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November 25th, 2015, 12:24 PM   #1
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Laplace transform

Trying to find the inverse of:
y'' + 100y = 1
y(0)=1
y'(0)=0
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November 25th, 2015, 05:19 PM   #2
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Why is the title "Laplace transform"?
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November 25th, 2015, 07:01 PM   #3
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And what do you mean by "inverse?"

I'm assuming that you need to solve the differential equation by Laplace transform?

If I did this correctly
$\displaystyle y''(t) + 100 y(t) = 1$; y(0) = 1; y'(0) = 0

Well, call L(y(t)) = Y(s). Then

$\displaystyle L(y''(t)) = s^2 Y(s) - s y(0) - y'(0)$

and
$\displaystyle L(1) = \frac{1}{s}$

So the Laplace transformation of the equation is
$\displaystyle s^2 Y(s) -s y(0) - y'(0) + 100 Y(s) = \frac{1}{s}$

or
$\displaystyle Y(s) = \frac{s^2 +1}{s(s^2 + 100)}$

Can you do the reverse transformation?

-Dan

PS: What is the LaTeX code for "L"?

Last edited by skipjack; November 25th, 2015 at 10:00 PM.
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November 25th, 2015, 10:05 PM   #4
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Wouldn't this do: $\mathscr{L}$?
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November 25th, 2015, 11:10 PM   #5
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Sorry wrote one part wrong
y (0)=0
Will this change it?
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November 26th, 2015, 12:37 AM   #6
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Question

Can't do reverse transformation, that's where I'm getting stuck.

Last edited by skipjack; November 26th, 2015 at 03:06 PM.
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November 26th, 2015, 12:24 PM   #7
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Quote:
Originally Posted by Dulla View Post
Can't do reverse transformation, that's where I'm getting stuck.
With your correction the Laplace transform gives
$\displaystyle Y(s) = \frac{1}{s(s^2 + 100)}$

This is much easier to do the inverse Laplace transform than your earlier problem. Put this into partial fractions:
$\displaystyle Y(s) = \frac{1}{s(s^2 + 100)} =\frac{A}{s} + \frac{Bs + C}{s^2 + 100}$

This gives A = 1/100, B = -1/100, C = 0. So
$\displaystyle Y(s) = \frac{1}{s(s^2 + 100)} = \frac{1}{100} ~\frac{1}{s} - \frac{1}{100}~\frac{s}{s^2 + 10^2}$

And the inverse Laplace transform is thus
$\displaystyle y(t) = \frac{1}{100} - \frac{1}{100} ~ \cos(10t)$

-Dan
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Last edited by skipjack; November 26th, 2015 at 03:07 PM.
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November 26th, 2015, 12:25 PM   #8
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Quote:
Originally Posted by skipjack View Post
Wouldn't this do: $\mathscr{L}$?
It'll do. Thanks!

-Dan
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November 26th, 2015, 12:56 PM   #9
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How do you get b = -1?
The way I was shown to do it was to let s be a number to cancel a and c out when looking for b, but for this I can't see how to do that.

Last edited by skipjack; November 26th, 2015 at 03:08 PM.
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November 26th, 2015, 03:01 PM   #10
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Quote:
Originally Posted by Dulla View Post
How do you get b = -1?
The way I was shown to do it was to let s be a number to cancel a and c out when looking for b, but for this I can't see how to do that.
There are other methods, but I like this one the best:
$\displaystyle \frac{1}{s(s^2 + 100)} =\frac{A}{s} + \frac{Bs + C}{s^2 + 100}$

Add the fractions on the RHS:
$\displaystyle \frac{1}{s(s^2 + 100)} =\frac{A(s^2 + 100)}{s(s^2 + 100)} + \frac{s(Bs + C)}{s(s^2 + 100)}$

$\displaystyle \frac{1}{s(s^2 + 100)} =\frac{A(s^2 + 100) + s(Bs + C)}{s(s^2 + 100)}$

$\displaystyle \frac{1}{s(s^2 + 100)} =\frac{(A + B)s^2 + Cs + 100A}{s(s^2 + 100)}$

Now compare the coefficients of like terms on both sides.
A + B = 0
C = 0
100A = 1

etc.

-Dan
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Last edited by skipjack; November 26th, 2015 at 03:09 PM.
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