# Homework Help: Inverse Laplace Transform

1. Nov 19, 2015

### Mark Brewer

1. The problem statement, all variables and given/known data
L-1{(2s2+3)/(s2+3s-4)2}

3. 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 eat

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

Any help would be appreciated!!!!!!!!!!

2. Nov 19, 2015

### axmls

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

3. Nov 19, 2015

### Mark Brewer

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

4. Nov 19, 2015

### Mark Brewer

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

5. Nov 19, 2015

### 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.)?

6. Nov 19, 2015

### Mark Brewer

A polynomial would then be formed, right?

7. Nov 19, 2015

### axmls

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.

8. Nov 19, 2015

### Mark Brewer

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