Solving Laplace Transform Problems
6.1
Here, we reviewed the basic logic and strategy of solving differential equations using
Laplace transforms. The idea is to transform the equation into something solvable. The
Laplace transform of a function is given by the improper integral:
Ł{ f ( t )} =

0
  st e ( )
Ex: Find the Laplace transform of the function f ( t ) = t .
 
0

 st te dt
   st u , e dt du

,
 
1 s e
 st
,
  st te dt

0

 st te dt
  t s e
 st  
1 s
  st e dt
 lim b


 
 b

1 sb 2 se s e sb
  s t e
 st
 

1 s
2
1
 st e , so :
2 s (0) s s e




1 s
2
0 (L'hôpital) 0
In practice, we don't often calculate these by hand, though I did a few elementary ones in class. We more frequently consult the table of standard Laplace transforms on p. 319.
-> Of particular importance here is the fact (which I proved in class) that the Laplace transform is a linear operator . Put simply, that means that when we want to 'Laplace' one side of an equation, we can 'go term-by-term and pull out the constants'. For example,
{
 
2 y
 
8 }

L {0} would be:
{
 
L y
 
L y

L {0} .
6.2
We began here by looking at another technical detail concerning the Laplace transform, which we already know is a linear operator. Theorem 6.2.2 (p. 315) gives a formula for transforming derivatives. So, as a result, we would have the following useful formulas:
 
{ ( )}
 f (0),
L f
 t
 2
{ ( )}
 sf (0)
 f

(0).
So now, I will state a general strategy for solving DE's using Laplace transforms:
(1) We take the Laplace transform of both sides of the DE, relying on the fact that it is a linear operator, and using the formulas above. We also, customarily, replace L { f ( t )} with
Y( s ). Then we solve the equation for Y( s ), which is usually algebra-intensive.

(2) We 'invert the transform'. This is something of an art - we have to manipulate the solution for Y( s ) into a recognizable transform off of the list of known ones. In this way, we 'recover' the original solution of the DE, y ( t ).
Let's start by doing some pure inversion problems:

4
( s

1)
3
Take a look down the list - there are no standard transforms with a cube on the denominator (though there's a bunch of squares). You will note however, that #11 on the list allows for an arbitrary power n + 1 on ( s - a ). If we picked n = 2, and a = 1, then we would seem to be in business. But there's one snag: the top would have to be n ! = 2! = 2 (instead of 4). That's ok, we just rewrite:
( )

2*
2
( s

1)
3

2

2 (1) t

, and so our 'inverted transform' would be:
( )

2
2 y t t e t
(4)

3 s s
2  
6
We once again peruse the list: there is nothing directly corresponding to Y ( s ). From the problem we worked in class, though, and from the obviously factorable denominator, we suspect that the use of partial fractions might help:
( s

3)( s

2)

 ( s

3
3)( s s
3 s

( 2) (

3),

2) s
A

3
 s
B

2
( s

3)( s

2), s

3 :
9

5 , s
 
2 :
  
5 ,
A

9
,
5
B

6
,
5

9 1
5 s

3

6
5 s
1

2
.
Now, we can see that these terms correspond to e at
on the list. Our inverse transform would be:

5
9 e
3 t 
5
6 e

2 t
(-2 on last exponent: s + 2 = s - - 2)
(6)

2 s

3 s
2 
4
This one could probably be done by partial fractions as well, but we notice transforms #7,8 on the list with s
2  a
2 in the denominator and wonder if we can do something with these. Let's try breaking up Y ( s ) over the common denominator and seeing what happens:

2 s
2 s


3
4
 s
2
2 s

4
 s
2
3

4

2 s
2 s

4
 s
2
3

4

The first term looks ready to go - it would be the transform of 2 cosh( 2t ). But the second one is a little problematic - we want to use sinh( at ) but that means we have to have a = 2 on the top - so just tweak it by rewriting the numerator.

2 s

3 s
2 
4

2 s
2 s

4
 s
2
3

4

2 s
2 s


3
4 2 s
2
2

4
, and we're ready to rip - the inverse transform is: y ( t ) = 2cosh(2 t ) -1.5sinh(2 t ).
(8)

8 s
2 
4 s

12
. I'm going to use partial fractions on this one as well - but we do s s
2 
4) need to keep in mind that the Quadratic factor in the denominator requires a linear numerator, i.e.

8 s
2 
4 s

12 s s
2 
4)
A Bs C s s
2 
4
,
8 s
2 
4 s

12

A s
2    s

0 :12

4 ,
A

3.
To find B and C, pick a couple of values for s to get a system of equations in B and C : s

1:16
  
)(1), 1
)(2), 4 B

2 C

12 s

2 : 36
 

2( B C 1)
4 B

2 C

12
2 B

10, B

5, C
 
4.
So, we have:

8 s
2  s s
2
4 s


4)
12
 
5 s s
2 s


4
4

3
1

5 s s
2 s

4

2 s
2
2

4
.
The inverse transform would be: y ( t ) = 3(1) + 5cos(2 t ) - 2sin(2 t ).
(10)

2 s

3

2 s

10
. The bottom doesn't factor - it has complex zeros. So we go s
2 down the list: nothing with three terms in the denominator. #'s 9 and 10 look like our best bet - we're going to have to complete the square, though, and do some manipulation:
 s
2
2
 s
2
 s
3

10

( s
2 
2
2 s s

3
 

( s

2 s
1)

2
3

3
2
. I'm going to factor 2 from the top, and then try to get it in the form of the transform for e at cos( bt ).

