MHB Partial Fractions for Cubic: Setup & Solve

AI Thread Summary
To separate the function s/(s+1)^3 for an inverse Laplace transform, the correct setup for partial fractions is A/(s+1) + B/(s+1)^2 + C/(s+1)^3. The user initially attempted a more complex decomposition but ended up with too many unknowns. By substituting values and simplifying, they found that A = 0, B = 1, and C = -1, leading to the correct partial fraction decomposition. The discussion clarifies the method for handling repeated linear factors in the denominator, emphasizing the need for specific forms based on the degree of the polynomial. Understanding these setups is crucial for successfully applying inverse Laplace transforms.
Dustinsfl
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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}
 
Last edited:
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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 $A=0$.
 
dwsmith said:
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.
 
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:
\[
\frac{A}{s+1} + \frac{B}{(s+1)^2} + \frac{C}{(s+1)^3}
\]
 
dwsmith said:
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:

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

ii) For repeated quadratic factors:

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

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

ii) For repeated quadratic factors:

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

I don't understand what you mean.
 
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