Solving 2nd Order Laplace DE: f(t)

[fredrick08]

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.

 

[fredrick08]

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?

 

[Mark44]

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).

 

[fredrick08]

ok so then F(s)=12e^-3s((-6/s)+(4/(s+1))+(2/(s-2)))
=invlaplace(12e^-3s)(-6H₀(t)+4e^-t+2e^2t)

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)+2e^2(t-3))

please help, I am not sure with these shifts... i think I am on right track, but it don't look right

 

[fredrick08]

any ideas?

 

[HallsofIvy]

ok so then F(s)=12e^-3s((-6/s)+(4/(s+1))+(2/(s-2)))
=invlaplace(12e^-3s)(-6H₀(t)+4e^-t+2e^2t)

You can't just take the Laplace transform on one factor! Multiply each term by 12e^-3s:
-72\frac{e^{-3s}}{s}+ 48\frac{e^{-3s}}{s+1}+ 24\frac{e^{-3s}}{s-2}
and use the fact that if F(s) is the Laplace transform of f(x), then e^{\alpha s}F(s) is the Laplace transform of f(x-\alpha).

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)+2e^2(t-3))

please help, I am not sure with these shifts... i think I am on right track, but it don't look right

 

[fredrick08]

ok then well how does
f(t)=-72H₃(t)+48H₃(t)e^-(t-3)+24H₃(t)e^2(t-3)
=H₃(t)(24e^2(t-3)+48e^-(t-3)-72) sound?

 

[HallsofIvy]

Is "H₃(t)" the same as "H(t-3)"?

 

[fredrick08]

yes in my book, it says H_a(t)=H₀(t-a)

 

[fredrick08]

so is that right?

 

[fredrick08]

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)+2e^2(t-3)-6) how that look?

 

[fredrick08]

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