The Laplace Transformation (after the French mathematician
Pierre Simon Marquis de Laplace) is a method for solving
differential equations, and the corresponding
initial and
boundary value problems.
Finding a solution of a differential equation using the Laplace
Transformation consists of three steps:
 Transformation of the equation into a subsidiary equation (moving
the equation from the so called tspace to
sspace)
 Solving the subsidiary equation by simple algebraic manipulations
 Transforming the subsidiary equation back into tspace
to obtain the solution of the given problem
The Laplace Transformation is widely used in
engineering, especially
in problems where the (
mechanical/
electrical)
driving force has
discontinuities. Another common use of the Laplace
Transformation is in
Process Control.
I will only describe the basic definition of the Laplace Transform,
some basic properties, and a basic example on the use of this method.
Consider a function f(t), defined for all t≥0. Multiply this
function by e^{st}, and integrate with respect to t from zero
to infinity. The resulting integral F(s) is given by:
∞
F(s) = ∫ e^{st} f(t) dt
0
The function F(s) of the variable s is called the
Laplace Transform, denoted by L(f). Furthermore, the original
function f(t) is called the
inverse transform, denoted by
L
^{1}(F):
∞
L(f) = ∫ e^{st} f(t) dt
0
f(t) = L^{1}(F)
Some simple general transforms are given in the following table. More
complex transforms can be found in books of mathematical tables.
f(t) L(f)
 
1 1/s
t 1/s^{2}
t^{n} n!/(s^{n+1}) n=1, 2, ...
e^{at} 1/(sa)
An important property of the Laplace Transform is its linearity:
L{af(t) + bg(t)} = aL{f(t)} + bL{g(t)}
In order to solve
differential equations using Laplace Transforms, the
transform of the
derivative of a function f(t) is needed. This is
given by:
L(f') = sL(f)  f(0)
This definition can be extended to transforms of derivatives of arbitrary higher
orders:
L(f^{n}) = s^{n}L(f)  s^{n1} f(0)  s^{n2} f'(0)  ...  f^{(n1)}(0)
Example: solve the initial value problem:
y" + 4y' +3y = 0 y(0) = 3, y'(0) = 1
 First set up the subsidiary equations, with Y(s) = L(y):
L(y') = sY  y(0) = sY  3
L(y") = s^{2}Y sy(0)  y'(0) = s^{2}Y  3s  1
s^{2}Y + 4sY +3Y = 3s + 1 + 4*3

Solve the subsidiary equation for Y:
(s + 3)(s + 1)Y = 3s + 13
3s + 13 2 5
Y =  =  + 
(s + 3)(s + 1) s+3 s+1

Solve the given problem by using the inverse transform:
L^{1}{1/(s + 3)} = e^{3t}
L^{1}{1/(s + 1)} = e^{t}
y(t) = 2e^{3t} + 5e^{t}
Of course, this differential equation would have been quite easy to
solve using a
substitution method. However, the Laplace Transformation
allows for a
systematic method to solve more complex differential
equations, and also
systems of differential equations.