
Differential Equations Ordinary and Partial Differential Equations Math Forum 
 LinkBack  Thread Tools  Display Modes 
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 timeywimey 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: $\mathscr{L}$?

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 Joined: May 2013 From: The Astral plane Posts: 2,305 Thanks: 962 Math Focus: Wibbly wobbly timeywimey stuff.  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 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,305 Thanks: 962 Math Focus: Wibbly wobbly timeywimey stuff.  
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 timeywimey stuff.  Quote:
$\displaystyle \frac{1}{s(s^2 + 100)} =\frac{A}{s} + \frac{Bs + C}{s^2 + 100}$ 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  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Relatioship between Laplace and Inverse Laplace Transform from tables  szz  Differential Equations  1  November 2nd, 2014 03:18 AM 
The Laplace Transform  mhhojati  Calculus  5  October 1st, 2014 05:18 AM 
Laplace Transform  wastedhippo  Calculus  3  September 29th, 2014 03:42 PM 
Laplace tranform and inverse of Laplace transform  Deiota  Calculus  1  April 28th, 2013 10:28 AM 
what is laplace transform of each of the following  rsoy  Calculus  2  March 24th, 2013 04:30 PM 