1. ## Inverse Laplace transform

Hi, i am currently studying laplace transforms and i am arriving at the wrong answer.
I have attached my working out aswell as the question.

I have a feeling that i am getting the partial fractions wrong, although i'm not sure how.
The correct answer in the textbook is:
(t/2) - (1/4) + (1/4)(e^(-2t))
please let me know if you need anything clarified as i know i do not write that neatly.

2. ## Re: Inverse Laplace transform

you have to break up the denominator into all it's factors

$(s^2 - 4)Y(s) = \dfrac{s-2}{s^2}$

$s^2-4=(s+2)(s-2)$

$Y(s) = \dfrac{s-2}{s^2(s+2)(s-2)} = \dfrac{1}{s^2 (s+2)}$

$\dfrac{1}{s^2 (s+2)} = \dfrac{A}{s} + \dfrac{B}{s^2} + \dfrac{C}{s+2}$

$1 = As(s+2) + B(s+2) + Cs^2 = s^2(A + C) + s(2A + B) + 2B$

$0=A+C$

$0 = 2A+B$

$1 = 2B$

$B=\dfrac 1 2$

$0=2A + \dfrac 1 2$

$A = -\dfrac 1 4$

$0=-\dfrac 1 4 + C$

$C=\dfrac 1 4$

$Y(s) = \dfrac{1}{2s^2} -\dfrac{1}{4s}+\dfrac{1}{4(2+s)}$

$y(t) = \dfrac{t}{2}-\dfrac{1}{4}+\dfrac{e^{-2 t}}{4}$

3. ## Re: Inverse Laplace transform

Thanks heaps, i did not think of adding 1/s - 2/s^2 into one fraction. as well as breaking up the denominator further.
will definitely keep this in mind for future problems.