Laplace Transforms Chapter 11 1

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.
Chapter 11
Laplace Transforms
------------------- .
Now you decide how to proceed.
1
74
Difficulties with example 2
Difficulties with example 3
Difficulties with example 4
Straight on, no difficulties so far
1
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Chapter 11 Laplace Transforms
.
Laplace transforms
1
Objective: In the first section the definition of Laplace transforms will be explained and
developed. It will prove useful, to copy all definitions and results into your notebook when going
through this and subsequent sections. Then they will be readily accessible to you and you will not
have to take recourse to the textbook for each and every single computation.
READ
11.1 Introduction
11.2 The Laplace transform definition
Textbook pages 321–322
When done
.
Example 2 deals with: y ' '5 y '4 y
------------------- 0
initial conditions: t
0
y0
0
y '0
3
2
75
Finding a solution always requires following these three steps:
Step 1: Apply the Laplace transform. Insert the initial conditions.
Step 2: Simplify the equation and solve for y (s ) or in another notation F(s).
Step 3: Apply the transform.
If needed, read again the first part of section 11.4 in the textbook.
.
The following shows all details for example 2 from the textbook.
3
Difficulties may arise with regard to rewriting y s 2
s 5s 4
Determine the roots in the denominator by solving the quadratic equation. That will enable us to
3
represent the expression y s as a product of linear factors.
2
s 5s 4
s1 ....................
s2 ...................
------------------- 2
76
Chapter 11 Laplace Transforms
.
Write down the definition of the Laplace transform:
2
L > f t @ = ………………………….
Being well versed in the notation greatly eases not only studying the textbook but to an even greater
extent eases later practice.
Write down the same definition using different notations
L > y t @ = …………..
L > f (x)@ = …………..
L > y x @ = …………..
------------------- .
s1
4
s2
1
3
76
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Therefore, the denominator can be written as a product of linear factors: s 2 5s 4
3
Thus, the transform reads: y s s 4 ˜ s 1
For this expression of y s the inverse transform can be found using the table: y
s 4 ˜ s 1
.........................
So we are done.
If, however, you prefer to use a more elementary approach, you may first split the expression for
y s into partial fractions and then find the inverse transforms for elementary expressions of the type
1
5s 11
, which are given by e at , of course. For the similar case of y s this has
s a ss 2 ˜ s 4
been demonstrated in frame 71, which you may like to refer to.
------------------- 3
77
Chapter 11 Laplace Transforms
.
f
³ f t ˜ e
L > f t @ =
s ˜t
dt
f
or F(s)
s ˜t
dt
y
or Y(s)
f
or F(s)
3
0
f
³ yt ˜ e
L > y t @ =
0
L > f (x)@ =
f
³ f x ˜ e
s˜ x
dx
0
L > y x @ =
f
³ yx ˜ e
s˜x
dx
y
or Y(s)
0
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Complete the definition
L-1 >F s @ =…………………………..
------------------- .
y
e t e 4t
4
77
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Again, please make a choice
Difficulties with example 3
Difficulties with example 4
Straight on
4
------------------- 78
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Chapter 11 Laplace Transforms
.
The symbol L 1>F s @ stands for the inverse Laplace transform.
L-1 f ( s )
4
F t or F(x) in case we use the variable x.
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Before going through examples that will show the extreme usefulness of Laplace transforms, you will
have to stand a certain number of dry spells, while you take in the rules for the transforms and copy
them into your notebook.
------------------- .
Example 3
5
78
Solve the following differential equation: y ' '8 y '17 y
given the initial conditions: t 0
y0 0
y '0 3
0
Step 1: Apply the Laplace transform to the differential equation according to the rules:
…………………….…………………..= 0
------------------- 5
79
Chapter 11 Laplace Transforms
.
11.1 Laplace transform of standard functions and general theorems
5
Objective: In this section you will acquaint yourself with the main facts on how to transform basic
functions like e at , sin Z t , C ˜ t , etc. Funnily enough, t n will have to wait for a while.
Please follow all arguments, and copy all calculations and results into your notebook.
Since this section is quite long we suggest you study it in parts. In this section we will use the
following notation: few denote the original function by y(t) or f(x).
Study
11.3 Laplace transform of standard functions
Theorem I: The shift Theorem
Textbook pages 322–324
When done
.
------------------- y s ˜ s 2 y0 y '0 sy0 8 y s ˜ s 8 y0 17 y s 0
6
79
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Step 2: Insert the initial conditions t
0, y0
0, y '0
2 and solve for y s .
y s ………………
------------------- 6
80
Chapter 11 Laplace Transforms
.
Given the constant function y(t)
Try to derive its Laplace transform y s on your own.
y s = …………………………
6
Solution found
Help needed
.
y s ------------------- 8
------------------- 7
3
s 2 8s 17
80
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
In the table we find the inverse transform for the following function:
y s 1
s a Z
2
2
1
y
Z
˜ e at ˜ sin Z t
Thus, we face the task to rearrange the given denominator so that the inverse transform according to
the rule above can be applied. We succeed by completing the square for s 2 8s and obtain
3
y s (s 4)2 1
So, using the notation from the rule, a
4 and Z 2
1
y t ...................................
------------------- 7
81
Chapter 11 Laplace Transforms
.
Given the constant function y(t) = C
Wanted: Laplace transform y (s ) .
You have to solve
7
f
y (s ) = ³ e st ˜ y t ˜ dt
0
For y(t) = C we insert C into the integral:
y s f
³e
st
˜ C ˜ dt
0
Factor the constant C, solve the integral, and insert the limits of integration:
y s f
³e
st
˜ C ˜ dt = ……………………………
0
------------------- .
y
3 ˜ e 4t ˜ sin t
8
81
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
We could have transformed the function y s differently. This is shown in example 2 in the textbook.
After determining the roots a and b of the denominator, we obtain the following expression:
3
y s s a ˜ s b Show me this transformation also
Difficulties with this example
Ready to go on
8
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Chapter 11 Laplace Transforms
.
f
G
y s C ˜ ³ e st dt
8
0
f
ª e st º
y s C «
»
¬ s ¼0
C
s
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Likewise we can determine the Laplace transform of an exponential function.
Given: y t e at
G
To do: Compute y :
G
y =……………………...........
Computation successfully done
Help needed
.
Given: y s ------------------- 10
------------------- 9
3
s 2 8s 17
82
Determine the roots of the denominator (solve the quadratic equation):
s1
....................
s2
..................
------------------- 9
83
Chapter 11 Laplace Transforms
.
Given: y t e at
G
Wanted: Laplace transform y .
9
Again, you must compute the Laplace transform by evaluating an integral.
f
G
L > y t @ y ³ e st ˜ y t ˜ dt
0
Insert y(t) as given:
f
G
L > y t @ y s ³ e st ˜ e at ˜ dt
0
By factoring t in the exponent, we get an integral, which we have already solved frequently.
