Using Laplace Transforms to Solve Differential Equations

1. Communicating Mathematics
Assesment 1
Using Laplace Transforms to Solve Dierential
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 dierential 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 dierential 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 dierent 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 variablet(usually time) to a
function of a complex variable s (frequency).
If we letf(t) be some function dened fort0, then fors0 the Laplace
transform offis
L(f) =
Z
1
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 Dierential Equations
When it comes to dierential 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, simplied 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 dierential 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 dier-
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 dierential equation or English interpretation.
This can make a lot of dierential equations much easier to solve, and is
applicable to dierential equations of any degree.
The following section shows how we can calculate the Laplace transform for
a dierential of any degree.
Proof forL

d
n
f
dt
n

To prove this we need to rst prove that
L(f
0
(0)) =f(0) +sL(f(t))
We must start by applying the Laplace integral transform tof
0
(t):
L(f
0
(t)) =
Z
1
0
e
st
f
0
(t)dt
Now we use integration by parts to calculate our integral.
2

4. Let
u= e
st
du=se
st
dt
dv= f
0
(t)dt
v= f(t)
Using these substitutions in the integral by parts formula, we get:
L(f
0
(t)) =

e
st
f(t)

1
0

Z
1
0
f(t)(se
st
dt)
=

f(t)
e
st

+s
Z
1
0
e
st
f(t)dt
=

f(t)
e
st

1
0
+L(f
0
(t)):
If we then apply the limits, we get :
L(f
0
(t)) =

f(1)
e
1

f(0)
e
0

+sL(f(t))
L(f
0
(t)) = sL(f(t))f(0):
This completes our proof forL(f
0
(t)) [4], if we now apply this to the second
derivative off
0
(t), we get:
L(f
00
(t)) = sL(f
0
(t))f
0
(0)
= s[L(f(t))f(0)]f
0
(0)
=s
3
L (f)s
2
L(f)sf(0)f
0
(0):
This follows forL(f
000
(t))
=s
3
L(f)s
2
f(0)sf
0
(0)f
00
(0):
3

5. By induction we can deduce that the general Laplace transform for a deriva-
tive of f
(n)
is:
L

d
n
f
dt
n

=s
n
Lf(t)s
n1
f(0)s
n2
f
0
(0):::f
n1
(0): [5]
Example
The following example illustrates how a Laplace transformation can be used
to solve a rst order dierential equation:
Consider the dierential equation
dy
dt
=ysin(t);
subject to the intital condition wherey= 1 whent= 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(y3sin(t)) =L(y)3L(sin(t))
By using standard Laplace transforms [3] again, we get:
L(y)3L(sin(t)) =L(y)

3
s
2
+ 1

Taking both the left and right hand sides gives:
sL(y)1 =

3
s
2
+ 1

From here the equation needs to be solved forL(y).
4

6. First we rearrange to give:
sL(y) L(y) = 1

3
s
2
+ 1

sL(y) L(y) =

s
2
2
s
2
+ 1

(s1)L(y) =

s
2
2
s
2
+ 1

L(y) =

s
2
2
(s
2
+ 1)(s1)

We have now obtained the Laplace transform of our original rst order
dierential 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 nd the inverse Laplace transform we must convert the Laplace transform
into partial fractions. This gives:
3
2

x
x
2
+ 1
+
3
x
2
+ 1


1
2(x1)
.
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
e
t
.
This is the solution to the dierential equation that we started o with.
5

7. Conclusion
Dierential equations can be applied throughout a variety of areas in mathe-
matics, physics, nance, statistics and so by extension, so can Laplace trans-
formations.By making dierential equations easier to solve, we can conclude
that the Laplace transform is a very important and powerful tool in mathe-
matics.
References
1.
11/08/2012
2.
23/10/2012
3.
dt-content-rid-46413641=courses=MM252YEAR2014=Standard20Laplace20
Transforms:pdf)08=11=2015(List of standard transforms in appendix)
4.
mathematics/laplace-transforms-derivatives) 06/11/2015
5. 8:pdf)15=06=2010
6.
Journal of Applied Physics) (2006)
6

8. Appendix
7