Finding patterns in differential equation coefficients Suppose I have a function, $b_N(x)$ which satisfies the following differential equations:
$$-Nb_N(x)^2=xb_N'(x)+\color{blue}{(N-x)}b_N(x)$$
$$2N^2b_N(x)^3=x^2b_N''(x)-\left\{x\left[(N-1)-x\right]+2x(N-x)\right\}b_N'+x\left[x+2\color{blue}{(N-x)^2}\right]b_N(x)$$
$$-6N^3b_N(x)^4=\alpha_1b_N'''(x)+\alpha_2b_N''(x)+\alpha_3b_N'(x)+\alpha_4b_N(x)$$
where
$$\alpha_1=x^3$$
$$\alpha_2=3x^2(N-x)+\left\{x^2\left[(N-1)-x\right]+2x^2(N-x)-2x^2\right\}$$
$$\alpha_3=\left\{3x(N-x)\left[(N-1)-x\right]+6x(N-x)^2\right\}+\left\{5x^2+2x(N-x)^2+x(N-1)+2x(N-x)\right\}$$
$$\alpha_4=x-7x(N-x)-6\color{blue}{(N-x)^3}$$
I'm trying to connect the coefficients of the third differential equation to the coefficients of the first and second order equations above.  It is clear that in each case the term of highest order (highest derivative) will have degree $m-1$ if $m$ is the power of the term on the LHS.  And the LHS will be $m!N^mb_N(x)^{(m+1)}$.  Now there seem to be a number of ways to go about finding a pattern in the coefficients of the higher order differential equation coefficients.  Looking at the differential equations above, all three involve successive powers of $(N-x)$.  For example, the $b_N$ terms in all three are in order
$$\color{blue}{(N-x)}$$
$$x\left[x+2\color{blue}{(N-x)^2}\right]$$
$$x-7x(N-x)-6\color{blue}{(N-x)^3}$$
Clearly a pattern that might repeat with successive iterations.  However, my first instinct when I began this problem was to simply distribute out and write each coefficient as a polynomial in standard form.  For example, the first and second order differential equations would then look like
$$-Nb_N(x)^2=xb_N'(x)+(x-N)b_N(x)$$
$$2N^2b_N(x)^3=\color{red}{1}x^2b_N''(x)+\left[\color{red}{3}x^2+(1-3N)x\right]b_N'(x)+\left[\color{red}{2}x^2+(1-4N)x+2N^2\right]b_N(x)$$
This also looks like there could be some pattern if we continued on to the third order and took the derivative of both sides.  For example, the leading coefficients from each polynomial coefficients in the second order equation are the coefficients of the terms from the expression
$$(x+1)(x+2)=\color{red}{1}x^2+\color{red}{3}x+\color{red}{2}$$
where 1 is the leading coefficient from $b_N''(x), 3$ is the leading coefficient from $b_N'(x),$ etc.  This is also true for the third order differential equation generated by $6N^3b_N(x)^4$ and the coefficents are generated by $(x+1)(x+2)(x+3)$.  My question is this.  Does it seem more advantageous to keep the original form where the coefficients are in terms of $(N-x)$, or $[(N-1)-x]$ or would it be easier to put all coefficients in standard polynomial form?  I've got a number of calculations for coefficients $\alpha_i$ but none really look like the first and second order equations.  Perhaps I'm also missing something.  