Lastly, we insert the limits of integration.
G
y
L > y t @
……………………………
In case of remaining difficulties consult the textbook.
------------------- .
s1
s2
4 i
4 i
10
83
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Thus y s 3
s 4 i ˜ s 4 i So we have a type of expression that we know how to handle.
3
y s s a ˜ s b 1
y
˜ e 4i t e 4i t ˜ 3
2i
Using Euler’s formula we obtain the result already known
y 3 ˜ e 4t ˜ sin t
Difficulties with example 4
Ready to go straight on
10
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Chapter 11 Laplace Transforms
.
f
y
s a ³ e dt
0
f
ª e s a t º
« s a »
¬
¼0
1
s a 10
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
The Laplace transforms of trigonometric functions are obtained by a direct approach, too.
Let us start with the sine function:
yt sin Z t
L >sin Z t @ ……………
Hint: Use Euler’s formula to express sin Z t by exponential functions.
Solution found
Help needed
.
Example 4 asks us to solve: y ' '6 y t
with initial conditions: t 0
y0 0
------------------- 13
------------------- 11
84
y '0 1
Applying the Laplace transform and rearranging according to the known procedures results in
y s 1
1
2 2
s 6 s s 6
2
Only the inverse transformations may pose a problem. But in the table we do
find an inverse transform for both expressions. Putting Z 2 6 we arrive at
1
1
sin 6 ˜ t y (t )
6 ˜ t sin 6 ˜ t
6
6 6
Simplify y t ...........................................
------------------- 11
85
Chapter 11 Laplace Transforms
.
The Laplace transform of the sine function y t sin Z t is to be found.
We already know the transformation of the exponential function
1
L e at y ( s )
sa
11
> @
Hint: You should have taken a note of this result before, in order to quickly have access to it.
We recall Euler’s formula
e iZ t
e iZ t
i sin Z t cos Z t
i sin Z t cos Z t
So, sin Z t can be expressed as a difference of exponential functions.
e iZ t e iZ t
…………
------------------- .
y t 1§
5
·
sin 6t ¸
¨t 6©
6
¹
12
85
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Go on
------------------- 12
86
Chapter 11 Laplace Transforms
.
e iZ t e iZ t
2i sin Z t
12
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
This implies sin Z t
1 iZ t iZ t
e e
2i
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Since we know the transform of an exponential function, the problem is almost solved. A little bit of
calculation is all that remains.
We know:
> @
1
sa
By inserting we obtain:
1ª 1
1 º
L >sin Z t @ y ( s )
«
2i ¬ s iZ s iZ »¼
L e at
y (s)
Using the common denominator for the fractions we get:
G
y s =…………………
------------------- .
You should now be in a position to solve the following DE.
y ' '6 y '8 y
4
Initial conditions: t
0
y0
13
86
0
y '0
0
Step 1: Obtain the Laplace transform of the differential equation
………………………………=………………………….
------------------- 13
87
Chapter 11 Laplace Transforms
.
L >sin Z t @ F ( s )
1 ª s i Z s iZ º
«
»
2 i «¬ s 2 Z 2
»¼
13
Or simply
ª Z º
L >sin Z t @ y s « 2
2»
¬s Z ¼
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
The Laplace transform of the cosine function can be arrived at in quite a similar way.
Try to solve the problem on your own.
L >cos Z t @
y s = ………………………………
Solution found
Help needed
.
s 2 y s sy0 y '0 6sy s 6 y0 8 y s ------------------- 18
------------------- 14
4
s
87
Please note: The constant 4 on the right side of the original DE needed to be transformed also.
Step 2: Insert the initial conditions t
0, y0
0, y '0
0 and solve for y s :
y s .....................................
------------------- 14
88
Chapter 11 Laplace Transforms
.
To be found: L >cos Z t @ .
Again, we start by expressing cos Z t as a sum or a difference of exponential functions using
Euler’s Formula.
14
Recall:
e iZ t = …………………………………..
e iZ t = ………………………………….
Therefore, the cosine function can alternatively be expressed as
cos Z t = ……………………………………
------------------- .
0 y s 4
s s 2 6s 8
15
88
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Decompose into partial fractions in order to find expressions that can easily be inverse transformed.
y s ............................
Solution successfully found
Help and explanation
15
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Chapter 11 Laplace Transforms
e iZ t
.
i sin Z t cos Z t
15
iZ t
i sin Z t cos Z t
1 iZ t
cos Z t
e e iZ t
2
e
>
@
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Now we can determine:
L >cos Z t @
y s …………………………….
Solution found
One more hint needed?
.
------------------- 18
------------------- 16
We must rewrite
4
y s 2
s s 6s 8
89
The bracketed expression in the denominator can be expressed as a product:
s 2 6s 8 s a ˜ s b
Hint: In this special case you can either guess the correct values or, more generally, you must solve
for the roots of the quadratic expression.
4
y s ss............ ˜ s............
------------------- 16
90
Chapter 11 Laplace Transforms
.
The task is to determine the Laplace transform of f t cos Z t .
L >cos Z t @
f
³e
st
16
cos Z t dt
0
We know already
1
L e at
sa
> @
We also know
cos Z t
>
1 iZ t
e e iZ t
2
@
By combining both results we obtain
ª1
º
iZ t
iZ t
L >cos Z t @ = L « e e
» ………………………..
¬2
¼
------------------- .
y s 4
ss 2 ˜ s 4 17
90
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Now the fraction can be split into partial fractions. In a later step the inverse transform will be
applied.
4
s s 2 ˜ s 4 A
B
C
s s 2 s 4 4
= ………………………………
ss 2 ˜ s 4
Solution found
Help
17
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Chapter 11 Laplace Transforms
.
L >cos Z t @
ª 1 iZ t
iZ t º
L « e e »
¬2
¼
1ª 1
1 º
2 «¬ s iZ s iZ »¼
17
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
After rearranging and using the common denominator we finally end up with:
L >cos Z t @
G
y s =……………………………..
------------------- .
Our approach for the decomposition into partial fractions is:
4
s s 2 ˜ s 4
18
91
A
B
C
s s 2 s 4 We must determine A, B, and C.
By using the common denominator of the partial fractions we obtain
4
(................................)
ss 2 ˜ s 4 ss 2 ˜ s 4
------------------- 18
92
Chapter 11 Laplace Transforms
.
L >cos Z t @
y s s
s Z2
18
2
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Compute the Laplace transform
y s = L >5 ˜ sin 4t @
………………
------------------- .
4
s s 2 ˜ s 4
A ˜ s 2 ˜ s 4 B ˜ s s 4 C ˜ ss 2
ss 2 ˜ s 4
19
92
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Both numerators must be equal.
