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in other words,

( √ ) ∫ √ ( )
e −a s 1 x 0 +i∞ e ts−a s a
−1
f(t) = L = ds = 1 − erf √ . (4.27)
s 2πi s 2 t
x 0 −i∞
The function

erfc(t) = 1 − erf(t)

is called the complementary error function, and so we have by (4.27)

( √ ) ( )
e −a s a
−1
L = erfc √ . (4.28)
s 2 t

5. Application of Laplace Transform

The various types of problems that can be treated with the Laplace transform

include ordinary and partial differential equations as well as integral and integro-
differential equations. In this chapter we delineate the principles of the Laplace
transform method for the purposes of solving differential equations.

Solving ordinary differential equations using Laplace

transforms

We can use the results of the previous sections to solve initial value problems for

differential equations, in particular, differential equations with specified initial
values.
Structure of solutions:

• Take the Laplace transform on both sides. You will get an algebraic equation
for Y (s);

• Solve this equation to get Y (s);

−1
• Take the inverse transform to get y(t) = L [Y (s)].

The derivative theorem opens up the possibility of utilizing the Laplace
transform as a tool for solving ordinary differential equations. Numerous
applications of the Laplace transform to ODE’s will be found in ensuing sections.

We illustrate the application of Laplace transforms to initial value problems in
the following examples.

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