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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

Joined: May 2013
From: The Astral plane

Posts: 2,305
Thanks: 962

Math Focus: Wibbly wobbly timey-wimey stuff.
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. Tags laplace, transform Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post szz Differential Equations 1 November 2nd, 2014 03:18 AM mhhojati Calculus 5 October 1st, 2014 05:18 AM wastedhippo Calculus 3 September 29th, 2014 03:42 PM Deiota Calculus 1 April 28th, 2013 10:28 AM r-soy Calculus 2 March 24th, 2013 04:30 PM

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