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 November 25th, 2015, 12:24 PM #1 Newbie   Joined: Nov 2015 From: Ireland Posts: 6 Thanks: 0 Laplace transform Trying to find the inverse of: y'' + 100y = 1 y(0)=1 y'(0)=0
 November 25th, 2015, 05:19 PM #2 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 Why is the title "Laplace transform"?
 November 25th, 2015, 07:01 PM #3 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,305 Thanks: 962 Math Focus: Wibbly wobbly timey-wimey stuff. 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.
 November 25th, 2015, 10:05 PM #4 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 Wouldn't this do: $\mathscr{L}$? Thanks from topsquark
 November 25th, 2015, 11:10 PM #5 Newbie   Joined: Nov 2015 From: Ireland Posts: 6 Thanks: 0 Sorry wrote one part wrong y (0)=0 Will this change it?
 November 26th, 2015, 12:37 AM #6 Newbie   Joined: Nov 2015 From: Ireland Posts: 6 Thanks: 0 Can't do reverse transformation, that's where I'm getting stuck. Last edited by skipjack; November 26th, 2015 at 03:06 PM.
November 26th, 2015, 12:24 PM   #7
Math Team

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Quote:
 Originally Posted by Dulla 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

Last edited by skipjack; November 26th, 2015 at 03:07 PM.

November 26th, 2015, 12:25 PM   #8
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Quote:
 Originally Posted by skipjack Wouldn't this do: $\mathscr{L}$?
It'll do. Thanks!

-Dan

 November 26th, 2015, 12:56 PM #9 Newbie   Joined: Nov 2015 From: Ireland Posts: 6 Thanks: 0 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.
November 26th, 2015, 03:01 PM   #10
Math Team

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Quote:
 Originally Posted by Dulla 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

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

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