Show that z(x) = (d^n)/(dx^n)l[(x-1)^(n+p)(x+1)^(n+q)r] is a…

Show that $z(x) = \frac{d^n}{dx^n}\Bigl[(x-1)^{n+p}(x+1)^{n+q}\Bigr]$ is a particular solution of the Jacobi-type equation $(E)\!: \qquad (1-x^2)y'' + \bigl[(p+q-2)x + (p-q)\bigr]y' + (n+1)(n+p+q)\,y = 0$.

Step 1: A first-order equation for the generating function

Set $F(x) = (x-1)^{n+p}(x+1)^{n+q}$, so that $z = F^{(n)}$. Differentiating $F$ by the product rule, \[F' = (n+p)(x-1)^{n+p-1}(x+1)^{n+q} + (n+q)(x-1)^{n+p}(x+1)^{n+q-1} = F\left[\frac{n+p}{x-1}+\frac{n+q}{x+1}\right].\]

Combining the bracket over the common denominator $x^2-1$ gives $\dfrac{n+p}{x-1}+\dfrac{n+q}{x+1} = \dfrac{(2n+p+q)x+(p-q)}{x^2-1}$, so $F$ satisfies the first-order equation

\[ (x^2-1)F' = \bigl[(2n+p+q)x+(p-q)\bigr]F. \tag{1} \]

Step 2: Promoting to a second-order equation for $F$

Differentiate (1) once more with respect to $x$. On the left, $\dfrac{d}{dx}\bigl[(x^2-1)F'\bigr] = 2xF' + (x^2-1)F''$; on the right, the product rule gives $(2n+p+q)F + \bigl[(2n+p+q)x+(p-q)\bigr]F'$. Equating and collecting the $F'$ terms,

\[ (x^2-1)F'' + \Bigl[2-(2n+p+q)\Bigr]x - (p-q)\Bigr]F' - (2n+p+q)F = 0. \tag{2} \]

Step 3: Applying $n$ derivatives via Leibniz's rule

Apply the operator $D^n=\dfrac{d^n}{dx^n}$ to (2) term by term. Since $(x^2-1)$ is a quadratic, only its $0$th, $1$st and $2$nd derivatives are nonzero ($x^2-1$, $2x$, $2$), and since $\bigl[2-(2n+p+q)\bigr]x-(p-q)$ is linear, only its $0$th and $1$st derivatives survive. The generalized Leibniz rule then gives

\[ D^n\bigl[(x^2-1)F''\bigr] = (x^2-1)F^{(n+2)} + 2nx\,F^{(n+1)} + n(n-1)F^{(n)}, \]

\[ D^n\Bigl\{\bigl[(2-(2n+p+q))x-(p-q)\bigr]F'\Bigr\} = \bigl[(2-(2n+p+q))x-(p-q)\bigr]F^{(n+1)} + n\bigl(2-(2n+p+q)\bigr)F^{(n)}, \]

\[ D^n\bigl[-(2n+p+q)F\bigr] = -(2n+p+q)F^{(n)}. \]

Step 4: Collecting coefficients

Summing the three lines, the coefficient of $F^{(n+1)}$ is $2n+(2-(2n+p+q)) = 2-(p+q)$, which combines with the surviving linear term to give $-\bigl[(p+q-2)x+(p-q)\bigr]$; the coefficient of $F^{(n)}$ is

\[ n(n-1) + n\bigl(2-(2n+p+q)\bigr) - (2n+p+q) = -(n+1)(n+p+q), \]

obtained by expanding $n^2-n+2n-2n^2-n(p+q)-2n-p-q = -n^2-n-n(p+q)-(p+q) = -(n+1)(n+p+q)$. Substituting these back, and writing $z=F^{(n)}$ so that $z''=F^{(n+2)}$, $z'=F^{(n+1)}$:

\[ (x^2-1)z'' - \bigl[(p+q-2)x+(p-q)\bigr]z' - (n+1)(n+p+q)z = 0. \]

Multiplying through by $-1$ reproduces $(E)$ exactly: $(1-x^2)z'' + \bigl[(p+q-2)x+(p-q)\bigr]z' + (n+1)(n+p+q)z = 0. \qquad \blacksquare$.

Step 5: Verification

Substituting $z=F^{(n)}$ with $F=(x-1)^{n+p}(x+1)^{n+q}$ into $(E)$ and simplifying symbolically confirms an identically vanishing residual for every tested combination, including non-integer $p,q$:

\[ (n,p,q) = \left(1,\tfrac12,\tfrac13\right),\ (2,0,0),\ (2,1,-1),\ \left(3,-\tfrac12,\tfrac14\right) \;\Longrightarrow\; \text{residual} \equiv 0.\ \checkmark \]

\[z(x) = \frac{d^n}{dx^n}\Bigl[(x-1)^{n+p}(x+1)^{n+q}\Bigr] \ \text{solves}\ (1-x^2)y''+\bigl[(p+q-2)x+(p-q)\bigr]y'+(n+1)(n+p+q)y=0\]