Solving 2nd Order Laplace DE: f(t)

In summary, the homework statement is f(t)-f'(t)-2f(t)=12H0(t-3), f(0)=f'(0)=0. The relevant equations are the laplace transform equations, and using partial fractions, the form for 1/[s(s+1)(s-2)] is found. Next, the inverse Laplace transforms are used to find f(t). Finally, the shift method is used to find H3(t).
  • #1
fredrick08
376
0

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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  • #2
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?
 
  • #3
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).
 
  • #4
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
 
  • #5
any ideas?
 
  • #6
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
 
  • #7
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?
 
  • #8
Is "H3(t)" the same as "H(t-3)"?
 
  • #9
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
?...
 

1. What is a second order Laplace differential equation?

A second order Laplace differential equation is a mathematical equation that involves a function and its derivatives with respect to the Laplace transform variable, usually denoted as s. It is a powerful tool in solving problems in engineering, physics, and other scientific fields.

2. How do you solve a second order Laplace differential equation?

The first step in solving a second order Laplace differential equation is to take the Laplace transform of both sides of the equation. This will convert the differential equation into an algebraic equation, which can then be solved for the function in terms of s. The inverse Laplace transform is then applied to get the solution in terms of the original variable, usually denoted as t.

3. What is the importance of solving second order Laplace differential equations?

Solving second order Laplace differential equations is important because it allows us to model and solve complex physical systems and phenomena. It is also a useful tool in control theory, where it is used to analyze and design systems such as electrical circuits and mechanical systems.

4. What are some common techniques used in solving second order Laplace differential equations?

There are several techniques that can be used to solve second order Laplace differential equations, such as the method of variation of parameters, the method of undetermined coefficients, and the method of Laplace transforms. Each technique has its own advantages and is useful in different situations.

5. Are there any limitations to using Laplace transforms to solve second order differential equations?

While Laplace transforms are a powerful tool, they do have some limitations. One limitation is that the initial conditions of the differential equation must be known in order to apply the Laplace transform. Additionally, some differential equations may not have a closed-form solution, making it difficult to solve using Laplace transforms.

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