( )

2
( s

 s

1)
2
3
2


3
2

2
(( s
 
5
2
)
( s

1)
2 
3
2
2
( s
(
 s
1)

2
1)

3
2

5
3 ( s

1)
3
2 
3
2
.

2
( s
(
 s
1)

2
1)

3
2

( s

1)
5
2 
3
2

Yikes! But we're done - the inverse transform would be:
( )

2 e
 t t

5
3 e
 t
Now, we'll try a full problem.
(16) y '' 2 ' 5 y

0; (0)

2, y

(0)
 
1.
We begin by transforming both sides, then solving for Y ( s ):

L y

L y

L {0}, ( )

{ }
2
( )
 sy (0)
 y

(0)

2( sY s
 y

Y s

0,
( s
2 
2 s

5) ( ) 2 s 1 2( 1) 0,
 s
2
2 s

1

2 s

5
.
Now, we need to invert the transform Y . This is a case similar to #10 - the denominator has complex zeros, and won't factor cleanly, so we complete the square on bottom:


2 s 1
 s
 

1
( s
2
2

1) 4

2
( s
 s
1)
2
2

2
2
2
( s
(
 s
1)

2
1)

2 2

1
2 ( s

1)
2
2 
2 2

2
(( s
 
1
2
)
( s

1)
2 
2
2

2
( s
(
 s
1)

2
1)

2
2

( s

1)
1
2 
2
2
The inverse transform, and solution, is:
( )

2 e
 t t

1
2 e
 t
6.3
Wherein we consider the infamous step function :

0, t
 c c  1, t
 c
These, as we will see, are valuable in solving differential equations with discontinuities, or what we call 'impulsive forcing'.
->From a problem-solving standpoint, we have two theorems in this section, numbered
6.3.1, and 6.3.2, which allow us to transform and invert functions involving step functions. Let's take a looks at the gist of them:

Th 6.3.1
If F ( s ) = L { f ( t )},
{ ( ) ( c
 c )}
 e
 cs
( ), and c
( ) (
 
L

1
{ e
 cs
( )}, ( )

L

1
F s
Th 6.3.2
Under approximately the same conditions:
{ ct
L e f t

F s c ), ct
( )

L

1
F s c )}
Ex : Invert the transform
So we want to invert: L

1

1
 e

3 t s
2

1
 e

3 t s
2 

L

1

1
 e
2 s s

3 t
2 

L

1 
L

1
 e s

3 t
2

The first term corresponds to the transform of t on the list. Taking a close look at the
1 second term, I think we can use theorem 6.3.1, with c = 3, F ( s ) = , and f ( t ) = t . The s
2 answer is c
( ) ( ) u t t

3
( )( 3) .
-> Note that the presence of e
-cs
is the 'tipoff' that we want to invert to a step function.
Then, we got to get it all straight - who's c , F ( s ), and f ( t ). Then invoke the theorem.
Let's also try transforming some step functions:
(10) u t
 u t
 u t is pretty easy, since { ( )}
 e
 cs s
:
Y ( s ) = e
 s s
(12)

2 e

3 s s

6 e

4 s s
( )
 
1
( )(

1)
In the second term, the 'c's match', c = 1. We also see that ( t - 1) is a 'translate' of the function t , whose Laplace transform we know. We use #13 on the list :
L { f ( t )} =
1
2
 e
 s
2 s s
(8) Is a little more tricky.
 
 t
0, t

1
2  
2
The structure of the function suggests that it is a step function, f namely:
( t ) = u t t
1
2  
2)
Now, I need to rewrite the polynomial so that the 'c's match'. I'm going to complete the square and see what I have: u t t
1
2
2 t 2)
 u t
1 t
2
2 t 1) 1) u t t
1
 2   u t t
1

1)
2  u t
1
( ).

Note that the 'c's match', c = 1, on the first term, and that ( t - 1)
2
is a 'translate' of the function t
2
, whose transform is
2 s
3

2
 s e e s
3 s
 s

. The transform of the function is
These have been some typical 'tricks' you use when transforming functions. Let's try inverting some as well:
(14)
 e

2 s s
2  
2
. I'm going to wing it a little: factor out the exponential, use partial fractions, and try to figure out what's going on.
 e

2 s s
2  
2
 e

2 s
1 s
2  
2
,
1 s
2  
2
 s
A

2
 s
B

1

(
 
B s

2),
1
, s
 
2, s

1,

3 A

1, A
 
1
3
3 B
  e

2 s e

2 s
1
3 s

2

1
3 s

1

1, B

1
3
At this point, you have to be a little careful, I recognize that
1 s

2 and
1 s

1 are the translates of e
-2 t
and e t
, respectively, but I have to remember that when I invert back to a step function (again, the tipoff that I'm going to do that is the presence of the e
cs
, where c = 2 here), that I have to 'translate' by c : y ( t ) =

1
3 u t e
2
( )

2( t

2) 
1
3 u t e
2
( )
( t

2)