Calculating the matrix exponential $e^{tA}$ for a $2\times2$ matrix $A\in M_2(\Bbb C)$. Let $$A=\begin{pmatrix}a & b\\ c & d\end{pmatrix}\in M_2(\Bbb C),$$ let $\delta=ad-bc$, let $\tau=a+d$, and fix some $\gamma\in\Bbb C$ such that $\gamma^2=\frac14(\tau^2-4\delta)$. 
We first assume that $\gamma\ne0$, so that $A$ is diagonalizable in the form $A=QBQ^{-1}$ where $$B=\begin{pmatrix}\lambda_1 & 0\\ 0 & \lambda_2\end{pmatrix}$$ and $Q\in\mathrm{GL}_2(\mathbb{C})$. Note that we can set $\lambda_1=\frac\tau2+\gamma$ and $\lambda_2=\frac\tau2-\gamma$. We can now calculate that $$e^{tA}=e^{tQBQ^{-1}}=Q e^{tB}Q^{-1}=Q\begin{pmatrix}e^{t\lambda_1} & 0 \\ 0 & e^{t\lambda_2}\end{pmatrix}Q^{-1}$$ however, this doesn't give me a particularly easy way of simplifying the resulting expression.
My textbook asks me to prove that for $\gamma\ne0$, $$e^{tA}=e^{t\tau/2}\left(\frac{\sinh(t\gamma)}{\gamma}A+\frac{\gamma\cosh(t\gamma)-2\tau\sinh(t\gamma)}{\gamma}I\right)$$ but I don't see how to get to this point.
 A: The matrix exponential $f(t)=\exp(tA)$ is the unique solution of
the differential equation
$$f'(t)=Af(t)\tag1$$
with the initial condition $f(0)=I$. All you need do to prove the
given formula is ensure it satisfies $(1)$ and that it is the identity
when $t=0$.
A: Note There is a typo in the expression you wish to prove.

The characteristic polynomial of $A$ is 
$$\chi_A(x) = \det(xI_2 - A) = x^2 - \tau x + \delta = \left(x - \frac{\tau}{2} + \gamma\right)\left(x - \frac{\tau}{2} - \gamma \right) = 0$$
By Cayley-Hamiltonian theorem, we have
$$\chi_A(A) = P_{+}P_{-} = 0
\quad\text{ where }\quad P_{\pm} \stackrel{def}{=} A - \left(\frac{\tau}{2} \mp \gamma\right)I_2$$
When $\gamma \ne 0$, the identity matrix can be decomposed as
$$I_2 = \frac{1}{2\gamma}(P_{+} - P_{-})$$
Now $P_{+}P_{-} = 0$ implies
$$\left(A - \left(\frac{\tau}{2} \pm \gamma\right)I_2 \right)P_{\pm} = 0
\quad\iff\quad AP_{\pm} = \left(\frac{\tau}{2} \pm \gamma\right)P_{\pm}
\quad\implies\quad
e^{tA} P_{\pm} = e^{t(\frac{\tau}{2} \pm \gamma)}P_{\pm}$$
We have
$$\begin{align}
e^{tA} = e^{tA} I_2 
&= \frac{1}{2\gamma}\left(e^{t(\frac{\tau}{2} + \gamma)}P_{+}
- e^{t(\frac{\tau}{2} - \gamma)}P_{-}\right)\\
&= \frac{e^{\frac{t\tau}{2}}}{2\gamma}\left[
e^{t\gamma}\left(
A - \left(\frac{\tau}{2} - \gamma\right)I_2\right)
-
e^{-t\gamma}\left(
A - \left(\frac{\tau}{2} + \gamma\right)I_2\right)
\right]\\
&= \frac{e^{\frac{t\tau}{2}}}{\gamma}
\left[
\sinh(t\gamma)A + \left( \gamma\cosh(t\gamma) - \frac{\tau}{2}\sinh(t\gamma) \right)I_2
\right]
\end{align}
$$
