Partial Fraction Decomposition

[hackedagainanda]

Homework Statement

Find the partial fraction decomposition of the rational=$\frac {1} {x^2 -c^2}$ with $c \neq {0}$

Relevant Equations

N/A

$\frac {1} {x^2 -c^2}$ with $c \neq {0}$

So the first thing I do is split the $x^2 -c^2$ into the difference of squares so $x +c$ and $x - c$

I then do $\frac {A} {x + c}$ $+$ $\frac {B} {x-c}$, and then let $x=c$ to zero out the expression. And that is where I am getting lost I don't see where to go from here, I don't understand where the $2c$ in the denominator is coming from.

The solution in the book is $\frac {1} {2c(x-c)}$ $-\frac {1} {2c(x+c)}$

 

[Frabjous]

You set the partial fraction equal to the original fraction and then solve for A and B.

 

[hackedagainanda]

You set the partial fraction equal to the original fraction and then solve for A and B.

I think I got it Ax + Bx = 0 and adding -Ac + Bc =1 A-A =0 add the B's and get 2B = 1 so B= 1/2 and A = -1/2, there is no x term in the numerator so we can move along to the variable c, -Ac = -1/2c and B = 1/2c so that lines up with the books answer.

Thanks for the suggestion and help, did I make any mistakes?

 

[Frabjous]

I think I got it Ax + Bx = 0 and adding -Ac + Bc =1 A-A =0 add the B's and get 2B = 1 so B= 1/2 and A = -1/2, there is no x term in the numerator so we can move along to the variable c, -Ac = -1/2c and B = 1/2c so that lines up with the books answer.

Thanks for the suggestion and help, did I make any mistakes?

I cannot tell exactly what you are doing. For future questions, you will need to be clearer.
Here’s how I do it.
The first equation can be written as (A+B)x=0. Since this has to hold for all values of x, this implies A=-B.
The second equation can now be rewritten
(B-A)c=1
2Bc=1
B=1/(2c) (notice that you are dividing by c, so that it cannot equal 0

 

[Mark44]

The first equation can be written as (A+B)x=0. Since this has to hold for all values of x, this implies A=-B.
The second equation can now be rewritten
(B-A)c=1
2Bc=1.
B=1/2c (notice that you are dividing by c, so that it cannot equal 0.

1/2c would normally be read as one half times c. To convey the idea that c is in the denominator, write 1/(2c) or $\frac 1 {2c}$.