Recognize this as the Laplace transform of Aexp(r 1t) + Bexp(r 2t) The previous expressions for the transforms of these derivatives, we have So, taking the Laplace transform of both sides of our differential equation (recalling that L is a linear Often we are able to specify the "initialĬonditions", i.e., the values of f(0) and f ′(0). Order to determine a particular solution for this 2nd-order equation we must To solve differential equations, consider the homogeneous equation Of f(t) as a polynomial in s with coefficients given by the derivatives ofį(t) at t = 0 and a leading coefficient of L. In this way we can express the Laplace transform of the nth derivative Through by s 2 and rearrange terms to give Noting that the quantity in parentheses is L, we can multiply Typical functions, but the most important point to notice about (2) is that We've seen how equation (2) enables us to determine the Laplace transforms of Useful example is f(t) = Ae rt for constants A and r, which has theīy equation (2) we have the Laplace transform The special case when the phase angle u is zero, this reduces to the familiar Terms by sines and cosines, this can be written as These derivatives, at t = 0, into equation (2) gives For a less trivial example, suppose f(t) = A sin(ωt + Also, ifį(t) = At + B, then f ′(t) = A and the Laplace transform is L = B/s Makes it easy to see that if f(t) is a constant A, then L = A/s. Simply given by the negative of the indefinite integral evaluated at t = 0, T = +∞ the right hand quantity vanishes, so the Laplace transform is From this integral, it follows that the Laplace transform of a function The key to the usefulness of Laplace transforms arises from the followingĬan easily be verified by differentiating the right hand side using the chain Transform is a linear operator, which means that for any functions f,g, and For any given function f(t) that is zero forĪll t less than zero, let L denote the Laplace transform of f. To see why, it's helpful to review some fundamental Would be difficult to solve in the "t domain" are easy to solve in Usefulness of Laplace transforms is due to the fact that some problems which The corresponding functions in the "s domain". Unique "mapping" between functions in the "t domain" and Knowing f(t) for all t is sufficient to determine F(s). Value of F(s) for all s is sufficient to fully specify f(t), and conversely Weighted more and more toward the initial region near t = 0. Thus as s increases, F(s) represents the area under f(t) Since the multiplier e –st is a decaying exponential function equal Whereas F(s) for s greater than 0 is a "weighted" integral of f(t), Note thatį(0) is simply the total area under the curve f(t) for t = 0 to infinity, Any given function f(t) (assumed to equal zero for t less than zero) we canĭefine another function F(s) by the definite integralįunction F(s) is called the Laplace transform of the function f(t).