# Laplace transform - stuck on the partial fraction

• May 4th 2009, 09:26 AM
mistykz
Laplace transform - stuck on the partial fraction
In solving a problem, I ended up having to take the inverse laplace of (e^(-2s))/((x^2)(x-1))
I know what formula I need to use, but for some reason I'm stuck and can't remember how to do partial fractions with a denominator like that (I separated them so the numerator = 1), can anyone help? (Thinking)
• May 4th 2009, 09:37 AM
Peritus
$
\frac{1}
{{x^2 \left( {x - 1} \right)}} = \frac{{ax + b}}
{{x^2 }} + \frac{c}
{{x - 1}}\quad \quad a,b,c \in Z

$

now all you have to do is find the constant coefficients, good luck.
• May 4th 2009, 09:46 AM
HallsofIvy
Quote:

Originally Posted by mistykz
In solving a problem, I ended up having to take the inverse laplace of (e^(-2s))/((x^2)(x-1))
I know what formula I need to use, but for some reason I'm stuck and can't remember how to do partial fractions with a denominator like that (I separated them so the numerator = 1), can anyone help? (Thinking)

That looks pretty standard to me. The partial fractions form is $\frac{1}{x^2(x-1)}= \frac{A}{x}+ \frac{B}{x^2}+ \frac{C}{x-1}$. Multiplying through by $x^2(x- 1)$ gives $1= Ax(x-1)+ B(x-1)+ Cx^2$. Taking x= 0, $1= -B$. Taking x= 1, $1= C$. Those are the only values of x that will reduce so simply but any other value for x will give a third equation. For example, taking x= 2, $1= A(2)(2-1)+ B(2-1)+ C(4)$. Since we know that B= -1 and C= 1, that becomes $2A- 1+ 4= 1$ or A= -1.

Another way to do any partial fractions, though harder since we are not taking advantage of those special values, x= 0 and x= 1, is to combine the fractions: $\frac{1}{x^2(x-1)}= \frac{A}{x}+ \frac{B}{x^2}+ \frac{C}{x-1}= \frac{A(x)(x-1)}{x^2(x-1)}+ \frac{B(x-1)}{x^2(x-1)}+ \frac{Cx^2}{x^2(x-1)}$ $\frac{0x^2+ 0x+ 1}{x^2(x-1)}= \frac{(A+ C)x^2+ (-A+ B)x- B}{x^2(x-1)}$.

Since this must be true for all x, we must have A+ C= 0, -A+ B= 0 and -B= 1. Again, we get B= -1, A= -1, and C= 1.

(Note that $\frac{ax+ b}{x^2}= \frac{ax}{x^2}+ \frac{b}{x^2}= \frac{a}{x}+ \frac{b}{x^2}$ so Peritus' form is the same as mine.)
• May 4th 2009, 01:14 PM
mistykz
ah, that's what i did wrong, i multiplied through by x, x^2, and (x-1)...didnt need that single x. oops! thanks!
• May 4th 2009, 04:28 PM
TheEmptySet
Quote:

Originally Posted by mistykz
In solving a problem, I ended up having to take the inverse laplace of (e^(-2s))/((x^2)(x-1))
I know what formula I need to use, but for some reason I'm stuck and can't remember how to do partial fractions with a denominator like that (I separated them so the numerator = 1), can anyone help? (Thinking)

Just FYI you can invert this with out partial fractions.

Remember that $\mathcal{L}^{-1}=\frac{F(s)}{s}=\int_{0}^{t}f(\tau)d\tau$

Where $\mathcal{L}(f(t))=F(s)$

Note that $\mathcal{L}^{-1}\left( \frac{e^{-2s}}{(s-1)}\right)=H(t-2)e^{t-2}$

Where $H(t)$ is the heaviside function

Now if we integrate this twice we will get the inverse transform

$\int_{0}^{t} H(\tau-2)e^{\tau -2}d \tau=\int_{2}^{t}e^{\tau-2}d\tau=e^{\tau-2}\bigg|_{2}^{t}=e^{t-2}-1$

Integrating one more time we get

$\int_{2}^{t} e^{\tau-2}-1 d\tau=e^{\tau - 2} -\tau \bigg|_{2}^{t}=e^{t-2}-t-(1-2)=1-t+e^{t-2}$

so

$f(t)=H(t-2)(1-t+e^{t-2})$