Expand the numerator on the right side and collect terms according to the different powers of s:
4 ..................................... ....................
------------------- 19
93
Chapter 11 Laplace Transforms
.
y (s)
5˜4
s 16
19
2
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Compute the Laplace transform y (s ) for y (t ) 5 ˜ cos 4t
G
y (s ) …………
------------------- .
4
s 2 >A B C @ s> 6 A 4 B 2C @ 8 A
20
93
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The equation can only be satisfied, if the right hand side is independent of both s and s 2 . This
amounts to both brackets being equal to 0.
From this we obtain the following equations to determine the unknown values of A, B, and C.
4 8A
0 A BC
0 6 A 4 B 2C
Determine A, B, and C
A ..............
B ...............
C ...............
Solution found
One last hint
20
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Chapter 11 Laplace Transforms
.
y (s)
5˜ s
s 16
20
2
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Given a linear function y t C ˜ t
To be found the Laplace transform:
L >C ˜ t @
y s G
y (s ) =……………………
Solution found
Hint needed
.
------------------- 22
------------------- 21
1
2
By inserting into the other equations
A BC 0
and
6 A 4 B 2C 0
we obtain
1
3 4 B 2C 0
BC 0
and
2
94
The first equation implies A
We rearrange and add
2 B 2C 1
4 B 2C 3
This implies
B ..................................
C
...................................
------------------- 21
95
Chapter 11 Laplace Transforms
.
How to determine the Laplace transform of a linear function is described in detail
in the textbook. Please recap section 11.3 and follow all computations
meticulously. If necessary, also look up the technique of integrating by parts.
21
In parallel to the procedure shown in the textbook compute
L >C ˜ t @ ……………….
------------------- .
A
1
2
B
1
C
1
2
22
95
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
From this we obtain the desired form of the Laplace transform
…. y s …………………………………….
------------------- 22
96
Chapter 11 Laplace Transforms
.
L >C ˜ t @
y
C
s2
22
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Given y t 5 ˜ t
Compute L > y t @
y ………………………………….
------------------- .
y s 1
1
1 1
2s s 2 2 s 4
23
96
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Step 3 requires us to obtain the inverse transform using the given table.
y t .........................................
------------------- 23
97
Chapter 11 Laplace Transforms
.
5
s2
y (s)
23
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Horizontal shift: If a function is shifted to the right by a units, its transform is
L > y (t a @ e as ˜ y ( s )
f
³ yt a ˜ e
s ˜t
dt
…………………………………
0
Hint: Substitute u for t-a and evaluate the integral with respect to u
------------------- .
y t 1 1 4t
e e 2t
2 2
24
97
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Let us turn to one last exercise which will be well known to you from physics -- the equation of
motion in the gravitational field near the earth’s surface. As is customary in physics, we denote time
by t and the derivative with respect to time by y . The direction of y is up. Then the acceleration is
directed down, i.e. negative.
y g
Laplace transform
…………………………………
Solve for L s L s .......................................
Inverse transform
y t …………………………….
------------------- 24
98
Chapter 11 Laplace Transforms
.
L > y t a @ e as ˜ y s because
f
f
0
a
s ˜t
³ yt a ˜ e dt
s ˜( u a )
du
³ yu ˜ e
f
e as ˜ ³ y u ˜ e s ˜u du
24
a
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Let us now start with the linear function y(t) = 5t, which is represented by a straight line through the
origin.
5
We already know its transform to be y
s2
Let us now look at the slightly more general linear function y(t) = 5t – 15
As can easily be seen, it is a straight line which is shifted to the right by 3 units, since
y t 5t 3 5t 15
Compute the transform of this function using the result on horizontal shifting:
L >5t [email protected] …………………………
------------------- .
s 2 ˜ L s sy0 y 0
g
s
25
98
g y 0 y0
2 s3
s
s
g 2
y t ˜ t y 0 ˜ t y0
2
This is the famous and well-known equation for a freely falling object.
L s ------------------- 25
99
Chapter 11 Laplace Transforms
.
L >5t [email protected] e 3 s ˜
5
s2
25
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Using the same argument we are in a position to transform any linear function.
Given y t a ˜ t b
Step 1: Rearrange the formula, so that you can apply the Laplace transform
y(t) = ………………………………
------------------- .
11.4 Solving simultaneous differential equations with constant coefficients
Read
26
99
11.5 Solving simultaneous differential equations with
constant coefficients
Textbook pages 330–331
------------------- 26
100
Chapter 11 Laplace Transforms
.
Step 1: The straight line is shifted to the right by
Step 2: The Laplace transform is y s b
units.
a
26
b
a a˜ f
˜e
.
s2
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Example: Given the straight line y t 5t 50
Rewrite the function so that the Laplace transform can be obtained
y t …………………
Now compute the Laplace transform
y = ………………….
------------------- .
I have understood the principles of finding solutions and
have followed all examples in the textbook
Explanations for example 1 in the textbook
Explanations for example 2 in the textbook
27
27
100
------------------- 138
------------------- 101
------------------- 121
Chapter 11 Laplace Transforms
.
y t 5t 50 5t 10 y s 27
5 10 s
˜e
s2
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Look up the shift theorem in the textbook.
Given a function y(t) and its transform y s We then know y s a = ………………….
------------------- .
The first example in the textbook was the following system of equations:
for t 0 : x0 y0 0
3x 2 x y 1
x 4 y 3 y 0
28
101
Step 1: Apply the Laplace transform to both equations, observe the initial conditions, and simplify all
expressions. Consult the textbook, if necessary.
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
------------------- 28
102
Chapter 11 Laplace Transforms
.
y s a L
>e
at
@
˜ y t 28
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Proof from the textbook understood
Additional explanations for the proof
.
------------------- 30
------------------- 29
1
s
s ˜ L >x @ 4 s 3 L > y @= 0
3s 2
L >x @ s L > y @
102
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Step 2: Now the system of equations must be solved for L >x @ and L > y @, consecutively. We start
with L >x @ and eliminate L > y @ . For this purpose let us multiply the first equation by 4 s 3 and the
second equation by s . That leads to
3s 2 L >[email protected]˜ 4s 3 s4s 3 L > y @ 4s 3
s
2
s L >x @
s4 s 3 L > y @= 0
Add the equations to obtain
…………………………..=………………………..
------------------- 29
103
Chapter 11 Laplace Transforms
.
To be shown: y s a >
@
L e at ˜ y t 29
We must determine the Laplace transform of the function
e at ˜ y t Assuming that the transform of y t is known to be y s ,
we recollect the definition of the Laplace transform
>
f
@ ³e
L e at ˜ y t at
˜ e st ˜ y t dt
0
The integrand can be simplified as follows
>
f
@ ³e L e at ˜ y t s a t
˜ y t dt
0
The integral evaluates to y s a , which is assumed to be known.
