How to compute $e^{At}$ with a $2\times2$ parametric matrix. I would like to calculate the exponential $e^{At}$ where $A$ is a $2\times2$ matrix:
$$
A=\left[\begin{array}{cc}
0 & 1 \\
a & b
\end{array}\right]
$$
and $a,b\in\mathbb R$. The calculus is based on the powers of this matrix, i.e.
$$e^{At}=I+tA+\frac{1}{2}t^2A^2+\frac{1}{3!}t^3A^3+ \cdots$$
But the power of $A$ seems to be very hard to unravel. For example:
$$A^2=\left[\begin{array}{cc}
a & b \\
ab & a+b^2
\end{array}\right], \quad A^3=\left[\begin{array}{cc}
ab & a+b^2 \\
a(a+b^2) & ab+b(a+b^2)
\end{array}\right], \quad \ldots
$$
Could you do better? Also, I wonder if anyone has already posed this problem by publishing (in literature) a few notes about it.

For general $A,$ the trick would be to diagonalize $A.$
This particular matrix is easier if you know the general formula for linear recurrences. $A$ has the property that $A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}y\\ax+by\end{pmatrix},$ so we can use it to define sequences $p_n,q_n$ as $$A^n\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}p_{n}\\p_{n+1}\end{pmatrix}\\A^n\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}q_{n}\\q_{n+1}\end{pmatrix}$$
where $p_0=1,p_1=0,q_0=0,q_1=1$ and $$p_{n+2}=bp_{n+1}+ap_n\\ q_{n+2}=bq_{n+1}+aq_n$$
The general closed formula for $p_n,q_n$, when $b^2+4a$ is not zero, so we don't have repeated roots, is in terms of $x_{+},x_-=\frac{b\pm \sqrt{b^2+4a}}2:$
$$q_n = \frac{1}{\sqrt{b^2+4a}}\left(x_+^n-x_-^n\right)\\
p_n = \frac{1}{\sqrt{b^2+4a}}\left(x_+x_-^n-x_-x_+^n\right).$$
So: $$\sum_{k=0}^{\infty}\frac1{k!}(At)^k\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}\sum_{k=0}^\infty\frac{q_{k}t^k}{k!}\\\sum_{k=0}^\infty\frac{q_{k+1}t^k}{k!}\end{pmatrix}=\frac{1}{\sqrt{b^2+4a}}\begin{pmatrix}e^{x_+t}-e^{x_-t}\\x_+e^{x_+t}-x_-e^{x_-t}\end{pmatrix}
$$
That should be the right column of $e^{At}.$ The left column is similarly calculated as:
$$\frac{1}{\sqrt{b^2+4a}}\begin{pmatrix}x_{+}e^{x_-t}-x_{-}e^{x_+t}\\a\left(e^{x_+t}-e^{x_-t}\right)\end{pmatrix}$$
This is all essentially equivalent to diagonalizing.
A bit more work will solve the case when $b^2+4a=0.$
