I've completed a question involving unit-step functions and would like verification on my working out and answers, particularly for the last part, as we are asked to express the solution explicitly without the use of any special functions, which I assume means to change the unit step functions into a piecewise function.
The question requires using Laplace Transforms to solve a second order ordinary differential equation.
After using Laplace transforms and employing both s-shifting and t-shifting theorem, my final answer for q(t) is as follows:
$q(t) = [\frac{1}{31}e^{-7(t-\pi)}sin(31(t-\pi))]u(t-\pi) + \frac{9}{31}e^{-7t}sin(31t)$
To change this into a piecwise function, I obtained the following, since the unit step function equals 0 before t = $\pi$, and equals 1 after t = $\pi$:
$$ q(t) = \left\{\begin{aligned} &\frac{9}{31}e^{-7t}sin(31t) &: 0 \leq t < \pi \\ &\frac{1}{31}e^{-7(t-\pi)}sin(31(t-\pi)) + \frac{9}{31}e^{-7t}sin(31t) &: t \ge \pi \end{aligned} \right.$$
I apologise for my formatting if it appears weird, first time using Mathjax :p Anyways, using trig addition formulae and expanding the exponent for the exponential term, I simplified this into:
$$ q(t) = \left\{\begin{aligned} &\frac{9}{31}e^{-7t}sin(31t) &&: 0 \leq t < \pi \\ &\frac{9-e^\pi}{31}e^{-7t}sin(31t) &&: t \ge \pi \end{aligned} \right.$$
I am asking this since in lectures and practice problems, we weren't asked to express the solution explicitly as a piecewise function instead of using unit step function.
Thanks in advance!