Laplace Help

JeremyK

New member
Joined
Mar 15, 2006
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2
I am having trouble with a couple of laplace transforms and I was hoping someone could get me past where I am stuck.

#1.
y'' + 4y' + 6y = 1 + e^(-t)
y(0) = 0
y'(0) = 0

Firth take the L of each item which winds up like this

s^2Y(s) + 4sY(s) + 6Y(s) = 1/s + 1/(s+1)

solve for y(s)

y(s) = (2s+1)/(s(s+1)(s^2+4s+6))
which simplifies to
(2s+1)/(s(s+1)(s^2+4s+6))

Then do partial fraction decomposition

A/s + B/(s+1) + (Cs+D)/(s^2+4s+6) = (2s+1)/(s(s+1)(s^2+4s+6))

I am not sure if I am setting that up right. I have worked that out and gotten my values of A=1/6 B= 2/3 C=-1/2 D=-3. However when I try to check my numbers by plugging those A through D values and say for instance s=3 back in to the left side of the above equation it doesn't work for s = 3 with the right side of the equation. This should check right? I can't seem to get past this partial fraction decomposition. Any help would be appreciated.

#2
Find the inverse laplace of the following equation

((s/2)+(5/3))/(s^2 + 4s + 6)

This problem is similar to #1 in that is has s^2 + 4s + 6. I can not figure out how to factor this guy. I have tried manipulating the problem but nothing I have done has helped in any apparent way. This is where I left off.

Find a common denomonator for the top which turns to (3s+10)/6 all over s^2 + 4s + 6. Then simplify to (3s+10)/(6s^2+24s+36). This does not seem to do anything.

Thanks for taking the time to read all that :)
Jeremy
 
#1) I hate to see perfect work like that go to waste on the algebra. Try D = -5/3.

#2) It's not just similar. It IS #1, the last piece, anyway. Factor? Why not Complete the Square?
 
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