Leibniz formula: Prove z=(d^n)/(dx^n) (x^2-1)^n satisfies (1-x^2)(d^2z)/(dx^2)-2x(dz/

[Faker97]

Hi guys, my first post on this forum. Not sure how to solve this problem and any help would be appreciated

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\(\displaystyle \mbox{Prove that }\, z\, =\, \dfrac{d^n}{dx^n}\, (x^2\, -\, 1)^n\, \mbox{ satisfies the differential equa}\mbox{tion}\)

. . . . .\(\displaystyle (1\, -\, x^2)\, \dfrac{d^2 z}{dx^2}\, -\, 2x\, \dfrac{dz}{dx}\, +\, n\, (n\, +\, 1)\, z\, =\, 0\)

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thanks.

https://gyazo.com/990ccebf2fe0bdb32589d6cecf60e176

 

[HallsofIvy]

The usual way to show that a given value (number or function) satisfies a given equation (algebraic or differential) is to put the value into the equation and see if it is satisfied!

Here, your differential equation is \(\displaystyle (1- x^2)\frac{d^2z}{dx^2}- 2x\frac{dz}{dx}+ n(n+1)z= 0\) and the purported solution is \(\displaystyle z= \frac{d^n}{dx^n}(x^2- 1)^n\).

So you want to show that \(\displaystyle (1- x^2)\frac{d^{n+ 2}}{dx^{n+2}}(x^2- 1)^n- 2x\frac{d^{n+ 1}}{dx^{n+1}}(x^2- 1)^n+ n(n+ 1)\frac{d^n}{dx^n}(x^2- 1)^n= 0\).

You might want to look at a few derivatives of \(\displaystyle (x^2- 1)^n\) to see if you can find a pattern.