>
@
Thus, it is proved that L e at ˜ y t y s a ------------------- .
>
L >x @ ˜ 3s 2 ˜ 4s 3 s 2
@
4s 3
s
30
103
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Now we must rearrange in order to obtain expressions that are amenable to the inverse transforms
that are shown on page 327 of the textbook. Let us start with the expression in brackets on the left
and expand the product:
L >[email protected] ˜ 12s 2 9s 8s 6 s 2
L >x ˜@ ˜ 11s 2 17 s 6
In order to obtain an expression of the type As a ˜ s b , we factor the constant 11 and solve the
quadratic equation. Thus we obtain
L >[email protected] ˜ 11 ˜ ........................ ˜ ..........................
4s 3
s
Further help
30
------------------- 105
------------------- 104
Chapter 11 Laplace Transforms
.
Given y t 3 ˜ sin 2t ˜ e 4t
Find the transform y s G
y s 30
………………………….
Solution found
Help needed
.
The task is to solve L >x @ ˜ 11s 2 17 s 6
------------------- 33
------------------- 31
4s 3
s
104
By factoring 11 we obtain
17
6 · 4s 3
§
L >x @ ˜ 11 ˜ ¨ s 2 s ¸
11
11 ¹
s
©
Let us now solve the quadratic equation
6·
§ 2 17
¨s s ¸ 0
11
11 ¹
©
s1
17 2
6
2
11
2 ˜ 11
17
2 ˜ 11
17
5
2 ˜ 11 2 ˜ 11
12
2 ˜ 11
17
289 264
2 ˜ 11
2 ˜ 112
6
11
17
2 ˜ 11
17
5
22
1
2 ˜ 11 2 ˜ 11
2 ˜ 11
Thus we can express
17
6·
§
L >[email protected] ˜ 11 ˜ ¨ s 2 s ¸ L >[email protected] ˜ 11 ˜ ................. ˜ ................
11
11 ¹
©
s2
25
2 ˜ 112
4s 3
s
------------------- 31
105
Chapter 11 Laplace Transforms
.
Given y t 3 sin 2t ˜ e 4t
Because of the damping factor e 4t we can use the shift theorem.
31
We start by computing the transform of
L >3 sin 2t @ …………………………
Hint: This result was already obtained. Possibly you may wish to consult your notes or the textbook.
..
L >3 sin 2t @
y s ……………………………………..
------------------- .
6·
§
L >x @ ˜ 11 ˜ s 1 ˜ ¨ s ¸ L >x @ ˜ s 1 ˜ 11s 6
11
©
¹
4s 3
s
32
105
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Solving for L >x @ this yields:
4s 3
1
L >x @
˜
s s 1 ˜ 11s 6 11
4s 3
§
s s 1 ˜ ¨ s 6·
¸
© 11 ¹
The fraction can be split into three partial fractions, whose denominators are given by the three linear
factors. After performing a slightly tedious calculation, which is prone to errors by miscalculation, we
obtain:
L >x @ ……………………………………..
Calculation happily done
------------------- 112
Help and explanation
------------------- Please note: The calculation could also be performed for the equivalent
expression
4s 3
L >x @
The result, of course, would be unchanged.
§
6 ··
§
11 ˜ ¨¨ s s 1 ˜ ¨ s ¸ ¸¸
© 11 ¹ ¹
©
32
106
Chapter 11 Laplace Transforms
.
L >3 sin 2t @
y s 3˜ 2
s2 4
32
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
However, we must determine L > y t @
>
L 3 sin 2t ˜ e 4t
@
We now use the shift theorem
>
L e at ˜ 3 sin 2t
>
@
y s a @
L e 4t ˜ 3 sin 2t = …………………………..
------------------- .
The following fraction must be decomposed:
4s 3
s s 1 ˜ 11s 6
33
106
We expect to decompose into partial fractions with the three linear factors as the denominators:
4s 3
A
B
C
ss 1 ˜ 11s 6 s s 1 11s 6
Now let us add the partial fractions using the common denominator, in order to gain equations for
determining A, B, and C.
4s 3
s s 1 ˜ 11s 6
..........................
s s 1 ˜ 11s 6
------------------- 33
107
Chapter 11 Laplace Transforms
.
>
L e 4t ˜ 3 sin 2t
@
3˜ 2
s 42 4
33
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Next task:
2 t
Given: y t e ˜ cos S t
>
2 t
Wanted: L e ˜ cos S t
@
……………………………
Solution found
Help needed
.
4s 3
ss 1 ˜ 11s 6 ------------------- 35
------------------- 34
A ˜ s 1 ˜ 11s 6 B ˜ s11s 6 C ˜ ss 1
ss 1 ˜ 11s 6
107
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
We multiply the expressions in the numerator and rearrange according to the powers of s:
4s 3
s s 1 ˜ 11 6 s 2 .................... s ˜ .................... ..........
ss 1 ˜ 11s 6 ------------------- 34
108
Chapter 11 Laplace Transforms
.
We use the same procedure as in the preceding exercise.
First, determine L >cos S t @ and, second, apply the shift theorem.
34
s
s S 2
L >cos S t @
2
The shift theorem tells us
>
@
at
L e ˜ y t y s a Thus
>
2 t
L e ˜ cos S t
@
…………………
------------------- .
4s 3
s s 1 ˜ 11s 6 s 2 11A 11B C s 17 A 6 B C 6 A
35
108
s s 1 ˜ 11s 6 ____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Comparing coefficients we obtain three equations for A, B, and C.
constant terms:
for s:
for s 2 :
3 6A
4 17 A 6 B C
0 11A 11B C
From this we easily determine
A …………
B …………
C
…………
Solution
Explicit calculation
35
------------------- 111
------------------- 109
Chapter 11 Laplace Transforms
.
>
L e
2t
˜ cos S t
@
s2
s 22 S 2
35
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Understanding and applying the linearity theorem should be quite obvious.
Obtain the Laplace transform of
y t 2 cos 2S t sin 2S t :
L >2 cos 2S t sin 2S t @ =……………………………….
Hint: All transforms have already been introduced. If necessary consult your notes or the textbook.
------------------- .
We obtained three equations containing three variables
1)
3 6A
2)
4 17 A 6 B C
3)
0 11A 11B C
Equation 1):
A
Equation 2) minus Equation 3):
4
4
Hence
Insert B into equation 3):
36
109
1
2
17
11
11B C 6 B C
2
2
3 5B
B ……………….
11 11
C
0
2 5
C
C
22 55
10
…………………
------------------- 36
110
Chapter 11 Laplace Transforms
.
L >2 cos 2S t sin 2S t @
2˜s
2S
2
2
2
s 2S s 2S 2 s 2S
2
s 2S 2
36
2
------------------- .
