How can partial fractions be used to solve inverse Laplace transforms?

[Mark Brewer]

Homework Statement

L^-1{(2s²+3)/(s²+3s-4)²}

The Attempt at a Solution

I factored the denominator

f(t)=(2s²+3)/((s-1)(s+4))²

now I've tried partial fractions to get

(2s²+3)/((s-1)(s+4))² = A/(s-1)² + B(s+4)²

(2s²+3)=A(s+4)² + B(s-1)²

by substitution, s=1 and s=-4

5=A(25)
A=1/5

35=B(25)
B=7/5

(1/5) 1/(s-1)² + (7/5) 1/(s+4)²

At this point I'm not sure if I am on the right track, but I did start to see some identities that may help.

1/5 L^-1{1/(s-1)²} +7/5 L^-1{1/(s+4)²}

I'm starting to see a pattern for n!\sⁿ⁺¹ and e^at

Am I on the right track, or did I go on a tangent?

Any help would be appreciated!

 

[axmls]

You need to take another look at your partial fraction expansion. How do you expand when you have double roots in the denominator?

 

[Mark Brewer]

I'm not sure if splitting the roots, or can I?

 

[Mark Brewer]

sorry, I'm not sure if I can split the roots to two rationals.

 

[axmls]

Note, when I say "double roots", I mean, how do you expand into separate terms when, for instance, one of the zeroes of the denominator is squared (or cubed, etc.)?

 

[Mark Brewer]

A polynomial would then be formed, right?

 

[axmls]

A polynomial would then be formed, right?

You may want to look back at your old notes regarding how to expand it using partial fractions. You're missing a couple of terms in your expansion.

 

[Mark Brewer]

Okay. Thank you, I'll reply as soon as I see my mistakes.