Question

I need to find the Laplace{(q'' + 110q' + 1000q)(θ(t-5))} with θ(t-a) as a Heaviside function. This is a part of a bigger problem but I don't know where to start with this part because of the notation. I am getting confused because I only know how to solve this in the form Laplace{ g(t-a)*θ(t-a) } but I don't understand how to manipulate the differential equation into that form.

Answer 1

θ(t - 5) = 1 if t ≥ 5 and 0 otherwise, so the Laplace transform of the second derivative term is

$\displaystyle \int_0^\infty q''(t) \theta(t-5) e^{-st} \, dt = \int_5^\infty q''(t) e^{-st} \, dt \\\\\\ = \int_0^\infty q''(t) e^{-st} \, dt - \int_0^5 q''(t) e^{-st} \, dt \\\\\\ = L_s\left\{q''(t)\right\} - \int_0^5 q''(t) e^{-st} \, dt$

Compute the remaining integral by parts.

$u = e^{-st} \implies du = -se^{-st} \, dt \\ dv = q''(t) \, dt \implies v = q'(t)$

$\implies \displaystyle \int_0^5 q''(t) e^{-st} \, dt = q'(t) e^{-st} \bigg|_0^5 + s \int_0^5 q'(t) e^{-st} \, dt = \dfrac{q'(5)}{e^{5s}} - q'(0) + s \int_0^5 q'(t) e^{-st} \, dt$

Integrating by parts again with similar choice of u and dv gives

$\displaystyle \int_0^5 q'(t) e^{-st} \, dt = \dfrac{q(5)}{e^{5s}} - q(0) + s \int_0^5 q(t) e^{-st} \, dt$

Recall that

$\displaystyle \int_0^\infty q''(t) e^{-st} \, dt = s^2 Q(s) - sq(0) - q'(0)$

$\displaystyle \int_0^\infty q'(t) e^{-st} \, dt = s Q(s) - q(0)$

where Q(s) is the Laplace transform of q(t).

It follows that

$L_s\left\{q''(t) \theta(t-5)\right\} = (s^2 Q(s) - s q(0) - q'(0)) - \left[\dfrac{q'(5)}{e^{5s}} - q'(0) + s \left(\dfrac{q(5)}{e^{5s}} - q(0) + s \int_0^5 q(t) e^{-st} \, dt\right) \right]$

or

$L_s\left\{q''(t) \theta(t-5)\right\} = s^2 Q(s) - \dfrac{sq(5)+q'(5)}{e^{5s}} - s^2 \int_0^5 q(t) e^{-st} \, dt$

Similarly, you can show that

$L_s\left\{q'(t) \theta(t-5)\right\} = sQ(s) - \dfrac{q(5)}{e^{5s}} - s\int_0^5 q(t) e^{-st} \, dt$

and

$L_s\left\{q(t) \theta(t-5)\right\} = Q(s) - \int_0^5 q(t) e^{-st} \, dt$

There's not much more one can say without knowing anything more about q(t) …