A
1
2
B
1
5
C
33
10
37
110
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Thus, the decomposition into partial fractions is complete:
L >x @
A
B
C
s s 1 11s 6 L >x @ ……………………….
------------------- 37
111
Chapter 11 Laplace Transforms
.
11.2 Laplace transform of derivatives
37
Objective: Obtain a good command of how to use two very useful theorems and, as a by-product,
transform old friends like t n .
This is the last section preparing you for highly useful applications of the Laplace transforms. Once
again, we advise you to take notes of all results, rules, and theorems; this will aid your memory.
Read the second part of
Section 11.3
Theorem II: Transforms of products t y(t)
Theorem III: Linearity
Theorem IV: Transforms of derivatives
Textbook pages 324–327
When done
.
L >x @
------------------- 1
1
33
1
˜
, or, equivalently L >x @
2 s 5s 1 10 11s 6
1 1
1
3
1
˜
˜
6·
2s 5 s 1 10 §
¨s ¸
11
©
¹
38
111
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Step 3:
Now we are in a comfortable position to do the inverse transform using the table on page 333 of the
textbook. We obtain
x ………………………….
------------------- 38
112
Chapter 11 Laplace Transforms
.
In this chapter we use the following abbreviations for denoting the values of the
original function y(t) or f t and its derivatives at t 0 :
y 0 ………………
38
y ' 0 ……………...
y ' ' 0 ……………..
Equivalently we use the notation
f 0 ……………..
f ' 0 …………….
f ' ' 0 ……………
------------------- .
6
x
1 1 t 3 11˜t
˜e ˜e
2 5
10
39
112
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Now we must find y. In the given system of equations we eliminate L >x @ .
We have
1
3s 2 L >[email protected]
+ s L >[email protected]
s
4 s 3 L > y @ 0
+ sL >x @
A convenient way to eliminate L >x @ is to multiply the first equation by s and the second one by
3s 2 , and then add the resulting equations.
That yields: ……………………..=…………………….
and
……………………..=…………………….
------------------- 39
113
Chapter 11 Laplace Transforms
.
y 0 y0
f 0
y ' 0
y '0
f ' 0
f '0
y ' ' 0
y ' '0
f ' ' 0
f ' '0
f0
39
The advantage of this shorthand notation will become apparent when we turn to solving differential
equations.
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Given: y t t ˜ e at
To be found: L > y t @ =………………………….
Solution found
Help needed
.
------------------- 43
------------------- 40
s 3s 2 ˜ L >x @
s 2 ˜ L > y @ 1
s 3s 2 ˜ L >x @ 4 s 3 ˜ 3s 2 ˜ L > y @ 0
113
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
We now add and obtain
>
L > y @ ˜ 4s 3 ˜ 3s 2 s 2
@
1
Multiplying and collecting powers this leads to
L > y @ ˜ 11s 2 17 s 6
1
The quadratic expression has already been factored in step 104 with the roots
6
s1 , s2 1
11
Therefore L > y @
1
.....................................
------------------- 40
114
Chapter 11 Laplace Transforms
.
In the textbook you find that the Laplace transform of a function t ˜ y t is
given by a derivative:
d
L >t ˜ y t @ > y s @ if we denote L > y t @ by y s ds
In our case we already know the Laplace transform y s for y t e at
40
………………………
If necessary, consult your notes or the textbook pages 332 and 334.
------------------- .
L >[email protected]
1
41
114
s 1 ˜ 11s 6
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
We decompose the fraction into two partial fractions and obtain
L > y @ …………………
Solution found
Help and detailed computation
41
------------------- 118
------------------- 115
Chapter 11 Laplace Transforms
.
The Laplace transform of y t e at is given by y s 1
sa
41
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
In order to determine the Laplace transform of t ˜ y t the theorem tells us to find the derivative of
y s with respect to s, and change the sign:
d
> y s @ L >t ˜ yt @
ds
In our case y s Thus, we get
1
sa
d
> y s @ ………………..
ds
------------------- Hint (redundant, we do hope):
.
1
sa
To decompose into partial fractions:
Assumption:
1
s 1 ˜ 11s 6
42
s a 1
1
115
s 1 ˜ 11s 6
A
B
s 1 11s 6
We use the common denominator
1
s 1 ˜ 11s 6
A11s 6 Bs 1
s 1 ˜ 11s 6
After multiplying and ordering according to powers of s we obtain
1
s 1 ˜ 11s 6
...................................
s 1 ˜ 11s 6
------------------- 42
116
Chapter 11 Laplace Transforms
.
1
s a 2
d
> y s @
ds
42
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
So we now know
d
> y s @ 1 2
ds
s a >
L t ˜ e at
@
Thus, we have identified the Laplace transform for t ˜ e at :
>
@
L t ˜ e at =…………………………
------------------- .
1
s 1 ˜ 11s 6
s>11A B @ 6 A B
s 1 ˜ 11s 6
43
116
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
This leads us to the equations for determining A and B
1 6A B
0 11A B
From this we obtain
A =............................ B =............................
------------------- 43
117
Chapter 11 Laplace Transforms
.
>
L t ˜ e at
@
1
s a 2
43
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Let us turn to another exercise. This time we use the notation f(t) and for the Laplace transform F(s)
Given:
f t t 3
To be found: F s ........................
------------------- .
A
1
5
B
11
5
44
117
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Thus we obtain
L > y @ ………………………
------------------- 44
118
Chapter 11 Laplace Transforms
.
f t t 3
F s 3!
s4
6
s4
44
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
> @
Easy, wasn’t it? L t ˜ t 2
> @
d
L t2
ds
d2
d2 1
t
>
@
L
(ds) 2
(ds ) 2 s 2
Next task
.
L >[email protected]
------------------- 1
11
1
1
1
1
, or, equivalently L > y @
6·
5s 1 5 11s 6 5s 1 5 §
¨s ¸
11
©
¹
45
118
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Step 3: Inverse transform
y t ………………………….
------------------- 45
119
Chapter 11 Laplace Transforms
.
Now this exercise will be slightly more demanding:
Given:
45
f t 3 sin 2t 2
Wanted: F s ……...........................
Solution of the somewhat tricky task found
Help needed
.
y t 6
1 t 1 11˜t
˜e ˜e
5
5
------------------- 51
------------------- 46
˜t ·
1 § t
¨ e e 11 ¸
¨
¸
5©
¹
6
119
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Most difficulties in algebraic operations arise by oversight or slips of the pen which afterwards can
hardly be discovered.
------------------- 46
120
Chapter 11 Laplace Transforms
.
The transform of sin 2 2t can neither be found in the table nor, as yet, in our notes.
