Find the inverse Laplace transform of F(s)=3s/(s^2-s-6).

Question

anonymous · Answer

complete the square:

$$F(s) = \frac{ 3s }{ (s - 0.5)^2 - 6.25 }$$

this now looks like cos.

anonymous · Answer

cosh*

anonymous · Answer

Is this the only way to do this?

anonymous · Answer

yes. there is an other way which will require partial fractions:

$$F(s) = \frac{ 3s }{ (s-3)(s+2) }$$

anonymous · Answer

That's what I did. What's next?

anonymous · Answer

$$\frac{ 3s }{ (s-3)(s+2)} = \frac{ A }{ s-3 }+\frac{ B }{ s+2 }$$

$$As + 2A + Bs -3B = 3s$$

solve for A and B. and plug them in.

The inverse is now two e^(at) terms

anonymous · Answer

What to do after that? How do you solve for A and B?

anonymous · Answer

Since adding fractions require equal denominators, we cross multiplied to get

As+2A+Bs−3B=3s 

which is (A+B)s + (2A-3B) = 3s + 0

A + B = 3
2A - 3B = 0

anonymous · Answer

I got it, thanks.

anonymous · Answer

A = 9/5
B = 6/5

$$F(s) = \frac{ 9/5 }{ s-3 } + \frac{ 6/5 }{ s+2 }$$

$$f(t) = 9/5 e^{3t} + 6/5 e^{-2t}$$

anonymous · Answer

glad i could help :)