# Inverse Laplace transform (Initial Value Problem)

• aero_zeppelin

## Homework Statement

I'm stuck trying to find out the inverse Laplace of F(s) to get y(t) (the solution for the differential equation):

Y(s) = 1 / [ (s-1)^2 + 1 ]^2

## The Attempt at a Solution

I tried using a translation theorem and then apply the sine formula, but the denominator is still all squared. I also tried partial fractions to expand Y(s) but I didn't get it right...

## Homework Statement

I'm stuck trying to find out the inverse Laplace of F(s) to get y(t) (the solution for the differential equation):

Y(s) = 1 / [ (s-1)^2 + 1 ]^2

## The Attempt at a Solution

I tried using a translation theorem and then apply the sine formula, but the denominator is still all squared. I also tried partial fractions to expand Y(s) but I didn't get it right...

Yes: show us what you did and what you got. How can we possibly help if we have no idea of what your problem is?

$[(s-1)+1]^2=[(s-1+i)(s-1-i)]^2$

Therefore, $$\left[\frac{1}{ (s-1-i)(s-1+i) }\right]^2=\left[\frac{1}{2i}\left(\frac{1}{s-1-i}-\frac{1}{s-1+i} \right ) \right ]^2$$

Can you continue it from there?

• 1 person
Yes: show us what you did and what you got. How can we possibly help if we have no idea of what your problem is?

My friend, that's the thing, I can't get past that. I stated the problem clearly: "I'm stuck trying to find out the INVERSE LAPLACE of the given Y(s) to get y(t)". There's no point in showing what I did before, it's not needed for what's after the equation I posted.

Anyway, I got it. It was just applying one of the formulas from the Laplace transform tables. I thought partial fractions or something else had to be done.