46
Therefore, let us try to rewrite the given expression sin 2 2t so that we can find the Laplace
transform with our given arsenal. We will use Pythagoras’ theorem and the addition formulae for
trigonometric functions so that the squared expression is eliminated.
sin 2 2t …………………………
Solution of this intermediate step found
Further hints needed
------------------- 50
------------------- 47
.
120
No difficulties with example 2 from the textbook
Show me a detailed walk through example 2 from the textbook
47
------------------- 138
------------------- 121
Chapter 11 Laplace Transforms
.
Recall Pythagoras’ theorem cos2 t sin 2 t 1 and
the addition formula cosu v cos u ˜ cos v sin u ˜ sin v
47
Now, in a seemingly round about way, let us first compute
cost t cos 2t =……………………..
------------------- .
Example 2: The following system of equations is to be solved
x 2 x y 1
x y 2 y 0
for t
0:
x0
1,
x0
y0
48
121
y 0
0
Step 1: Applying the Laplace transform and inserting the initial conditions results in:
…………………………………=………………………………
…………………………………=……………………………….
------------------- 48
122
Chapter 11 Laplace Transforms
.
cos 2t
cos 2 t sin 2 t
48
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Because of cos 2 t 1 sin 2 t , we can eliminate cos 2 t from the expression above.
cos 2t
…………………
------------------- .
s
2
2 L >x @
-5 L > y @
1
s
s
s L >x @ s 2 2 L > y @ x0
49
122
1
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Step 2: Solve for L >x @ by eliminating L > y @ .
For this we multiply the first equation by s 2 2 and the second by s. Then by adding
L > y @ is eliminated.
We obtain
L >x @ =………………………..=………………………….
------------------- 49
123
Chapter 11 Laplace Transforms
.
cos 2t 1 2 sin 2 t
49
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
So we are in a position to rewrite sin 2 t as follows
sin 2 t =……………………..…
------------------- .
>
L >x @ ˜ s 2 2 s 2
2
@
s2 2
s s 2 2 s
s
50
123
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
We rearrange the right hand side and use the common denominator s. On the left side we expand the
square:
L >x @ ˜ >[email protected]
........................................
s
------------------- 50
124
Chapter 11 Laplace Transforms
.
1
>1 cos 2t @
2
sin 2 t
50
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
The original task was to obtain the Laplace transform of the function f t 3 sin 2 2t .
Using the above result we get:
1
f t 3 ˜ 1 cos 4t 2
ª3
º
This expression is now accessible to transformation: L « 1 cos 4t » …………..………………
¬2
¼
------------------- .
>
L >x @ ˜ s 4 5 s 2 4
@
s 4 4s 2 2
s
51
124
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
This leads to
s 4 4s 2 2
L >x @
s s 4 5s 2 4 We wish to decompose into partial fractions, which will yield expressions that can then be inversely
transformed. For this we must expand the bracketed expression into a product of the type
s 2 a ˜ s 2 b .
For finding a and b we solve the equation s 4 5s 2 4 for s 2 . With regard to s 2 it is just a quadratic
equation. The zeroes are not difficult to find:
s1
2
………
2
s2 ………
Hence
s 4 5s 2 4 s 2 ............. ˜ s 2 ............
Solution found
Show me the solution of the quadratic equation
51
------------------- 126
------------------- 125
Chapter 11 Laplace Transforms
.
ª3
º
L « 1 cos 4t »
¬2
¼
3 1 3
s
˜ ˜
2 s 2 s 2 42
3 ª s 2 16 s 2 º
2 «¬ s s 2 16 »¼
3 ª1
s º
2
«
2 ¬ s s 4 2 »¼
3˜8
s s 2 16 51
24
s s 16
2
------------------- .
To be solved for s 2 : s 4 5s 2 4
0
52
125
We solve the quadratic equation:
5
25
r
4
2
4
s2
s1
2
………….
2
………….
s2
s 4 5s 2 4
s
2
5
9
r
2
4
5 3
r
2 2
......................... ˜ s 2 ........................
------------------- 52
126
Chapter 11 Laplace Transforms
.
Let us do two more exercises using the table of inverse transformations.
Given the Laplace transform F s Find the inverse transform
4
s s 4
2
f (t ) …………………….
Solution found
Help needed
.
s1
2
1
s2
s 4 5s 2 4
s
2
2
52
------------------- 54
------------------- 53
4
126
1 ˜ s 2 4
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Now the Laplace transform can be expressed as L >x @
s 4 4s 2 2
s s2 1 ˜ s2 4
Next, we decompose into partial fractions:
s 4 4s 2 2
s s2 1 ˜ s2 4
A B1 B2 s C1 C2 s
2
2
s
s 1
s 4
Express the right side as a single fraction with the original common
denominator
s 4 4s 2 2
s s2 1 ˜ s2 4
.................................................................
s s2 1 ˜ s2 4
------------------- 53
127
Chapter 11 Laplace Transforms
.
4
s s 4
We need to find the inverse transform of: F s 53
2
In the table you find that
given F s 1
s s Z 2 f t the inverse transform is
2
Constant factors stay unchanged. If you identify Z 2
4 and Z
1
Z2
1 cos Z t 2 , then you obtain:
f t ............................
------------------- .
s 4 4s 2 2
s s2 1 ˜ s2 4
A s 4 5s 2 4 B1 ˜ s s 2 4 B2 s 2 s 2 4 C1s s 2 1 C2 s 2 s 2 1
s s2 1 ˜ s2 4
54
127
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Clear the fractions. On the right side expand all products and order them according to powers of s.
s 4 4s 2 2 ………………………
------------------- 54
128
Chapter 11 Laplace Transforms
.
f t 4
1 cos 2t 1 cos 2t
4
54
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
One last task.
2
s 4s 3
Given F s 2
Then the inverse transform is f t =……………………..
Solution found
Help
.
s 4 4s 2 2
------------------- 57
------------------- 55
s 4 A B2 C2 s 3 B1 C1 s 2 5 A 4 B2 C2 s 4 B1 C1 4 A
128
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Since the coefficients of like powers of s must be equal, we obtain five equations for
determining A, B1 , B2 , C1 , and C2 .
Solving for each unknown, consecutively, we get:
A =……..
B1 =……
B2 =……
C1
. ......
Solution found
Help and explicit calculation
55
C2
…….
------------------- 130
------------------- 129
Chapter 11 Laplace Transforms
.
We must transform
2
F s 2
s 4s 3
55
In the table you can find
1
F s s a ˜ s b which leads to the inverse transform
f t >
1
e at ebt
ab
@
In order to factor the denominator let us find the roots of the quadratic expression:
s1
s2
..............
.................
Thus, the denominator can be written as a product:
s 2 4 s 3 (............................) ˜ (................................)
------------------- .
