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Laplace Transforms

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


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) = ?


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,

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]

             = ------------
               (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)

         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!   
Associated Topics:
College Calculus

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