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{\Large\bf The Inverse Laplace Transform}

\vspace{.3in}

The inverse Laplace transform of the function Y(s)
is the unique function y(t) that is continuous on
[0,infty) and satisfies L[y(t)](s)=Y(s). If all
possible functions y(t) are discontinous one can select
a piecewise continuous function to be the inverse
transform.

We will use the notation 

$$
L^{-1}[Y(s)](t)
$$

or Li[Y(s)](t) to denote the inverse Laplace transform of Y(s).

\vspace{.3in}

{\bf Determining the Inverse Laplace Transform}

\vspace{.2in}

The Laplace transform Y(s) of a function y(t) defined on
[0,infty) is defined by an integral. It turns out that
formula for determining y(t) given Y(s) also involves
an integral. The integral is complex valued integral, and
its evaluation is beyond the scope of this course.  For
this reason we take a more pedestrian approach in
computing the inverse transform. We will use tables and a
few tricks.

There are only small number of functions listed in the
table. The set of invertible functions can be increased
by exploiting linearity.

\vspace{.3in}

{\bf Linearity of the Inverse Transform} 

\vspace{.2in} 

Let y_1(t) and y_2(t) be the inverse Laplace transforms
of Y_1(s) and Y_2(s), respectively, and let c be a constant.
We have

$$
L^{-1}[cY_1(s)](t)=cL^{-1}[Y_1(s)](t)=cy_1(t)
$$

$$
L^{-1}[Y_1(s)+Y_2(s)](t)=L^{-1}[Y_1(s)](t)+L^{-1}[Y_2(s)](t)=
y_1(t)+y_2(t)
$$

As a corollary, we have a third formula:

$$
L^{-1}[c_1Y_1(s)+c_2Y_2(s)](t)=c_1L^{-1}[Y_1(s)](t)+c_2L^{-1}[Y_2(s)]=
c_1y_1(t)+c_2y_2(t)
$$

Here are several examples. What is the inverse transform of
7/s^3? From the table above we know that the inverse of
2/s^3 is y(t)=t^2. Hence 

$$
L^{-1}[7/s^3](t)=\frac{7}{2}L^{-1}[2/s^3](t)=\frac{7}{2}t^2
$$

Consider the Laplace transform Y(s)=(3s+4)/(s^2+4). Using the
table above, we know that the inverse of s/(s^2+4) is cos(2t)
and that the inverse of 2/(s^2+4) is sin(2t). Hence:

$$
L^{-1}[\frac{3s+4}{s^2+4}](t)=3L^{-1}[\frac{s}{s^2+4}](t)+
2L^{-1}[\frac{2}{s^2+4}](t)=3\cos 2t+2\sin 2t.
$$

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