# Math Help - Differential equation

1. ## Differential equation

I must show that the polynomial $H_{n}$ satisfies a diferential equation.I used some notation(1,2..) so that I don't rewrite the same thing again. By diferentiating $H_{n}(x) = -xH_{n-1}(x) - (n - 1)H_{n-2}(x) (1)$and
using induction on n,I'm supposed to show that, for n >= 1,
$H'_n(x) = -nH_{n-1}(x)$(2)
I have to use (2) to express $H_{n-1}$ and $H_{n-2}$ in terms of derivatives of $H_{n}$, and substitute these into (1) to show that
$H''_{n} - xH'_{n} + nH_{n} = 0$(3)
for n>= 0. Now let $\phi'_n(x) = exp(-(x^2)/4 )H_{n}(x).$ Using (3) I must show that
$\phi'_n(x) +(n+1/2-(x^2)/4)\phi_n(x)=0$

Also H_n is a polynomial defined as follows:
$H_{n}(x) for n = 0,1,2.... :$ $H_{0}(x) =1$ and $H_{1}(x) = -x$; then, for n >=2, H_{n} is defined by the recurrence
$H_{n}(x) = -xH_{n-1}(x) - (n - 1)H_{n-2}(x):$ (1)

Attempt at a solution

I need help with the entire problem mainly because for example $H'_n(x) = -nH_{n-1}(x)$ I can only prove it for the base case when but fot the general I get $H'_n(x) = -H_{n-1}(x)-xH_{n-1}(x)-(n - 1)H_{n-2}(x)=-H_{n-1}(x)-xH'_{n-1}(x)-nH'_{n-2}(x)-H'_{n-2}(x)$= $-H_{n-1}(x)-xnH'_{n-1}(x)+xH'_{n-2}(x)-n^2H'_{n-3}(x)+2nH'_{n-3}(x)-nH'_{n-3}(x)+2H'_{n-3}(x)$......and I do not know what do do next...if it isn't obvious I just substituted in the relation that I have to prove since I'm assuming that everything from n-1 is true......
Also for the other parts I still need help in showing (3) and (4).Thanks in advance

2. Originally Posted by StefanM
I must show that the polynomial $H_{n}$ satisfies a diferential equation.I used some notation(1,2..) so that I don't rewrite the same thing again. By diferentiating $H_{n}(x) = -xH_{n-1}(x) - (n - 1)H_{n-2}(x) (1)$and
using induction on n,I'm supposed to show that, for n >= 1,
$H'_n(x) = -nH_{n-1}(x)$(2)
I have to use (2) to express $H_{n-1}$ and $H_{n-2}$ in terms of derivatives of $H_{n}$, and substitute these into (1) to show that
$H''_{n} - xH'_{n} + nH_{n} = 0$(3)
for n>= 0. Now let $\phi'_n(x) = exp(-(x^2)/4 )H_{n}(x).$ Using (3) I must show that
$\phi'_n(x) +(n+1/2-(x^2)/4)\phi_n(x)=0$

Also H_n is a polynomial defined as follows:
$H_{n}(x) for n = 0,1,2.... :$ $H_{0}(x) =1$ and $H_{1}(x) = -x$; then, for n >=2, H_{n} is defined by the recurrence
$H_{n}(x) = -xH_{n-1}(x) - (n - 1)H_{n-2}(x):$ (1)

Attempt at a solution

I need help with the entire problem mainly because for example $H'_n(x) = -nH_{n-1}(x)$ I can only prove it for the base case when but fot the general I get $H'_n(x) = -H_{n-1}(x)-xH_{n-1}(x)-(n - 1)H_{n-2}(x)=-H_{n-1}(x)-xH'_{n-1}(x)-nH'_{n-2}(x)-H'_{n-2}(x)$= $-H_{n-1}(x)-xnH'_{n-1}(x)+xH'_{n-2}(x)-n^2H'_{n-3}(x)+2nH'_{n-3}(x)-nH'_{n-3}(x)+2H'_{n-3}(x)$......and I do not know what do do next...if it isn't obvious I just substituted in the relation that I have to prove since I'm assuming that everything from n-1 is true......
Also for the other parts I still need help in showing (3) and (4).Thanks in advance
I'm having a really hard time reading your post, but I think you are talking about Hermite polynomials.