x f^{'}(x)\\ &= g{'''}(u) - 3 \big(g{''}(u) - g^{'}(u)\big) - g'(u) \\ &= g{'''}(u) - 3 g{''}(u) + 2 g^{'}(u). \end{align*} Plug these expression in the latter differential equation for $f$ to obtain the following differential equation for $g$ \begin{align*} & x^3 f^{'''}(x) + 4 x^2 f^{''}(x) + 2 x f^{'}(x) \\ &= g{'''}(u) - 3 g{''}(u) + 2 g^{'}(u) + 4\big(g{''}(u) - g^{'}(u) \big) + 2 g'(u) \\ &= g{'''}(u) - 3 g{''}(u) + 2 g^{'}(u) + 4 g{''}(u) - 4 g^{'}(u) + 2 g'(u)\\ &= \frac{192}{1-2e^u} \end{align*} which reduces to the following simple differential equation $$g{'''}(u) + g{''}(u) = \frac{192}{1-2e^u}$$ which can be solved explicitly. Simply substitute $h(u) = g''(u)$ and obtain $$h{'}(u) + h(u) = \frac{192}{1-2e^u}.$$ Its general solution is \begin{align*} h(u) = g''(u) & = e^{-u}\left(A + \int{\frac{192 e^u}{1-2e^u} du }\right)\\ & = e^{-u}\left(A - \frac{192}{2} \int{\frac{ (-2) e^u}{1-2e^u} du }\right) \\ & = e^{-u}\left(A - 96 \log{(2e^u-1)}\right) \\ & = A e^{-u} - 96 \, e^{-u} \log{(2e^u-1)}. \end{align*} Integrating once more gives us \begin{align*} g'(u) & = B + A \int e^{-u} du - 96 \int e^{-u} \log{(2e^u-1)} du\\ & = B - A e^{-u} - 96 \int e^{-u} \log{(2e^u-1)} du \end{align*} We need to solve $$\int e^{-u} \log{(2e^u-1)} du.$$ \begin{align*} \int e^{-u} \log{(2e^u-1)} du &= -\int \log{(2e^u-1)} d(e^{-u})\\ &= - e^{-u} \log{(2e^u-1)} + \int e^{-u} d\big( \log{(2e^u-1)} \big)\\ &= - e^{-u} \log{(2e^u-1)} + \int e^{-u} \frac{2e^u}{2e^u-1} du\\ &= - e^{-u} \log{(2e^u-1)} + 2 \int \frac{1}{2e^u-1} du \\ &= - e^{-u} \log{(2e^u-1)} - 2 u + 2 \log(2e^u-1)\\ &= - e^{-u} \log{(2e^u-1)} + 2 \log(2-e^{-u}) \end{align*} So $$g'(u)=B-Ae^{-u} + 96 \, e^{-u} \log{(2e^u-1)} - 192 \,\log(2-e^{-u}) $$ Since $u = \log{x}$ and $xf'(x) = g'(u)$ we obtain \begin{align*} f'(x) &= \frac{1}{x} g'(\log{x}) = \frac{1}{x}\left( B-\frac{A}{x} + \frac{96}{x} \log{(2x-1)} - 192 \,\log\Big(2-\frac{1}{x}\Big)\right)\\ &= \frac{B}{x} -\frac{A}{x^2} + \frac{96}{x^2} \log{(2x-1)} - \frac{192}{x} \,\log\left(2-\frac{1}{x}\right) \end{align*} Recall that $f(x) = P''(x)$, which means that $f'(x) = P'''(x)$ \begin{align*} P^{'''}(x) &= \frac{1}{x}\left( B-\frac{A}{x} + \frac{96}{x} \log{(2x-1)} - 192 \,\log\Big(2-\frac{1}{x}\Big)\right)\\ &= \frac{B}{x} -\frac{A}{x^2} + \frac{96}{x^2} \log{(2x-1)} - \frac{192}{x} \,\log\left(2-\frac{1}{x}\right) \end{align*} Now, one can integrate repeatedly the formula for $P^{'''}(x)$ three times to obtain a general expression for $P(x)$ depending on five independent constant parameters $A, B, C, D, E, F$ which need to be determined by first plugging the expression for $P(x)$ into the original differential equation, and second by taking into account the requirements that $\int_{-\infty}^{\infty} P^{'}(x) dx = 1$ and $0 \leq P(x) \leq 1$.