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Leibniz's Formula - Differential equation

Use Leibniz's formula to establish that:

$Z_{n} \equiv \frac{\mathrm{d^n} }{\mathrm{d} x^n} e^{\frac{-x^2}{2}}$

is a solution of the differential equation:

$\frac{\mathrm{d^2Z_n} }{\mathrm{d} x^2} + x\frac{\mathrm{dZ_n} }{\mathrm{d} x} + (n+1)Z_n = 0$

I have no idea how to go about this, besides the following:

$\frac{\mathrm{d^n} }{\mathrm{d} x^n} e^\frac{-x^2}{2} = \frac{\mathrm{d^n} }{\mathrm{d} x^n} xe^\frac{-x^2}{2} /x = x(e^\frac{-x^2}{2}/x)^{(n)} +n(e^\frac{-x^2}{2}/x)^{(n-1)}$

I then differentiated this once to get dZn/dx, and then again for d^2, and substituted the results into the equation - however I got nothing out of doing so. Any help please? :frown:

:frown:

OP

Original post by TeeEm

http://www.madasmaths.com/archive/maths_booklets/advanced_topics/odes_series_solutions.pdf

Thanks, sorry but I still can't quite find my way around this. The section for Leibniz all seem to want differentiating the entire DE n times, but I then end up with differentiating Zn, which is a differential in itself...

Original post by angrybirdzzz

Thanks, sorry but I still can't quite find my way around this. The section for Leibniz all seem to want differentiating the entire DE n times, but I then end up with differentiating Zn, which is a differential in itself...

What Leibniz is really good for is finding the both derivative of a product if 2 functions. Your issue here is that exp(-x^2/2) isn't a product of 2 functions, o Leibniz doesn't apply directly. It's derivative, however....

OP

Original post by DFranklin

What Leibniz is really good for is finding the both derivative of a product if 2 functions. Your issue here is that exp(-x^2/2) isn't a product of 2 functions, o Leibniz doesn't apply directly. It's derivative, however....

Cheers, I actually just figured the derivative was a good starting point as well! :biggrin:

:biggrin:

username1162972

Original post by angrybirdzzz

Use Leibniz's formula to establish that:

$Z_{n} \equiv \frac{\mathrm{d^n} }{\mathrm{d} x^n} e^{\frac{-x^2}{2}}$

is a solution of the differential equation:

$\frac{\mathrm{d^2Z_n} }{\mathrm{d} x^2} + x\frac{\mathrm{dZ_n} }{\mathrm{d} x} + (n+1)Z_n = 0$

I have no idea how to go about this, besides the following:

$\frac{\mathrm{d^n} }{\mathrm{d} x^n} e^\frac{-x^2}{2} = \frac{\mathrm{d^n} }{\mathrm{d} x^n} xe^\frac{-x^2}{2} /x = x(e^\frac{-x^2}{2}/x)^{(n)} +n(e^\frac{-x^2}{2}/x)^{(n-1)}$

I then differentiated this once to get dZn/dx, and then again for d^2, and substituted the results into the equation - however I got nothing out of doing so. Any help please? :frown:

:frown:

Helped a natsci do this.

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