# Laplace transform of proper rational function

• ChiralSuperfields
ChiralSuperfields
Homework Statement
Relevant Equations
##\frac{2s + 1}{((s - 1)(s - 1)} = \frac{A(s - 1) + B(s - 1)}{(s - 1)(s - 1)}##
For this problem (b),

The solution is,

However, I don't understand how they got their partial fractions here (Going from step 1 to 2).

My attempt to convert into partial fractions is:

##\frac{2s + 1}{(s - 1)(s - 1)} = \frac{A(s - 1) + B(s - 1)}{(s - 1)(s - 1)}##

Thus,

##2s + 1 = A(s - 1) + B(s - 1)##
##2s + 1 = (A + B)s - A - B##

##2 = A + B##
##1 = - A - B##
##-1 = A + B##

However, ##2 ≠ -1##

Does someone please know how they got their partial fractions expression in step 2?

Thanks!

I disagree with their first step. ##\frac {2s+1}{s^2-2s+1} = \frac {2s}{(s-1)^2} + \frac{1}{(s-1)^2}##.
I don't know why they are adding 1 to the denominators.
CORRECTION: The text in the OP is a typo and the actual problem has denominator ##s^2-2s+2##. Thanks, @Orodruin .

Last edited:
ChiralSuperfields
FactChecker said:
I disagree with their first step. ##\frac {2s+1}{s^2-2s+1} = \frac {2s}{(s-1)^2} + \frac{1}{(s-1)^2}##.
I don't know why they are adding 1 to the denominators.
Because the actual problem has the denominator ##s^2 - 2s + \color{Red}{2}## … the 1 in the solution’s first expression is a typo.

ChiralSuperfields and FactChecker
Apart from that, your partial fraction ansatz is equivalent to assuming
$$2s + 1 = (A+B)(s-1)$$
This obviously will not work. You should check how to do partial fractions when the denominator is a square (or higher power). Even if it will not be useful here.

ChiralSuperfields
Orodruin said:
is a typo
Typos seem quite overabundant in the solutions of the OP’s threads. If this is the main mode of learning, they may want to consider a different source.

ChiralSuperfields
There are no tricky partial fractions involved in this problem. Every step is a simple rearrangement of the fractions.

ChiralSuperfields
I mean, you can do partial fractions, but the roots of the denominator are complex and the approach in the solution is much more straightforward.

ChiralSuperfields

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