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### Laplace Transforms

```
Date: 06/21/97 at 01:35:22
From: Kenneth
Subject: Laplace Transform

Hi!

I have a little trouble understanding the linearity of the Laplace
transform.  Defined as "the Laplace transform is a linear operation;
that is, for any functions f(t) and g(t) whose Laplace transforms
exist and any constants a and b."  I do not understand the meaning.
Could you rephrase this so it's easier to understand? What's the point
of this theorem?

If f(t) = tsinwt, L(f) = ?

Solution:

f(0) = 0
f'(t) = sin wt + wt cos wt, f'(0)=0

What's next?

Thank you very much for your time and help,
Kenneth
```

```
Date: 06/21/97 at 11:21:33
From: Doctor Anthony
Subject: Re: Laplace Transform

A linear operator is similar to, say, differentiation or integration;
that is:

d/dx[f(x) + g(x)] = f'(x) + g'(x) and d/dx(a*f(x) = a*f'(x)

A non-linear operator would be squaring, i.e., (a+b)^2 is NOT equal
to a^2 + b^2.

To find the Laplace transform of xsin(wx) we evaluate:

L[xsin(wx)] = INT(0 to inf.)[e^(-px)xsin(wx)dx]

2wp
= ------------
(p^2 + w^2)^2

To solve f'(x) = sin(wx) + wxcos(wx), we write down the transforms
term by term, to get:

-y(0) + p*y(p) = w/(p^2+w^2) + w[(p^2-w^2)/(p^2+w^2)^2  and y(0) = 0

w            w(p^2-w^2)
y(p) =  ---------  +  --------------
p(p^2+w^2)     p(p^2+w^2)^2

At this stage we use partial fractions to separate the terms in the
denominators.  I will work though the first one:

1               A         Bp + C
--------    =    ------  +  --------
p(p^2+w^2)          p        p^2 + w^2

1     =   A(p^2 + w^2) + Bp^2 + Cp

p = 0 gives 1 = Aw^2, so A = 1/w^2. Also A+B = 0, so B = -1/w^2 and
C = 0 because there is no term in p on the lefthand side.

1                1                 1
-----------   =  ---------   -   -------------
p(p^2+w^2)        w^2*p           w^2(p^2+w^2)

and
w              1                w
---------    =   -------   -   -------------
p(p^2+w^2)        w*p            w^2(p^2+w^2)

We now use the inverse transforms to write down the result:

= 1/w  - (1/w^2)sin(wx)

The second fraction is:

wp^2 - w^3         wp               w^3
------------- =  ----------   -  -----------
p(p^2+w^2)^2     (p^2+w^2)^2      p(p^2+w^2)^2

The first of these is (1/2)x*sin(wx) and the second one requires a
further application of partial fractions.

1              A         Bp + C         Dp + E
--------   =    ------   + --------    + ----------
p(p^2+w^2)^2       p        (p^2+w^2)     (p^2+w^2)^2

Some rather tedious algebra and a table of inverse transforms will
allow you to complete the solution of the differential equation. The
use of these transforms depends very much on not having 'p' functions
with unknown inverses. It is a good idea to collect as many inverses
as you can find from a variety of textbooks.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus

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