Compare coefficients for like powers of s
s 4 4s 2 2
56
129
s 4 A B2 C2 s 3 B1 C1 s 2 5 A 4 B2 C1 s 4 B1 C1 4 A
Coefficients of
Coefficients of
Coefficients of
Coefficients of
Coefficients of
s4 :
s3 :
s2 :
s1 :
s0 :
1
A B2 C2 0
B1 C1
4
5 A 4 B2 C 0
4 B1 C1
2
4A
Solve this set of linear equations, starting with A from the last equation.
A =……..
B1 =……
B2 =……
C1
. ......
C2
…….
------------------- 56
130
Chapter 11 Laplace Transforms
.
s1
3
s2
1
56
So we rewrite the quadratic expression as a product of linear factors:
s 2 4s 3
s 3 ˜ s 1
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Given
F s 1
s a ˜ s b >
1
e at ebt
ab
@
the inverse transform is
f t the inverse transform is
f t =…………….……………….
Given
F s 2
s 3 ˜ s 1
------------------- .
A
1
2
B1
0
B2
1
3
C1
0
C2
1
6
57
130
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Thus L >x @
s 4 4s 2 2
can be written as a sum of three fractions
s s 4 5s 2 4
L >x @ =………+……………+……………….
------------------- 57
131
Chapter 11 Laplace Transforms
.
f t >
2 3t
e et
2
@
57
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
More exercises and their solutions can be found in the textbook.
Please keep in mind: Easy exercises are fun. But the important thing is to tackle the hard ones.
------------------- .
L >x @
s
1 1 s
1
2s 3 s 2 1 6 s 2 4 58
131
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Step 3: In order to determine the inverse transform we use the table in the textbook on page 332
and 333 and obtain
xt …………………………..
------------------- 58
132
Chapter 11 Laplace Transforms
.
In the next section we will finally encounter applications. Certain algebraic transformations
of fractions, i.e. the decomposition into partial fractions, will prove especially useful.
58
This technique was already dealt with in the textbook, in a previous section on integration by partial
fractions.
If you do not feel familiar enough with it, go back to section 6.5.7 in the textbook pages 170–174 and
also to frames 117–138 in the accompanying chapter in the study guide.
After studying the textbook go to
Partial fractions well known
.
x(t )
------------------- 61
------------------- 59
1 1
1
cos t cos 2t
2 3
6
132
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
So the first part of example 2 is done. It still remains to determine L > y @ and y t .
Recall that the transformed system of equations was:
s
2
2 L >x @ -s L > y @
s L >x @ s 2 2 L > y @
1
s
s
1
In order to eliminate L >x @ , multiply the first equation by s and the second by s 2 2 and add
them. That leads to
L > y @ =………………………..
------------------- 59
133
Chapter 11 Laplace Transforms
.
Just to make sure, split the following expressions into partial fractions:
1.) Roots real and unequal:
5 x 11
f1 x 2
x 6x 8
59
f1 x …………………………
Roots real and repeated:
1
1
f 2 x = ……………………..
x 3x 2 4 ( x 1) ˜ ( x 2) 2
Roots real and complex
2 x 2 13 x 20
x( x 2 4 x 5)
f3 x ……………………………………
------------------- .
L >[email protected]˜ s 2 2 s 2
2
1
60
133
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
We expand the square and solve for L > y @ :
L > y @ =………………………..
------------------- 60
134
Chapter 11 Laplace Transforms
.
5 x 11
5 x 11
1
9
x 6 x 8 x 2 ˜ x 4 2x 2 2 x 4
1
1
1
f 2 x 9( x 1) 3( x 2) 9( x 2) 2
4
2x 3
f3 x 2
x x 4x 5
……………………………………………………………………….
f1 x 60
2
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
If you succeeded, you know the basic rules for finding partial fractions and you may proceed to the
next section.
61
------------------- If, however, you encountered difficulties, it makes sense to return to the textbook and study the
relevant pages 170–174, work through the relevant frames in the study guide and do some examples.
You will be happier knowing the basics well when entering subsequent sections.
Afterwards
.
L >[email protected]
s
4
61
------------------- 1
5s 2 4 134
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
The bracketed expression has already been factored in the preceding frames 125–127 resulting in
s 2 1˜ s 2 4. Next on the agenda is the decomposition into partial fractions:
1
s 1˜ s 2 4
A
B
2
s 1 s 4
1
s2 1 ˜ s2 4
A s2 4 B s2 1
s2 1 ˜ s2 4
2
2
s2 A B 4 A B
s2 1 ˜ s2 4
From this we deduce in the usual way
A B 0
4A B 1
A =……………
B =……………
------------------- 61
135
Chapter 11 Laplace Transforms
.
11.3 Solution of linear differential equations with constant coefficients
61
After having successfully gone through all preliminary steps, which might have been somewhat
depletive in places, you now encounter applications and you will reap the benefits. Follow all the
examples arduously and if necessary, consult the table on page 275.
You will learn to transform a set of linear differential equations into a set of algebraic equations and
after solving these to use the inverse transform, thus obtaining a solution of the original DEs.
.
READ
11.4 Solution of linear differential equations
with constant coefficients
Textbook pages 328–329
When done
.
A
1
3
------------------- B
1
3
62
135
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By inserting we obtain L > y @
1
1
2
3s 1 3s 4
2
Step 3: Find the inverse transform using the table on page 333 of the textbook:
y t ………………………….
------------------- 62
136
Chapter 11 Laplace Transforms
.
62
All examples meticulously done and understood
Hints and detailed calculations of the examples
.
y t ------------------- 86
------------------- 63
1
1
sin t sin 2t
3
6
136
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Just one more exercise. The solution should not be difficult for you after working through the
preceding examples.
The solution follows the now customary scheme consisting of three steps:
- Laplace Transformation
- Solve for L >x @ and L > y @ and rearrange into a convenient expression for applying the inverse
transform.
- Perform the inverse transformation
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137
Chapter 11 Laplace Transforms
.
One slight irritation may arise from the notation used in this section:
the function to be determined is denoted by y t and, consequently, its derivatives by y' t and y ' ' t . This is in parallel to the notation f t , f ' t , and f ' ' t that was used in some
previous sections.
Difficulties with the first example
Difficulties with the other example
.
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Given two simultaneous differential equations with constant coefficients:
4 x y x 1
4 x 4 y y 1
for t
0 : x0
y0
63
137
0
xt ………………………….
Solution found
Stepwise solution showing intermediate results
64
------------------- 142
------------------- 138
Chapter 11 Laplace Transforms
.
The first example deals with the differential equation y '4 y
The initial conditions are given as t = 0, y0 5
e 2t
64
In a first step we perform the Laplace transform of the differential equation and recall the rules of
transformation
L > y '@ s ˜ y s y0
L >[email protected]
> @
L e at
y s 1
sa
The transform of the given differential equation is:…………………………………..
Please, check your results of Laplace transforming each member of the given equation using the
tables on page 332 and 333.
