TUGAS PERSAMAAN DIFERENSIAL

OLEH:

Kelompok III

BAIQ ZILALIN AZZIMA E1R 012 006

FEBRI ARIANTI E1R 012 011

FITRIA WINDIARNI E1R 012 014

PROGRAM STUDI PENDIDIKAN MATEMATIKA

FAKULTAS KEGURUAN DAN ILMU PENDIDIKAN

UNIVERSITAS MATARAM

2014

7. 3 OPERATIONAL PROPERTIES I

Introduction: it is not convenient to use Definition 7.1 each time we wish to

find the Laplace transform of a function . For example, the integration by parts

involved in evaluating, say L is formidable to say the least. In this

section and the next we present several labor-saving operational properties of the

Laplace transform that enable us to build up a more extensive list of transforms

without having to resort to the basic definition and integration.

7.3.1 TRANSLATION ON THE S-AXIS

A TRANSLATION Evaluating transforms such as L and L

is straightforward provided that we know (and we do) L and L .

In general, if we know the Laplace transform of a function. L , it

is possible to compute the Laplace transform of an exponential multiple of

that is, L with no additional effort other than translating, or

shifting, the transform to . This result is known as the first

translation theorem or first shifting theorem.

PROOF The proof is immediate, since by Definition 7.1

L

If we consider a real variable, then the graph of is the graph of

shifted on the -axis by the amount units to the left. See Figure 7.10.

FIGURE 7.10 Shift on s-axis

First Translation Theorem

If L and a is any real number, then L

Theorem 7.6

For emphasis it is sometimes useful to use the symbolism

L L

Where means that in the Laplace transform of we replace

the symbol wherever it appears by .

Example 1 Using The First Translation Theorem

Evaluate L L .

Solution The Result Follow From Theorem 7.1 And 7.6

L L

L L

INVERSE FORM OF THEOREM 7.6 To compute the inverse of , we

must recognize , find by taking the inverse Laplace transform of ,

and then multiply by the exponential function . This procedure can be

summarized symbolically in the following manner:

L -1

L -1

Where L -1

.

The first part of the next example illustrates partial fraction

decomposition in the case when the denominator of contains repeated

linear factors.

Example 2 Partial Fraction: Repeated Linear Factors

Evaluate L -1

,

- L

-1 {

}

Solution A repeated linear factor is a term , where is a real

number and is a positive integer . Recall that if appears in the

denominator of rational expression, then the assumed decomposition

contains partial fractions with constant numerators and denominators

Hence with and we write

By putting the two terms on the right-hand side over a common denominator,

we obtain the numerator , and this identity yields

and . Therefore

and

L -1 ,

- L -1 ,

- L -1,

-

Now

is

shifted three units to the right. Since L -1 ,

- , it

follows from (1) that

L -1,

- L -1,

- .

Finally, is L -1 ,

-

To start, observe that the quadratic polynomial has no real

zeros and so has no real linear factors. In this situation we complete the

square:

Our goal here is to recognize the expression on the right-hand side as some

Laplace transform in wich has been replaced throughout by . What

we are trying to do is analogous to working part of Example 1 backwards.

The denominator in is already in the correct form-that is, with

replaced by . However, we must fix up the numerator by manipulating

the constants:

.

Now by termwise division, the linearity of L -1, parts and of

Theorem 7.4, and finally ,

L -1{

}

L -1,

-

L -1,

-

L -1,

-

L -1,

-

Example 3 An Initial-Value Problem

Solve

Solution Before transforming the DE, note that its right-hand side is similar to

the function in part of Example 1. After using linearity, Theorem 7.6,and

the initial conditions, we simply and then solve for L

L L L L

[ ]

The first term on the right-hand side was already decomposed into individual

partial fraction in in part of Example 2:

Thus L -1 ,

- L -1,

-

L -1,

-

From the inverse form of Theorem 7.6, the last two terms in are

L -1,

- and L -1,

-

.

Thus is

Example 4 . An Initial Value Problem

Solve

Solution L L L L L

[ ]

Since the quadratic term in the denominator does not factor into real linear

factors, the partial fraction decomposition for is found to be

Moreover, in preparation for taking the inverse transform, we already

manipulated the last term into the necessary form in part of example 2. So

in view of the results in and we have the solution

L

-1 ,

-

L

-1

L

-1

L

-1

7.3.2 TRANSLATION ON THE T-AXIS

UNIT STEP FUNCTION In engineering one frequently encounters functions that

are either off or on. For example, an external force acting on a mechanical

system or a voltage impresseed on a circuit can be turned off after a period of

time. It is convenient, then to define a special function that is the number 0

(off) up to a certain time and then the number 1(on) after that time. This

function is called the unit step function or the Heavside function.

Definition 7.3 Unit Step Function

The unit step function U is defined to be U , <

Notice that we define U only on the nonnegative t-axis, since this is all

that we are concerned with in the study of the Laplace transform. In a broader

sense U for < The graph of U is given in Figure 7.11

FIGURE 7.11 Graph of unit step function

When a function defined for is multiplied by U the unit

step function turns off a portion of the graph of that function. For example,

consider the function . To turns off a portion of the graph of

on, say, the interval < we simply form the product U

See figure 7.12. In general, the graph of U is (off) for <

and is the portion of the graph of (on) for

FIGURE 7.12 Function is U

The unit step function can also be used to write piecewise-defined

function in a compact form. For example, if we consider the intervals