Before the example it has been stated that in a system described by $$ Q(D)y(t) = P(D)x(t), \quad (1.) \iff \\ (D^N + a_1D^{N-1} + ... + a_{N-1}D + a_N)y(t) = (b_{N-M}D^M + b_{N-M+1}D^{M-1} + ...+b_{N-1}D + b_N)x(t) $$
if $N = M$, setting $y(t) = h(t)$ = impulse response, we will get
$$
h(t) = b_0 \delta(t) + \text{characteristic modes}
$$
If $N > M$, $b_0 = 0$ so we get
$$ h(t) = \text{characteristic modes}, \quad (2.)$$
Where the characteristic modes has been defined as the exponential terms/functions in the solution to the differential equation $Q(\lambda) = 0$. Then comes an example (Example 2.3) of a method, "impulse matching", to find the unkown coefficients of the $N$ characteristic modes in $h(t)$ in equation (2.), which I do not understand.
Find the impulse response $h(t)$ for a system specified by $$(D^2 + 5D + 6)y(t) = (D+1)x(t), \quad (2.20)$$ In this case, $b_0$ = 0. Hence, $h(t)$ consists of only the characteristic modes. The characteristic polynomial is $\lambda^2 + 5\lambda + 6 = (\lambda + 2)(\lambda + 3)$. The roots are $-2$ and $-3$. Hence, the impulse response $h(t)$ is $$ h(t) = (c_1e^{-2t} + c_2e^{-3t})u(t) , \quad (2.21)$$ letting $x(t)$ = $\delta(t)$ and $y(t) = h(t)$ in Eq. (2.20), we obtain $$ \ddot h(t) +5\dot h(t) + 6h(t) = \dot \delta(t) + \delta(t), \quad (2.22)$$ Recall that initial conditions $h(0^-)$ and $\dot h(0^-)$ are bot zero. But the application of an impulse at $t=0$ creates new initial conditions at $t=0^+$. Let $h(0^+) = K_1$ and $\dot h(0^+) = K_2$. These jump discontinuities in $h(t)$ and $\dot h(t)$ at $t=0$ result in impulse terms $\dot h(t) = K_1 \delta(t)$ and $\ddot h(0) = K_1 \dot \delta(t) + K_2 \delta(t)$ on the left-hand side. Matching the coefficients of impulse terms on both sides of Eq. (2.22) yields $$ 5K_1 + K_2 = 1, \quad K_1 = 1 \implies K_1=1, K_2=-4$$ We now use these values $h(0^+) = K_1 = 1$ and $\dot h(0^+) = K_2 = -4$ in Eq. (2.21) to find $c_1$ and $c_2$. Setting $t=0^+$ in Eq. (2.21), we obtain $c_1 + c_2 = 1$. Also setting $t=0^+$ in $h(t)$, we obtain $-2c_1 - 3c_1 = -4$. These two simultaneous equations yield $c_1 = -1$ and $c_2 = 2$. Therefore $$ H(t) = (-e^{-2t} + 2e^{-3t})u(t) $$
I do not understand the part in boldface. Why does the jump discontinuities result in impulse terms and what is the logic behind the calculation of the impulse terms from the jump discontinuities?