Solving 2nd Order Laplace DE: f(t)

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    2nd order Laplace
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Homework Help Overview

The problem involves solving a second-order differential equation using the Laplace transform, specifically f"(t) - f'(t) - 2f(t) = 12H0(t-3) with initial conditions f(0) = f'(0) = 0. Participants are discussing the application of the Laplace transform and the process of finding the inverse transform.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to manipulate the Laplace transform of the given differential equation and are exploring the use of partial fractions. There are questions about the validity of certain steps, particularly regarding the inverse Laplace transform and the handling of exponential shifts.

Discussion Status

The discussion is ongoing, with participants sharing their attempts at manipulating the equation and expressing uncertainty about the correctness of their approaches. Some guidance has been offered regarding the use of partial fractions and the interpretation of shifted functions, but no consensus has been reached on the final form of the solution.

Contextual Notes

Participants are grappling with the implications of the Heaviside function and its shifts, as well as the proper application of the inverse Laplace transform in the context of the problem's constraints.

fredrick08
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Homework Statement


f"(t)-f'(t)-2f(t)=12H0(t-3), f(0)=f'(0)=0

relevant equations, are the laplace transform equations..
H0(t-a)=e-as/s

The Attempt at a Solution


LT:s2F(s)-sF(s)-2F(s)=12e-3s/s
=>F(s)=12e-3s/s(s+1)(s-2)

ok now from here, I am lost, i can't do partial fractions can I? and i need to solve for f(t) which is inverse laplace of F(s), please can someone help me out, because I am not sure if can just take the inv laplace of that. please.
 
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i tried doing parfrac, but I am pretty sure u cant, unless u multiply the e^-3s back at the end of it... but i get a ridiculous answer... please anyone have any ideaS?
 
F(s)=12e-3s/s(s+1)(s-2) = 12e-3s * 1/[s(s+1)(s-2)]
Use partial fractions to find a different form for 1/[s(s+1)(s-2)]. Then you'll have F(s) as the sum of three expressions, each with a factor of 12e-3s. Then do your inverse Laplace transforms to get f(t).
 
ok so then F(s)=12e-3s((-6/s)+(4/(s+1))+(2/(s-2)))
=invlaplace(12e-3s)(-6H0(t)+4e-t+2e2t)

then this i think its s-shift or t-shift thing I am not sure about... its something like

H3(t)(-6H0(t-3)+4e-(t-3)+2e2(t-3))

please help, I am not sure with these shifts... i think I am on right track, but it don't look right
 
any ideas?
 
fredrick08 said:
ok so then F(s)=12e-3s((-6/s)+(4/(s+1))+(2/(s-2)))
=invlaplace(12e-3s)(-6H0(t)+4e-t+2e2t)
You can't just take the Laplace transform on one factor! Multiply each term by 12e-3s:
[tex]-72\frac{e^{-3s}}{s}+ 48\frac{e^{-3s}}{s+1}+ 24\frac{e^{-3s}}{s-2}[/tex]
and use the fact that if F(s) is the Laplace transform of f(x), then [itex]e^{\alpha s}F(s)[/itex] is the Laplace transform of [itex]f(x-\alpha)[/itex].

then this i think its s-shift or t-shift thing I am not sure about... its something like

H3(t)(-6H0(t-3)+4e-(t-3)+2e2(t-3))

please help, I am not sure with these shifts... i think I am on right track, but it don't look right
 
ok then well how does
f(t)=-72H3(t)+48H3(t)e-(t-3)+24H3(t)e2(t-3)
=H3(t)(24e2(t-3)+48e-(t-3)-72) sound?
 
Is "H3(t)" the same as "H(t-3)"?
 
yes in my book, it says Ha(t)=H0(t-a)
 
  • #10
so is that right?
 
  • #11
no its not right because i did parfrac on 12=... lol so if the method is ok the correct answer should be.

H3(t)(4e-(t-3)+2e2(t-3)-6) how that look?
 
  • #12
?...
 

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