Using laplace transforms to solve a piecewise defined function initial value problem

Solution 1:

You can define a piecewise function using the Heaviside Unit Step Function and then take the Laplace Transform of that.

In general, if:

$$f(t)= \left\{\begin{array} fg(t) &0 \le t < a\\h(t) &t \ge a \end{array}\right.$$

then:

$$f(t) = g(t) - g(t)u(t-a) + h(t) u(t-a)$$

In your example, we have:

$$\tag 1 f(t) = 1 - (1)u(t- \pi) + (0) u(t- \pi) = 1 - 1~u(t- \pi)$$

Of course, you can do this other ways and here is an example (use the definition straight off), Laplace transform of unit step function.

The Laplace Transform of $(1)$ is given by:

$$\mathscr{L} (1 - 1~u(t-\pi)) = \dfrac{1}{s} - \dfrac{e^{-\pi s}}{s} = \dfrac{1 - e^{-\pi s}}{s}$$

The Laplace Transform of the other part with initial conditions yields:

Putting it all together yields:

$$y(s) = \dfrac{1 - e^{-\pi s}}{s(s^2 + 16)}$$

I will assume you can take it from here.