Show that y = (d^(n-1))/(dx^(n-1))l[(1-x^2)^((2n-1)/(2))r]…

Show that $y = \frac{d^{\,n-1}}{dx^{\,n-1}}\Bigl[(1-x^2)^{\frac{2n-1}{2}}\Bigr]$ is a particular solution of $(x^2-1)y'' + xy' - n^2y = 0$, for $n\in\mathbb{N}$ and $|x|\lt 1$.

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

Set $F(x) = (1-x^2)^{n-\frac12}$, so that $y = F^{(n-1)}(x)$ is the quantity to be examined. Differentiating $F$ directly, \[F'(x) = \left(n-\frac12\right)(1-x^2)^{n-\frac32}(-2x) = -(2n-1)\,x\,(1-x^2)^{n-\frac32}.\]

Since $(1-x^2)^{n-3/2} = (1-x^2)^{n-1/2}/(1-x^2) = F(x)/(1-x^2)$, this can be rewritten as

\[ F'(x) = -\frac{(2n-1)x}{1-x^2}\,F(x), \]

which is equivalent to the first-order linear equation

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

Step 2: Differentiating $(1)$ $n$ times via Leibniz's rule

Apply the operator $D^n = d^n/dx^n$ to both terms of $(1)$, using the general Leibniz rule $D^n[uv] = \sum_{j=0}^n \binom{n}{j}u^{(j)}v^{(n-j)}$. Only finitely many derivatives of the polynomial factors survive: for $u=1-x^2$, $u^{(0)}=1-x^2$, $u^{(1)}=-2x$, $u^{(2)}=-2$, and $u^{(j)}=0$ for $j\geq3$; for $u=(2n-1)x$, $u^{(0)}=(2n-1)x$, $u^{(1)}=2n-1$, and $u^{(j)}=0$ for $j\geq2$.

Applying Leibniz's rule to the first term of $(1)$, with $v=F'$ so that $v^{(n-j)}=F^{(n-j+1)}$: $D^n\bigl[(1-x^2)F'\bigr] = (1-x^2)F^{(n+1)} - 2n\,x\,F^{(n)} - n(n-1)\,F^{(n-1)}$.

and to the second term, with $v=F$: $D^n\bigl[(2n-1)xF\bigr] = (2n-1)\,x\,F^{(n)} + n(2n-1)\,F^{(n-1)}$.

Step 3: Assembling and simplifying

Since $D^n$ of the zero right-hand side of $(1)$ is still zero, adding the two expansions above gives

\[ (1-x^2)F^{(n+1)} + \bigl[-2n+(2n-1)\bigr]x\,F^{(n)} + \bigl[-n(n-1)+n(2n-1)\bigr]F^{(n-1)} = 0. \]

The bracketed coefficients collapse sharply: the $xF^{(n)}$ coefficient is $-2n+2n-1=-1$, and the $F^{(n-1)}$ coefficient is $n\bigl[-(n-1)+(2n-1)\bigr]=n\cdot n=n^2$. Hence

\[ (1-x^2)F^{(n+1)} - x\,F^{(n)} + n^2F^{(n-1)} = 0. \tag{2} \]

Step 4: Identifying $y$ and its derivatives

Set $y = F^{(n-1)}(x)$, so that $y' = F^{(n)}$ and $y''=F^{(n+1)}$. Equation $(2)$ becomes

\[ (1-x^2)y'' - xy' + n^2y = 0, \]

and multiplying through by $-1$ yields exactly the required equation: $(x^2-1)y'' + xy' - n^2y = 0$.

Step 5: Verification

Substituting $y=D^{n-1}\bigl[(1-x^2)^{n-1/2}\bigr]$ into $(x^2-1)y''+xy'-n^2y$ for $n=1,2,3,4$ and expanding the required derivatives explicitly confirms the residual is identically $0$ in every case, consistent with the general argument above.

\[y = \frac{d^{\,n-1}}{dx^{\,n-1}}\bigl[(1-x^2)^{n-\frac12}\bigr] \ \text{ solves } \ (x^2-1)y'' + xy' - n^2y = 0\]