I do not know about *the* general method, but here's a recipe for $\delta = 0$.

One can view this as a 3-term recurrence relation, and consider the corresponding orthogonal polynomials. Favard's Theorem guarantees such polynomials exist, and in fact Riesz Representation Theorem upgrades the orthogonality to the usual one if you ensure $\alpha_3 n + \alpha_4 < 0$. This transfers the problem to a problem to a problem about asymptotics of orthogonal polynomials, which can be done using multiple methods (WKB, Riemann-Hilbert, amongst many others).

**Example:**

Consider $\alpha_1 = 0, \ \alpha_2 = 2, \ \alpha_3 = -2, $ and $\alpha_4 = 2$. Then, assuming no algebra mistakes, we have (I re-index so I can use Wikipedia)
$$
a_{n + 1} = 2a_n - 2na_{n - 1}, \ a_0 = 1, \ a_2 = 2
$$
This recurrence is satisfied by Hermite polynomials $H_n(x)$ (evaluated at $x = 1$), which satisfy
$$
\int_{-\infty}^{\infty} x^k H_n(x) \ e^{x^2} = 0 \quad \text{ for } k = 0, 1, ..., n-1.
$$
and
$$
H_{n + 1}(x) = 2xH_n(x) - 2nH_{n - 1}(x)
$$
Now, again the wikipidia page tells me that (this I did not check too carefully, but I know asymptotic expansions of Hermite Polynomials exist)
$$
a_n = H_n(1) \sim e^{1/2} \dfrac{2^n}{\sqrt{\pi}} \Gamma\left( \dfrac{n + 1}{2} \right) \cos \left( \sqrt{2n} - \dfrac{n \pi}{2} \right).
$$

Now, in the general situation, a glance at Chihara's book "Introduction to Orthogonal Polynomials" (Pages 215 - 217) tells me that for recurrences of the form you give will often be handled using Hermite and Charlier polynomials (maybe others if you get funny about what $x$ value you consider).

P.S. I am new to this, so if this needs downgraded to a comment or is somehow not what you wanted, my bad!

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