The step function $H(t)$ is always 0 until a time point $t=0$, it becomes a constant and then stay at that constant forever.

\begin{align}H(t)=\left\{\begin{aligned} 0 &\, &t \lt 0\\ 1 &\, &t \ge 0 \end{aligned}\right.\end{align}

And we can shift the $H(t)$ by $T$ time, that is $H(t-T)$.

The delta function $$\delta (t) is always 0, except for a single time point $t=0$, it has a value 1. Thus delta function is all in one instant, an impulse. Also, the delta function can be shifted by $T$ time, that is $\delta (t-T)$.

Although this is not a continuous function, this is what we do in real life. Because the deposits are always made at some specific instants.

Well, the derivative of $\delta(t)$ isn't quite legitimate at the jump point, but fortunately we have the tool called integration. And you may already spotted that, the step function is actually the integral of the delta function.

$$\int_{-\infty}^{\infty}\delta (t)dt=\left[H(t)\right]^{\infty}_{-\infty}=1-0=1$$

For a more general case of integral $\delta (t)$, let's say we integrate it with $f(t)$,

$$\int_{-\infty}^{\infty}\delta (t)f(t)dt=f(0)$$

Because $\delta (t)$ is 0 everywhere except for the point $t=0$, and $\delta (0)=1$ so the result we get $f(0)$.

As for the shifted delta function, the example will be

$$\int_{-\infty}^{\infty}\delta (t-T)e^tdt=e^T$$

What if it appears in the first order differential equation? Let's try $\frac{dy}{dt}=ay+\delta (t-T)$ with $y(0)=0$.

The solution is simple

\begin{align}y(t)=\left\{\begin{aligned}0 & \, &-\infty < t \le T\\ e^{a(t-T)} & \, & t \ge T \end{aligned}\right.\end{align}

Why? If you interpret this as a bank account, it becomes super clear. In the beginning, you have no money in the account, $y(0)=0$, and you deposit 1 pound at $t=T$. So from time point $T$ to now, the total balance with interest rate $a$ is $e^{a(t-T)}$.