# 2nd order laplace de

1. Mar 13, 2009

### fredrick08

1. The problem statement, all variables and given/known data
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

3. 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, im lost, i cant 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 im not sure if can just take the inv laplace of that. please.

Last edited: Mar 13, 2009
2. Mar 13, 2009

### fredrick08

i tried doing parfrac, but im pretty sure u cant, unless u multiply the e^-3s back at the end of it... but i get a ridiculous answer... plz anyone have any ideaS?

3. Mar 14, 2009

### Staff: Mentor

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. Mar 14, 2009

### fredrick08

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 im not sure about... its something like

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

plz help, im not sure with these shifts... i think im on right track, but it dont look right

5. Mar 14, 2009

### fredrick08

any ideas?

6. Mar 14, 2009

### HallsofIvy

Staff Emeritus
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 $f(x-\alpha)$.

7. Mar 14, 2009

### fredrick08

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. Mar 14, 2009

### HallsofIvy

Staff Emeritus
Is "H3(t)" the same as "H(t-3)"?

9. Mar 14, 2009

### fredrick08

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

10. Mar 14, 2009

### fredrick08

so is that right?

11. Mar 14, 2009

### 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)+2e2(t-3)-6) how that look?

12. Mar 14, 2009

?....