How to solve second degree recurrence relation? For first degree recurence relation it is as simple as $f(n)=a^n\cdot f(0)+b\dfrac{a^n-1}{a-1}$.
But how do you solve second degree?
For example
$$f(n)=\begin{cases}
1,&\text{for }n=1\\
2,&\text{for }n=2\\
-3f(n-1)+4f(n-2),&\text{for }n>2\;.
\end{cases}$$
I tried googling "How to solve second degree recurrence relation?" but it gave me solutions only for first degree and some other random stuff.

Or use generating functions. Rewite the recurrence as:
$$
f(n + 2) = - 3 f(n + 1) + 4 f(n) \qquad f(1) = 1, f(2) = 2
$$
Define:
$$
F(z) = \sum_{n \ge 0} f(n + 1) z^n
$$
By the properties of ordinary generating functions (see e.g. Wilf's "generatingfunctionology"):
$$
\frac{F(z) - f(1) - f(2) z}{z^2} = - 3 \frac{F(z) - f(1)}{z} + 4 F(z)
$$
My own tame computer algebra system gives:
$$
F(z) = \frac{6}{5} \cdot \frac{1}{1 - z} + \frac{1}{5} \cdot \frac{1}{1 + 4 z}
$$
This expands as two geometric series:
$$
\begin{align*}
f(n) &= \frac{6}{5} + \frac{1}{5} (-4)^{n - 1} \\
     &= \frac{6}{5} - \frac{(-4)^n}{20}
\end{align*}
$$