 A: There's a pattern but it's somewhat complicated. Let's write your relation as
$$A_m b^{m+1} = a_{0,m} b + a_{1,m} b' + \cdots + a_{m,m} b^{(m)}$$
where $A_m$ is a constant and $a_{i,j}$ are functions of $x$. Taking the derivative of the $LHS$ gives
$$(A_m b^{m+1})' = A_m (m+1) b^m b'$$
We already know that
$$A_1 b^2 = a_{0,1}b + a_{1,1}b'$$
so we solve this for $b'$
$$b' = \frac{A_1 b^2 - a_{0,1}b}{a_{1,1}}$$
Substituting this back into the $LHS$ expression gives
$$LHS = A_m (m+1) b^m \frac{A_1 b^2 - a_{0,1}b}{a_{1,1}} = \frac{(m+1)A_1 A_m}{a_{1,1}}b^2 - \frac{a_{0,1}}{a_{1,1}}(m+1)A_m b^{m+1}$$
Here we can use the $A_m$ relation. So
$$a_{1,1} LHS = (m+1)A_1A_m b^{m+2} - a_{0,1}(m+1)(a_{0,m} b + a_{1,m} b' + \cdots + a_{m,m} b^{(m)})$$
Therefore we can write
$$(m+1)A_1 A_m b^{m+2} = a_{0,1}(m+1)(a_{0,m} b + a_{1,m} b' + \cdots + a_{m,m} b^{(m)}) + a_{1,1}RHS$$
The derivative of the $RHS$ is
$$RHS' = (a_{0,m}' b + a_{1,m}' b' + \cdots + a_{m,m}' b^{(m)}) +( a_{0,m} b' + a_{1,m} b'' + \cdots + a_{m,m} b^{(m+1)})$$
Combining terms gives
$$RHS' = a_{0,m}' b + (a_{1,m}'+a_{0,m})b' + \cdots (a_{m,m}'+a_{m-1,m})b^{(m)} + a_{m,m} b^{(m+1)}$$
Finally
$$(m+1)A_1 A_m b^{m+2} = a_{0,1}(m+1)(a_{0,m} b + a_{1,m} b' + \cdots + a_{m,m} b^{(m)}) + a_{1,1}(a_{0,m}' b + (a_{1,m}'+a_{0,m})b' + \cdots (a_{m,m}'+a_{m-1,m})b^{(m)} + a_{m,m} b^{(m+1)})$$
and combining terms yields
$$\begin{align*}A_{m+1} b^{m+2} = &((m+1)a_{0,1}a_{0,m} + a_{1,1}a_{0,m}')b 
\\&+(a_{0,1}(m+1)a_{1,m}+a_{1,1}(a_{1,m}'+a_{0,m})) b' + 
\\&+\cdots  
\\&+(a_{0,1}(m+1) + a_{1,1}(a_{m,m}'+a_{m-1,m}))b^{(m)}  + 
\\&+a_{1,1}a_{m,m} b^{(m+1)}
\\&= a_{0,m+1}b + a_{1,m+1}b' + \cdots + a_{m,m+1}b^{(m)} + a_{m+1,m+1} b^{(m+1)}
\end{align*}$$
So we can read off the coefficient recurrence relations directly as
$$\begin{align*}
A_{m+1} &= (m+1)A_1 A_m\\
a_{0,m+1} &= (m+1)a_{0,1}a_{0,m} + a_{1,1}a_{0,m}'\\
a_{i,m+1} &= (m+1)a_{0,1}a_{i,m}+a_{1,1}(a_{i,m}'+a_{i-1,m}),\quad \text{for all } 1\leq i \leq m\\
a_{m+1,m+1} &= a_{1,1}a_{m,m} b^{(m+1)}
\end{align*}$$
Finally, from your equations
$$A_{1} = -n, \quad a_{0,1} = n-x, \quad a_{1,1} = x$$
The recurrence relations become
$$\begin{align*}
A_{m+1} &= -n(m+1) A_m\\
a_{0,m+1} &= (m+1)(n-x)a_{0,m} + xa_{0,m}'\\
a_{i,m+1} &= (m+1)(n-x)a_{i,m}+x(a_{i,m}'+a_{i-1,m}),\quad \text{for all } 1\leq i \leq m\\
a_{m+1,m+1} &= x a_{m,m} b^{(m+1)}
\end{align*}$$
I could go further and try to solve these recurrences if I have time.
