Laplace Transform Notational Question

[Saladsamurai]

Homework Statement

Given: $$\frac{d^2y}{dt^2} + 12\frac{dy}{dt} + 32y = 32u(t)$$

and that all initial conditions are zero, use the Laplace transform to find y(t)

This is an example problem from my text. They start the solution by taking the Laplace of both sides assuing y'(0) = 0 & y(0) = 0 :

$$s^2Y(s) + 12sY(s) + 32Y(s) = \frac{32}{s}$$

I am just a little confused why $L[32y] = Y(s)$ but $L[32u(t)] = 32/s$

y is a function of time, so why is y(t) treated differently from u(t)?

I know this is probably a stupid question...sorry


~Casey

 

[Saladsamurai]

Sorry! I think I got it! I just found in my notes that some texts denote the 'unit step function' u(t). Where u(t) = 1 for t >0 and u(t) = 0 for t<=0.

Poop!