Quadratic Partial Fraction Decomposition

Question

I am trying to find the inverse laplace transform of $(s^2+4) \over (s-2)(s+2)$.

The solution is $ {2\over(s-2)} - {2\over(s+2)} + 1 $.

But I can't figure out how to break it up so I can find the solution algebraically.

i.e $ (s^2+4)/((s-2)(s+2)) = As+B/(s-2)+C/(s+2) $ etc. What terms should I use?

Do the long division first $(s^2+4)/(s^2-4)=1+8/(s^2-4)$. Now apply partial fractions to the second term. — David Mitra, Mar 15 '14 at 10:37
That first summand should be $A$, not $As$, because the numerator is only a quadratic. If it were cubic, you would need an $A$ term and a $Ds$ term. Normally partial fractions is taught so that you do long division first and then work on the remainder; there wouldn't be a term involving $A$ at all if you had done that. — , Mar 15 '14 at 11:11

score 2 · Accepted Answer · answered Mar 15 '14 at 10:48

2

Write ${s^2+4 \over (s-2)(s+2)} = A + {B\over(s-2)} + {C\over(s+2)}$

Multiply through by $(s-2)(s+2)$ and equate coefficients.

Since the Laplace transform is linear (I believe), you can use the known ones for constants and $ 1/s $.

Alternatively, just write $$ {s^2+4 \over (s-2)(s+2)} = {(s^2-4)+8 \over s^2-4} = 1 + {8 \over s^2-4}$$

and we know that (correct me if I'm wrong) the inverse Laplace transform of $ 1/(s^2 - b^2) $ is ${1 \over b}\sinh(bt)$.

From this, we get the final answer along with the inverse Laplace transform of 1.

Hope this helps! :)

answered Mar 15 '14 at 10:48

Sam OT

i'd like to up vote but have less than 15 rep. – martinap Mar 15 '14 at 10:58
Haha, no worries. Happy to help. :) Perhaps when you get 15 rep, you can upvote it? (I'm just starting out on SE, so the more rep the better!) – Sam OT Mar 15 '14 at 12:15
Personally I prefer the systematic method. In the other proof given, you have to spot that $(s-2)+(s+2) = 4$ (which is obvious, of course, but not obvious that you need to use it. Also, did you want the (inverse) Laplace transforms, or were you ok with that bit? – Sam OT Mar 15 '14 at 12:16
i'm good thanks. – martinap Mar 16 '14 at 10:57

score 1 · Answer 2 · answered Mar 15 '14 at 10:51

$$\begin{align}\frac{s^2+4}{s^2-4}&=\frac{s^2-4+8}{s^2-4}\\&= 1+\frac{8}{s^2-4}\\&=1+2\frac{4}{(s+2)(s-2)}\end{align}$$

Obviously $4=(s+2)-(s-2)$

substituting, we have $$1+2\frac{(s+2)-(s-2)}{(s+2)(s-2)}$$

$$1+2\frac{(s+2)}{(s+2)(s-2)}-2\frac{(s-2)}{(s+2)(s-2)}$$

Cancelling,

$$\require{cancel}{1+2\frac{\cancel{(s+2)}}{\cancel{(s+2)}(s-2)}-2\frac{\cancel{(s-2)}}{(s+2)\cancel{(s-2)}}}$$

And thus, we have

$$1+\frac{2}{s-2}-\frac{2}{s+2}$$

2 Answers2