Finding impulse response $h(t)$ by impulse matching; problem understanding Example 2.3 in "Linear systems and signals", Lathi 3rd edition
From Eq. $(2.21)$ you can see that the impulse response $h(t)$ jumps from zero to some non-zero value at $t=0$. The derivative of a jump is a Dirac delta impulse, so there's a Dirac delta impulse in the derivative of the impulse response. Also the derivative of $h(t)$ jumps from zero to some possibly non-zero value at $t=0$, so there's also a Dirac delta impulse in the second derivative of $h(t)$.
A slightly different way to see this is the following. Write $h(t)$ as
$$h(t)=f(t)u(t)\tag{1}$$
with
$$f(t)=c_1e^{-2t}+c_2e^{-3t}\tag{2}$$
The derivative of $h(t)$ is then
$$h'(t)=f'(t)u(t)+f(0)\delta(t)\tag{3}$$
because $u'(t)=\delta(t)$ and $f(t)\delta(t)=f(0)\delta(t)$. The second derivative is
$$h''(t)=f''(t)u(t)+f'(0)\delta(t)+f(0)\delta'(t)\tag{4}$$
From the given difference equation we can match the coefficients of $\delta(t)$ and $\delta'(t)$ on both sides of the equation:
$$\begin{align}f'(0)+5f(0)&amp;=1\\
f(0)&amp;=1\end{align}\tag{5}$$
Using $(2)$ this results in
$$\begin{align}3c_1+2c_2&amp;=1\\c_1+c_2&amp;=1\end{align}\tag{6}$$
with the solution
$$\begin{align}c_1&amp;=-1\\c_2&amp;=2\end{align}\tag{7}$$
From $(1)$ and $(2)$ we obtain
$$h(t)=\left(-e^{-2t}+2e^{-3t}\right)u(t)\tag{8}$$
Also take a look at this answer to a related question.