1. ## Partial Fraction

havin trouble solving this partial fraction:

$1/s^2-2s+2$

i think i can break it up into:

$1/(s-1)^2+1$

havin trouble solving this partial fraction:

$1/s^2-2s+2$

i think i can break it up into:

$1/(s-1)^2+1$
is it: $\frac{1}{s^2}-2s+2$ or $\frac{1}{s^2-2s+2}$?

3. ## correction

sorry there should be partheses around the entire bottom function

havin trouble solving this partial fraction:

$1/s^2-2s+2$

i think i can break it up into:

$1/(s-1)^2+1$
You have,
$\frac{1}{s^2-2s+2}$

Express as,
$\frac{1}{s^2-2s+1+1}$

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

Now, I presume you have an inverse Laplace transform for $s>1$.

In that case if you have,
$\frac{1}{s^2+1}$ $s>0$
Then, that is the function,
$f(x)=\sin x$
Since you have a substitution you use the shift rule.
Thus,
$f(x)=e^x\sin x$
If the unique inverse LaPlace Transform.

havin trouble solving this partial fraction:

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

i think i can break it up into:

$\frac{1}{(s-1)^2+1}$
Do you want to break this up into partial fractions or just transform into a
standard form of LT that can be inverted by table look-up?

My table of LT's has:

$
\frac{1}{(s-b)^2+a^2}=\mathcal{L}\ \frac{e^{bt}\sin(at)}{a}
$

which should take case of your last form.

However we can do the same thing with partial fractions if you will:

$
\frac{1}{s^2-2s+2}=\frac{1}{ (s-(1+\bold{i})) (s-(1-\bold{i}))}=
\frac{\bold{i}}{2(s-(1-\bold{i})) } - \frac{\bold{i}}{2(s-(1+\bold{i})) }
$

RonL

Havin trouble solving this partial fraction: $\frac{1}{s^2-2s+2}$

i think i can break it up into: $\frac{1}{(s-1)^2+1}$. . Yes, you can . . . but why?

Partial Fractions are used when the denominator can be factored.
. . Like: . $\frac{1}{x^2+2x-15}\:=\:\frac{1}{(x-3)(x+5)}$ or: . $\frac{1}{x^3+4x} \:=\:\frac{1}{x(x^2 +4)}$

As Captain Black pointed out, it can be factored: . $(s - 1 - i)(s - 1 + i)$
. . but I doubt that you want to introduce complex numbers.

If this is an integration problem:

. . $\int\frac{ds}{s^2 - 2x + 2} \;=\;\int\frac{ds}{(s-1)^2+1} \;=\;\tan^{-1}(s-1) + C$

7. Originally Posted by Soroban

Partial Fractions are used when the denominator can be factored.
. . Like: . $\frac{1}{x^2+2x-15}\:=\:\frac{1}{(x-3)(x+5)}$ or: . $\frac{1}{x^3+4x} \:=\:\frac{1}{x(x^2 +4)}$

As Captain Black pointed out, it can be factored: . $(s - 1 - i)(s - 1 + i)$
. . but I doubt that you want to introduce complex numbers.

If this is an integration problem:

. . $\int\frac{ds}{s^2 - 2x + 2} \;=\;\int\frac{ds}{(s-1)^2+1} \;=\;\tan^{-1}(s-1) + C$
The variable $s$ points in the direction of a Laplace transform problem, not an intrgration.

RonL