How to solve the recurrence relation $d_n = a \sqrt{n+1}d_{n+1} + b \sqrt{n} d_{n-1}$ (when $n \in \mathbb { Z } ^ { + } )$, and $d_0 := a d_1$ How to solve the recurrence relation
$d_n = a \sqrt{n+1}d_{n+1} + b \sqrt{n} d_{n-1}$ (when $n \in \mathbb { Z } ^ { + } )$,
and $d_0 := a d_1$ to find the general form of $d_n$?
I was able to solve for the special case when either $a$ or $b$ are zero (dividing both sides by $\sqrt{n!}$), but not in the general case.
Update: It is interesting to note that the $d_n$'s are related to Hermite polynomials (see answers below). $d_n$'s are proportional to the coefficients of a squeezed state of Harmonic oscillator in the basis of Eigenfunctions of the Hamiltonian (That's how I obtained the recurrence relation). The eigenfunctions of Hamiltonian are also themselves related to the Hermite polynomials.
 A: Let $d_n=c_n\sqrt{n!}$, then we have $c_n=a(n+1)c_{n+1}+bc_{n-1}$ (when $n>0$) and $c_0=ac_1$.
Then $f(x)=\sum\limits_{n=0}^{\infty}c_n x^n$ satisfies $f(x)=af'(x)+bxf(x)$, solved by $f(x)=c_0 e^{(x-bx^2/2)/a}$.
It remains to expand it into a power series (in $x$). With a known formula in mind, this is easy.
A: Let us write the recurrence relationship under the form :
$$b \sqrt{n} d_{n-1} - d_n + a \sqrt{n+1}d_{n+1}=0\tag{1}$$
What will be presented here is not a solution by itself. It is an heuristic way (using linear algebra) to understand that Hermite polynomials (as established in the brilliant solution by @Metamorphy) are involved in the process.
I'll present it for the simple case $a=1$, $b=1$.
Consider the following symmetric tridiagonal $n \times n$ matrix :
$$M_n:=\begin{pmatrix}0&\sqrt{\frac12}&&&\\\sqrt{\frac12}&0&\sqrt{\frac22}&&\\&\sqrt{\frac22}&\ddots&\ddots&\\&&\ddots&\ddots&\sqrt{\frac{n-1}2}\\&&&\sqrt{\frac{n-1}2}&0\end{pmatrix}$$
as given in the solution to Proving that all the real roots of Hermite polynomials are in $(-\sqrt{4n+1}, \sqrt{4n+1})$.
Its characteristic polynomial can be shown to be 
$$\det(M_n-xI):=\dfrac{(-1)^n}{2^n}H_n(x)\tag{2]$$ 
where $H_n$ is the $n$-th Hermite polynomial (see proof below).
What is the connection with the question of the OP ? 
The recurrence relationship (1) between $d_{k-1},d_{k},d_{k+1}$ for $k=1,\cdots n-2$ can be seen as expressing that (for any fixed $n$) :
Vector $(d_0,d_1,\cdots,d_{n-1})^T$ is an eigenvector of $M_n$ associated with eigenvalue $1/\sqrt{2}$...
One could object that it is not true for the last line. But in fact, one can consider that the recurrence relationship stops to be effective at rank $n$ without "polluting" previous values. 
Proof of (2) : Expansion of the characteristic polynomial of $M_n$ along its last column gives : 
$$\det(M_n-xI)=-x \det(M_{n-1}-xI) +(-\sqrt{\tfrac{n-1}{2}})^2\det(M_{n-2}-xI) \ \iff$$ 
$$\chi_n(x)=-x \chi_{n-1}(x)+\tfrac{n-1}{2}\chi_{n-2}(x)$$
$$ \ \text{with} \ \chi_1(x)=-x, \ \chi_2(x)=x^2-\tfrac12.$$
allowing the comparison with the recurrence relationship for Hermite polynomials :
$$H_{n+1} (x)=2xH_n (x)-2nH_{n-1} (x).$$
(end of proof).
Connected as well : 
The roots of Hermite polynomials are all real?
https://math.stackexchange.com/q/1629234
