Consider the forced LTI differential equation
\begin{eqnarray}
\frac{d^n}{dt^n} y(t) + a_{n-1} \frac{d^{n-1}}{dt^{n-1}}y(t) + \ldots +a_2\frac{d^2}{dt^2}y(t)+ a_1 \frac{d}{dt}y(t) + a_0 y(t) = f(t)
\end{eqnarray}
Our aim is to construct $y(t)$ fot arbitrary $f(t)$. Assume that the impulse response $h(t)$ of this LTI differential equation is known:
\begin{eqnarray}
\frac{d^n}{dt^n} h(t) + a_{n-1} \frac{d^{n-1}}{dt^{n-1}} h(t) + \ldots +a_2\frac{d^2}{dt^2}h(t)+ a_1 \frac{d}{dt}h(t) + a_0 h(t) = \delta(t)
\end{eqnarray}
or,
\begin{eqnarray}
\left(\frac{d^n}{dt^n} + a_{n-1} \frac{d^{n-1}}{dt^{n-1}} + \ldots +a_2\frac{d^2}{dt^2}+ a_1 \frac{d}{dt} + a_0 \right) h(t) = \delta(t)
\end{eqnarray}
Because the differential equation is time invariant, a time shift in its forcing (input) $\delta(t)$ will generate exactly the same time shift in its response (output) $y(t)$. Assuming a time shift of $s$
\begin{eqnarray}
\left(\frac{d^n}{dt^n} + a_{n-1} \frac{d^{n-1}}{dt^{n-1}} + \ldots +a_2\frac{d^2}{dt^2}+ a_1 \frac{d}{dt} + a_0 \right) h(t-s) = \delta(t-s)
\end{eqnarray}
Multiply both sides with $f(s)$. As $s$ and $t$ are different, time derivatives will not affect $f(s)$.
\begin{eqnarray}
\left(\frac{d^n}{dt^n} + a_{n-1} \frac{d^{n-1}}{dt^{n-1}} + \ldots +a_2\frac{d^2}{dt^2}+ a_1 \frac{d}{dt} + a_0 \right) h(t-s)f(s) = \delta(t-s)f(s)
\end{eqnarray}
Let us integrate both sides in $s$:
\begin{eqnarray}
\int_{s=-\infty}^{\infty}\left[ \left(\frac{d^n}{dt^n} + a_{n-1} \frac{d^{n-1}}{dt^{n-1}} + \ldots +a_2\frac{d^2}{dt^2}+ a_1 \frac{d}{dt} + a_0 \right) h(t-s)f(s)\right]ds = \int_{s=-\infty}^{\infty}\delta(t-s)f(s) ds
\end{eqnarray}
Again, as $s$ and $t$ are independent, the integration in s can jump over the derivatives in $t$:
\begin{eqnarray}
\left(\frac{d^n}{dt^n} + a_{n-1} \frac{d^{n-1}}{dt^{n-1}} + \ldots +a_2\frac{d^2}{dt^2}+ a_1 \frac{d}{dt} + a_0 \right) \left[ \int_{s=-\infty}^{\infty} h(t-s)f(s) ds \right] = \int_{s=-\infty}^{\infty}\delta(t-s)f(s) ds
\end{eqnarray}
Also, remembering the definition of impulse/dirac delta function tells us that
\begin{eqnarray}
\int_{s=-\infty}^{\infty}\delta(t-s)f(s) ds = f(t)
\end{eqnarray}
Substituing this into the above result gives
\begin{eqnarray}
\left(\frac{d^n}{dt^n} + a_{n-1} \frac{d^{n-1}}{dt^{n-1}} + \ldots +a_2\frac{d^2}{dt^2}+ a_1 \frac{d}{dt} + a_0 \right) \left[ \int_{s=-\infty}^{\infty} h(t-s)f(s) ds \right] = f(t)
\end{eqnarray}
We can immediately deduced that
\begin{eqnarray}
y(t) = \int_{s=-\infty}^{\infty} h(t-s)f(s) ds
\end{eqnarray}
which is the solution. This integral is known as the convolution integral.
which can be factorized as
\begin{eqnarray}
(D-\alpha_1)^{m_1} \ldots (D-\alpha_{k-1})^{m_{k-1}}(D-\alpha_k)^{m_k} y =f(t)
\end{eqnarray}