- Thread starter
- #1

I am trying to separate out

\[

\frac{s}{(s+1)^3}

\]

for an inverse Laplace transform.

How does one setup up partial fractions for a cubic? I know for a square I would do

\[

\frac{A}{s+1} + \frac{Bs+C}{(s+1)^2}

\]

I tried doing

\[

\frac{A+Bs}{(s+1)^2} + \frac{Cs^2+Ds+E}{(s+1)^3}

\]

which led to

\[

s^2(B+C) + s(A+B+D) + A + E = s

\]

Therefore, let \(A = B = 1\). Then \(C = E = -1\) and \(D = -1\).

\[

\frac{1}{s+1} - \frac{s^2+s+1}{(s+1)^2} = \frac{2}{s+1} + \frac{1}{(s+1)^3} - \frac{1}{(s+1)^2}

\]

but the answer is

\[

\frac{1}{(s+1)^2} - \frac{1}{(s+1)^3}

\]

How can I solve this?

[HR][/HR]

I now tried

\[

\frac{A}{s+1} + \frac{Bs+C}{(s+1)^2} + \frac{Ds^2 + Es + F}{(s+1)^2}

\]

which led to

\begin{align}

A + B + D &=0\\

2A + B + C +E &= 1\\

A + C + F &= 0

\end{align}

\[

\frac{s}{(s+1)^3}

\]

for an inverse Laplace transform.

How does one setup up partial fractions for a cubic? I know for a square I would do

\[

\frac{A}{s+1} + \frac{Bs+C}{(s+1)^2}

\]

I tried doing

\[

\frac{A+Bs}{(s+1)^2} + \frac{Cs^2+Ds+E}{(s+1)^3}

\]

which led to

\[

s^2(B+C) + s(A+B+D) + A + E = s

\]

Therefore, let \(A = B = 1\). Then \(C = E = -1\) and \(D = -1\).

\[

\frac{1}{s+1} - \frac{s^2+s+1}{(s+1)^2} = \frac{2}{s+1} + \frac{1}{(s+1)^3} - \frac{1}{(s+1)^2}

\]

but the answer is

\[

\frac{1}{(s+1)^2} - \frac{1}{(s+1)^3}

\]

How can I solve this?

[HR][/HR]

I now tried

\[

\frac{A}{s+1} + \frac{Bs+C}{(s+1)^2} + \frac{Ds^2 + Es + F}{(s+1)^2}

\]

which led to

\begin{align}

A + B + D &=0\\

2A + B + C +E &= 1\\

A + C + F &= 0

\end{align}

Last edited: