Sometimes you want to model a short impulse:
This can be modeled by an inhomogenous equation of the form $$ D^2 u(t) + \omega^2 u(t) = g(t). $$ where $g(t)$ models the force (which varies over time). Sometimes we don't care about the precise shape of the function $g(t)$, but only the 'total effect' of the force given, then it is useful to use the delta function to model it.
Consider the following sequence of functions $$ g_n(t) = n \cdot 1_{[0,1/n]}(t). $$ where $1_{[a,b]}(t)$ means the value of the function is $1$ if $t \in [a,b]$ and $0$ otherwise. We have that $$ \int g_n(t) dt = 1 $$ for all $n$, but the graph of the function is getting narrower and taller.
We also have the property that, suppose we have a smooth function $f(t)$, then we have $$ \lim_{n \to \infty} \int g_n(t) f(t) dt = f(0). $$ consider the case that $f(t) = 1$ or $t$.
We will write $$ \delta(t) = \lim_{n \to \infty} g_n(t). $$ which is meaningful inside an integral.
Consider the equation $$ y'(t) = \delta(t), \quad t \geq 0 $$ and with initial condition $y(t)=0$ for $t<0$.
We can solve it by integration: for $b > 0$, we have $$ y(b) = y(-\epsilon) + \int_{-\epsilon}^b y'(t) dt = 0 + 1 = 1. $$ where $\epsilon$ is any positive number.
The shape of $y(t)$ is a 'step function', $$ y(t) = \begin{cases} 0 & t < 0 \cr 1 & t > 0 \end{cases} $$ the value of $y(0)$ is undefined (and doesn't matter)
Consider the equation $$ (d/dt)^2 y(t) = \delta(t), \quad t \geq 0 $$ and with initial condition $y(t)=0$ for $t<0$.
Let $g(t) = y'(t)$, then g(t)$ satisfies $$ (d/dt) g(t) = \delta(t) $$ which is just the case in Ex 1. We see $g(t) = 1$ only if $t>0$. Then, we get, for $t>0$ $$ y(t) = y(0) + \int_0^t y'(s) ds = \int_0^t 1 ds = t. $$
Let $P(D)$ be a differential operator, where $D=d/dt$ and $P(x)$ is a degree $n$ polynomial. Suppose we are facing an equation of the type $$ P(D) y(t) = g(t) $$ with some homogeneous boundary condition (meaning the function $y(t)$ vanishes on the boundary of the domain).
Suppose we know the solution for $$ P(D) G(t; s) = \delta(t-s). $$
Then, we can solve the original equation by doing an integral $$ y(t) = \int G(t; s) g(s) ds. $$ Indeed, we have $$ P(D) y(t) = \int P(D) G(t; s) g(s) ds = \int \delta(t-s)g(s) ds = g(t). $$
Consider the domain being $[0,\infty)$, and we have $$ d/dt y(t) = g(t), \quad g(t)=1_{[1,2]}(t). y(0) = 0$$ In this case, we can first solve for the Green's function $$ d/dt G(t; s) = \delta(t-s), \quad G(0;s)=0$$ we get $ G(t;s) = 1 $ for $t>s$, and 0 else.
Thus, we can get get $$ y(t) = \int g(s) G(t;s) ds = \int_{1}^2 1_{t>s} ds. $$ if $t>2$, then $y(t) = \int_1^2ds = 1$, if $1<t<2$, we get $\int_1^t ds = (t-1)$. if $t<1$, we get $y(t)=0$.
Consider the equation that, for $x \in [0,2]$ $$ (d/dx)^2 y(x) + y(x) = \delta (x-1) $$ with boundary condition $y(0) = y(2) = 0. $
If we integrate this equation over the interval $(1-\epsilon, 1+\epsilon)$, across the location of the delta function, we find that $$ \lim_{\epsilon} y'(1+\epsilon) - y'(1-\epsilon) = 1 $$ (the term $\int_{(1-\epsilon, 1+\epsilon)} y(x) dx = O(\epsilon) \to 0$) hence we had a discontinuity of the slope of $y(x)$ when $x=1$. We may write down the general solution over the interval $(0,1)$ that vanishes on $x=0$ $$ y_-(x) = a \sin(x) , \quad x \in [0,1] $$ and the general solution over $x \in (1,2)$ that vanishes on $x=2$ $$ y_+(x) = b \sin(x-2), \quad x \in [1,2] $$ then we can solve the condition that $$ y_-(1) = y_+(1), \quad y_-'(1) - y'_+(1) = 1 $$ this gives $$ a \sin(1) = b \sin(-1), \quad a\cos(1) - b \cos(-1) = 1 $$ so $a=-b$ and $a = 1 / (2 \cos(1))$.