I was given a differential equation to solve using Laplace transformation. and I got a term that had : $-8H(t-\pi)(sint)$
The question asks to rewrite the solution without the use of the heaviside function. How do I go about doing this? I know it has to be in intervals (i.e. $t\lt \pi$ etc).
The diff eq. is:
$$\ddot y+y=sin2t+(-7+cost)\delta(t-\pi) $$
where $y(0)=0$, $\dot y(0)=1$
I think you have a sign error in your function, it should be:
$$\tag 1 f(t) = 8 ~\sin(t)~u(t-\pi)$$
We want to write the function in $(1)$ without the use of the unit step function.
For $(1)$, the unit step function is equal to zero for all times less than $\pi$ and is then just the sine function for times greater than $\pi$.
If we were to plot $(1)$, we would get:
We can write $(1)$ as a piece-wise function without the unit step function as:
$$f(t) = \left\{\begin{array}{ll} ~~~~0 & : t \lt \pi\\ 8 \sin t & :t \ge \pi \end{array} \right.$$
Plot this function and verify it produces the same plot as above.