------------------- .
Given
4 x y x 1
4 x 4 y y 1
for t
0 : x0
y0
65
138
0
Step 1: Determine the Laplace transforms and insert the initial conditions:
………………………………………….
………………………………………….
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139
Chapter 11 Laplace Transforms
.
s ˜ y s y0 4 y s 1
s2
initial condition
y0
65
5
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
As a second step we solve the equation for y s and insert y0
5.
y s =………………………………
------------------- .
4 s L >x @ s L > y @ L >x @
4 s L >x @ 4 s L > y @ L > y @
1
s
1
s
66
139
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Step 2: Solve for L >x @ :
L >x @ =…………………………
------------------- 66
140
Chapter 11 Laplace Transforms
.
y s 5
1
s 4 s 2 ˜ s 4
66
We are now in a position to perform the third step, finding the inverse transform. It is obtained by
consecutively transforming both expressions, yielding y1 and y2 . The complete inverse transform
y then is given by the sum y1 y2 .
Consult the table on page 332 and 333 in the textbook and identify the inverse transforms of
5
y1 ...................
y1 s s4
1
y2 ....................
y 2 s s 2 ˜ s 4
Solution found
Help needed
.
L >x @
------------------- 69
------------------- 67
3s 1
1· §
1·
§
s ˜ 12¨ s ¸ ˜ ¨ s ¸
6
2¹
©
¹ ©
140
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Decompose into partial fractions in order to obtain expressions which are easily inverse transformed:
L >x @ =………………………
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141
Chapter 11 Laplace Transforms
.
In the table on page 333 you see the inverse transforms for general classes of functions.
1
5
the table shows y s , its inverse transform being e at .
Rather than y1 s s4
sa
Since the constant factor 5 remains unchanged by the transform. If we identify a 4 ,
5
to be y1 =……………………….
we can easily find the inverse transform for
s4
------------------- .
L >x @
1 3
1
1
1
˜
˜
1· 4 §
1·
s 4 §
¨s ¸
¨s ¸
6
2¹
©
¹
©
67
68
141
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Step 3: The inverse transformation results in
xt ………………………………
------------------- 68
142
Chapter 11 Laplace Transforms
.
y1
5 ˜ e 4t
68
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It now remains to determine the inverse transform y2 for y2 s From the table we know that for y s Inserting a
y2
2 and b
4 we get:
1
s a ˜ s b 1
s 2 ˜ s 4
the inverse transform is y
1
e at ebt a b
....................................
------------------- .
t
t
3 1 xt 1 e 6 ˜ e 2
4
4
69
142
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Now we must determine L > y @ and y.
The Laplace transform has already been performed in frame 139 with the following result:
4 s L >x @ s L > y @ L >x @
4 s L >x @ 4 s L > y @ L > y @
1
s
1
s
Now compute
y t ………………………….
Solution happily found
Stepwise solution showing intermediate results
69
------------------- 147
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Chapter 11 Laplace Transforms
.
y2
>
1 2t
e e 4t
2
@
69
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We already know: y1
5 ˜ e 4t
Now combine both expressions to obtain the complete inverse transform:
y
y1 y2
.......................................
------------------- .
The following must be solved for L > y @
4 s L >x @ s L > y @ L >x @
4 s L >x @ 4 s L > y @ L > y @
70
143
1
s
1
s
By eliminating L >x @ in the usual fashion we obtain
L > y @ ……………………
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144
Chapter 11 Laplace Transforms
.
y
5 ˜ e 4t >
1 2t
e e 4t
2
@
1 2t 9 4t
e e
2
2
70
In the textbook the inverse transform is arrived at in a slightly different way.
Show me how the expression from the textbook is arrived at
Continue with the next exercise
.
L >[email protected]
------------------- 71
------------------- 74
1
1· §
1·
§
s ˜ 12¨ s ¸ ˜ ¨ s ¸
6
2¹
©
¹ ©
144
____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Decomposition into partial fractions produces expressions that are easily inversely transformed:
L >[email protected]
1
1· §
1·
§
s ˜ 12¨ s ¸ ˜ ¨ s ¸
6¹ ©
2¹
©
A
B
C
1· §
1·
12s §
¨s ¸ ¨s ¸
6¹ ©
2¹
©
Determine the unknown values
A =……………..
B =……………..
C =……………
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145
Chapter 11 Laplace Transforms
.
5
1
5s 11
. Use the common denominator to get y s s 2 ˜ s 4 s 4 s 2 ˜ s 4 In order to arrive at possibly simpler expressions let us decompose into partial fractions:
5s 11
A
B
s 2 ˜ s 4 s 2 s 4
Given y s 71
We must determine A and B. First express everything with the common denominator:
5s 11
As 4 A Bs 2 B
s 2 ˜ s 4 s 2 ˜ s 4
Clearing the fractions leads to: 5s 11 A B ˜ s 4 A 2 B
Determine A and B by comparing coefficients (equality holds for all different values of s).
5 A B
11 4 A 2 B
A ....................................
B .......................................
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If decomposing into partial fractions proved difficult for you, go back to the textbook again
to follow the solution of the exercise above.
.
A = –12
B=
3
2
C =
1
2
145
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Thus
L >[email protected]
1
=……………………
1· §
1·
§
s ˜ 12¨ s ¸ ˜ ¨ s ¸
6¹ ©
2¹
©
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146
Chapter 11 Laplace Transforms
.
1
2
A
B
9
2
72
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Thus, the Laplace transform becomes:
y s 5s 11
s 2 ˜ s 4
1
9
2s 2 2s 4
Again, we quickly obtain the inverse transformation using the table in the textbook on page 332
and 333.
y
......................................
------------------- .
1 3
1
1
1
L >[email protected] 1· 2 §
1·
s 2§
¨s ¸
¨s ¸
6
2¹
©
¹
©
73
146
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Now we can determine the inverse transform with a little help from the table, as nobody can keep all
transforms in memory. The result is:
y t ………………………..
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147
Chapter 11 Laplace Transforms
.
y
1 2t 9 4t
˜e e
2
2
73
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This coincides with the result previously obtained in this study guide.
There usually exist various different ways to rearrange the expression y (s ) or F s . Experience,
combined with some educated guessing, will guide you to finding expressions which are amenable to
inverse transformations. One suitable approach often is the decomposition into partial fractions,
which was dealt with in the textbook.
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Please continue on page 1
(bottom half)
.
y t 1 t
t
3 6 1 2
˜e e
2
2
147
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Thus, we have determined the complete solution of the system of equations as follows
t
t
·
1§ xt 1 ¨¨ e 6 e 2 ¸¸
2©
¹
y t 1 e
t
6
e
t
2
You have successfully completed this somewhat demanding chapter, and you may be proud of your
stamina!
of this chapter.
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