Derive the Differential Equation for Laguerre polynomials

Question

How can i derive this linear differential equation

$$ xD^2L^{(\alpha)}_n (x)+(1+\alpha -x)DL^{(\alpha)}_n (x)+nL^{(\alpha)}_n (x)=0 $$

from the following recurrence relations

$$\frac{d}{dx}L^{(\alpha)}_n (x)=\frac{1}{x}[ nL^{(\alpha)}_n (x)-(\alpha +n) L^{(\alpha)}_{n-1} (x) ]$$ $$\frac{d}{dx}L^{(\alpha)}_n (x)=\frac{1}{x}[ (x-\alpha-n-1)L^{(\alpha)}_n (x)+(n+1) L^{(\alpha)}_{n+1} (x) ]$$

My work:

$D[xL^{(\alpha)}_n (x)]=D[ (x-\alpha-n-1)L^{(\alpha)}_n (x)+(n+1) L^{(\alpha)}_{n+1} (x) ]$

$xD^2 L^{(\alpha)}_n (x)+DL^{(\alpha)}_n (x)=(-\alpha -n)L^{(\alpha)}_n (x)+(x-\alpha -n-1) DL^{(\alpha)}_n (x)+(n+1)DL^{(\alpha)}_{n+1} (x)$

$xD^2 L^{(\alpha)}_n (x)+DL^{(\alpha)}_n (x)=(-\alpha -n)L^{(\alpha)}_n (x)+(x-\alpha -n-1) DL^{(\alpha)}_n (x)+\frac{(n+1)}{x}[(n+1)L^{(\alpha)}_{n+1} (x)-(\alpha +n+1)L^{(\alpha)}_n]$

$xD^2 L^{(\alpha)}_n (x)+(1+\alpha -x)DL^{(\alpha)}_n (x)=(-\alpha -n)L^{(\alpha)}_n (x)+(-n-1) DL^{(\alpha)}_n (x)+\frac{(n+1)}{x}[(n+1)L^{(\alpha)}_{n+1} (x)-(\alpha +n+1)L^{(\alpha)}_n]$

$xD^2 L^{(\alpha)}_n (x)+(1+\alpha -x)DL^{(\alpha)}_n (x)+nL^{(\alpha)}_n (x)=-\alpha L^{(\alpha)}_n (x)-(n+1) DL^{(\alpha)}_n (x)+\frac{(n+1)^2}{x}L^{(\alpha)}_{n+1} (x)-\frac{(n+1)(\alpha +n+1)}{x}L^{(\alpha)}_n$

Now $ L.H.S=-\alpha L^{(\alpha)}_n (x)-\frac{(n+1)}{x}[(x-\alpha-n-1)L^{(\alpha)}_n (x)+(n+1) L^{(\alpha)}_{n+1} (x) ]+\frac{(n+1)^2}{x}L^{(\alpha)}_{n+1} (x)-\frac{(n+1)(\alpha +n+1)}{x}L^{(\alpha)}_n$

$ L.H.S=-\alpha L^{(\alpha)}_n (x)-\frac{(n+1)(x-\alpha-n-1)}{x}L^{(\alpha)}_n (x)-\frac{(n+1)(\alpha +n+1)}{x}L^{(\alpha)}_n$

$$ L.H.S = (-\alpha -n-1)L^{(\alpha)}_n$$

But is must be zero

please,some help .

Answer 1

By comparison with the page below, the discrepancy comes from the coefficient of term underlined in red.