Partial Fractions for Cubic: Setup & Solve

[Dustinsfl]

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?

---

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}

 

[polygamma]

You have too many unknowns on the right side of the equation.Assume $ \displaystyle \frac{s}{(s+1)^{3}} = \frac{A}{s+1} + \frac{B}{(s+1)^{2}} + \frac{C}{(s+1)^{3}}$Multiply both sides by $(s+1)^{3}$ and let $s=-1$.

Then $-1 = C$.Now subtract the partial fraction from both sides.

$ \displaystyle \frac{s}{(s+1)^{3}} + \frac{1}{(s+1)^{3}}= \frac{1}{(s+1)^{2}} = \frac{A}{s+1} + \frac{B}{(s+1)^{2}}$.Multiply both sides by $(s+1)^{2}$ and let $s=-1$.

Then $1 = B$.Subtract the partial fraction from both sides.

$\displaystyle \frac{1}{(s+1)^{2}} - \frac{1}{(s+1)^{2}} = 0 = \frac{A}{s+1}$.

Then $A=0$.

 

[Chris L T521]

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?

---

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}

Let $\dfrac{P(x)}{Q(x)}$ be a rational function, Whenever you have a repeated linear factor $(a_1x+b_1)^n$ in your denominator $Q(x)$, you need to use the following decomposition rule:
\[\frac{P(x)}{Q(x)} = \frac{A_1}{a_1x+b_1}+\frac{A_2}{(a_1x+b_1)^2}+ \ldots+\frac{A_n}{(a_1x+b_1)^n}.\]
In your case, the decomposition you want will be of the form
\[\frac{s}{(s+1)^3} = \frac{A}{s+1}+\frac{B}{(s+1)^2} + \frac{C}{(s+1)^3}\]
Can you take things from here?

Edit: Ninja'd by Random Variable.

 

[Dustinsfl]

I have one question. Why, when we have a square, do we do:
\[
\frac{A}{s + 1} + \frac{Bs+C}{(s + 1)^2}
\]
but for the cubic, we had to do:
\[
\frac{A}{s+1} + \frac{B}{(s+1)^2} + \frac{C}{(s+1)^3}
\]

 

[MarkFL]

I have one question. Why, when we have a square, do we do:
\[
\frac{A}{s + 1} + \frac{Bs+C}{(s + 1)^2}
\]
but for the cubic, we had to do:
\[
\frac{A}{s+1} + \frac{B}{(s+1)^2} + \frac{C}{(s+1)^3}
\]

We don't...only when we have quadratic factors in the denominator do we use linear numerators. If the degree of $P(x)$ is less than $2n$ then we may state the following:

i) For non-repeated quadratic factors:

\(\displaystyle \frac{P(x)}{\prod\limits_{k=1}^n\left(a_kx^2+b_kx+c_k \right)}=\sum_{k=1}^n\left(\frac{A_kx+B_k}{a_kx^2+b_kx+c_k} \right)\)

ii) For repeated quadratic factors:

\(\displaystyle \frac{P(x)}{\left(ax^2+bx+c \right)^n}=\sum_{k=1}^n\left(\frac{A_kx+B_k}{\left(ax^2+bx+c \right)^k} \right)\)

 

[Dustinsfl]

We don't...only when we have quadratic factors in the denominator do we use linear numerators. If the degree of $P(x)$ is less than $2n$ then we may state the following:

i) For non-repeated quadratic factors:

\(\displaystyle \frac{P(x)}{\prod\limits_{k=1}^n\left(a_kx^2+b_kx+c_k \right)}=\sum_{k=1}^n\left(\frac{A_kx+B_k}{a_kx^2+b_kx+c_k} \right)\)

ii) For repeated quadratic factors:

\(\displaystyle \frac{P(x)}{\left(ax^2+bx+c \right)^n}=\sum_{k=1}^n\left(\frac{A_kx+B_k}{\left(ax^2+bx+c \right)^k} \right)\)

I don't understand what you mean.