This document discusses using Laplace transforms to solve differential equations. It begins with an introduction to Laplace transforms and their inventor Pierre-Simon Laplace. The author then explains that Laplace transforms can be used to transform differential equations into simpler algebraic equations that are easier to solve. As an example, the document walks through using Laplace transforms to solve a first order differential equation. It concludes that Laplace transforms are an important mathematical tool for solving differential equations.

1. Communicating Mathematics
Assesment 1
Using Laplace Transforms to Solve Differential
Equations
By George Stevens
Due: 11th of October 2015

2. Abstract
Laplace transforms are a type of mathematical transform, with a diverse
range of applications throughout mathematics,physics and engineering. The
Laplace transformation is named after its discoverer, the French mathemati-
cian Pierre-Simon Laplace, who has been hailed as the “French Newton” for
his work in mathematics and physics. His discoveries followed from work
done by Euler and Lagrange, using integrals as solutions to differential equa-
tions, eventually leading to the current Laplace transform.[6]
In this paper I am going to be focusing one of the most important mathemat-
ical applications, using Laplace transforms to solve differential equations.
What is a Laplace Transform?
In mathematics, there is many cases where a problem cannot be easily solved
in a particular function space, however the same problem may be much eas-
ier to solve in a different function space. When this is the case, the initial
function can be transformed through a variety of methods in order to make
the problem easier to solve. A Laplace transform is an integral transform
which transforms a function of a positive real variable t (usually time) to a
function of a complex variable s (frequency).
If we let f(t) be some function defined for t ≥ 0, then for s ≥ 0 the Laplace
transform of f is
L(f) =
∞
0
e−st
f(t)dt.
Once we have solved the Laplace transform for the problem in the new func-
tion space, we would take the inverse Laplace transform of the solution to
obtain a solution in the original space.
Applications to Differential Equations
When it comes to differential equations, taking the Laplace transform of the
equation turns the equation subject to the initial conditions, to a simpler,
algebraic equation which we can solve to get a transformed solution. We can
then take the inverse transform of this to obtain the solution to the initial
1

3. problem.[1] The following analogy is a nice, simplified way to visualise how
a Laplace transform works:
“Suppose that you come across a poem written in English of whose meaning
you don’t understand. However suppose that you know a French-speaking
gentleman who is a master of interpreting poems. So you translate the poem
into French and send it to the French gentleman. The French gentleman
writes a perfectly good interpretation of the poem in French and sends this
back to you where you translate it back into English and you have the mean-
ing of the poem.”[2]
Let the English version of the poem be some differential equation and the
English interpretation be the solution to this particular equation. We can
then say that the translation to French would be like taking the Laplace
transform of that equation, so then interpreting the poem in French would
be the same as getting the solution to the Laplace transform of the differ-
ential equation. Finally by translating the interpretation back to English
we would essentially be taking the inverse Laplace transform of the solution,
gaining the solution to our differential equation or English interpretation.
This can make a lot of differential equations much easier to solve, and is
applicable to differential equations of any degree.
The following section shows how we can calculate the Laplace transform for
a differential of any degree.
Proof for L
dn
f
dtn
To prove this we need to first prove that
L (f (0)) = −f(0) + sL (f(t))
We must start by applying the Laplace integral transform to f (t).
L(f (t)) =
∞
0
est
f (t)dt
Now we use integration by parts to calculate our integral.
2

4. Let
u = e−st
du = −se−st
dt
dv = f (t)dt
v = f(t)
Using these substitutions in the integral by parts formula, we get:
L(f (t)) = e−st
f(t)
∞
0
−
∞
0
f(t)(−se−st
dt)
=
f(t)
est
+ s
∞
0
e−st
f(t)dt
=
f(t)
est
∞
0
+ L(f (t)).
If we then apply the limits, we get :
L(f (t)) =
f(∞)
e∞
−
f(0)
e0
+ sL(f(t))
L(f (t)) = sL(f(t)) − f(0).
This completes our proof for L (f (t)) [4], if we now apply this to the second
derivative of f (t), we get:
L(f (t)) = sL(f (t)) − f (0)
= s [L(f(t)) − f(0)] − f (0)
= s3
L − (f)s2
L(f) − sf(0) − f (0).
This follows for L(f (t))
= s3
L(f) − s2
f(0) − sf (0) − f (0).
3

5. By induction we can deduce that the general Laplace transform for a deriva-
tive of f(n)
is :
L
dn
f
dtn
= sn
Lf(t) − sn−1
f(0) − sn−2
f (0) − ... − fn−1
(0). [5]
Example
The following example illustrates how a Laplace transformation can be used
to solve a first order differential equation:
Consider the differential equation
dy
dt
= y − sin(t),
subject to the intital condition where y = 1 when t = 0.
First we take the Laplace transform of both sides.
Left-hand side
By using standard Laplace transforms [3] we get:
L
dy
dt
= −y(0) + sL(y)
= sL(y) − 1.
Right hand side
L(y − 3sin(t)) = L(y) − 3L(sin(t))
By using standard Laplace transforms [3] again, we get:
L(y) − 3L(sin(t)) = L(y) −
3
s2 + 1
Taking both the left and right hand sides gives:
sL(y) − 1 =
3
s2 + 1
From here the equation needs to be solved for L(y).
4

6. First we rearrange to give:
sL(y) − L(y) = 1 −
3
s2 + 1
sL(y) − L(y) =
s2
− 2
s2 + 1
(s − 1)L(y) =
s2
− 2
s2 + 1
L(y) =
s2
− 2
(s2 + 1)(s − 1)
We have now obtained the Laplace transform of our original first order
differential equation, however to solve the equation fully, we must perform
an inverse Laplace transform so that it can be solved in the initial function
space.
To find the inverse Laplace transform we must convert the Laplace transform
into partial fractions. This gives:
3
2
x
x2 + 1
+
3
x2 + 1
−
1
2(x − 1)
.
By looking at the standard Laplace transforms once again [3], we can take
our partial fractions back to our original function by substituting them for
the following:
3
2
cos(t) +
3
2
sin(t) −
1
2
et
.
This is the solution to the differential equation that we started off with.
5

7. Conclusion
Differential equations can be applied throughout a variety of areas in mathe-
matics, physics, finance, statistics and so by extension, so can Laplace trans-
formations.By making differential equations easier to solve, we can conclude
that the Laplace transform is a very important and powerful tool in mathe-
matics.
References
1. copper.hat (http://math.stackexchange.com/users/27978/copper-hat)
11/08/2012
2. Jp McCarthy (http://math.stackexchange.com/users/19352/jp-mccarthy)
23/10/2012
3. Paul Harris (https://studentcentral.brighton.ac.uk/bbcswebdav/pid-2396791-
dt-content-rid-46413641/courses/MM252Y EAR2014/Standard20Laplace20
Transforms.pdf)08/11/2015(List of standard transforms in appendix)
4. Romel Verterra (http://www.mathalino.com/reviewer/advance-engineering-
mathematics/laplace-transforms-derivatives) 06/11/2015
5. Youri Shestopalov (http://www2.kau.se/yourshes/AB28.pdf)15/06/2010
6. William B Johnson (Transform method for semiconductor mobility,
Journal of Applied Physics) (2006)
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8. Appendix
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