havin trouble solving this partial fraction:
$\displaystyle 1/s^2-2s+2$
i think i can break it up into:
$\displaystyle 1/(s-1)^2+1$
You have,
$\displaystyle \frac{1}{s^2-2s+2}$
Express as,
$\displaystyle \frac{1}{s^2-2s+1+1}$
Thus,
$\displaystyle \frac{1}{(s-1)^2+1}$
Now, I presume you have an inverse Laplace transform for $\displaystyle s>1$.
In that case if you have,
$\displaystyle \frac{1}{s^2+1}$ $\displaystyle s>0$
Then, that is the function,
$\displaystyle f(x)=\sin x$
Since you have a substitution you use the shift rule.
Thus,
$\displaystyle f(x)=e^x\sin x$
If the unique inverse LaPlace Transform.
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:
$\displaystyle
\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:
$\displaystyle
\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})) }
$
(please check the arithmetic)
RonL
Hello, Chadly724!
Havin trouble solving this partial fraction: $\displaystyle \frac{1}{s^2-2s+2}$
i think i can break it up into: $\displaystyle \frac{1}{(s-1)^2+1}$. . Yes, you can . . . but why?
Partial Fractions are used when the denominator can be factored.
. . Like: .$\displaystyle \frac{1}{x^2+2x-15}\:=\:\frac{1}{(x-3)(x+5)}$ or: .$\displaystyle \frac{1}{x^3+4x} \:=\:\frac{1}{x(x^2 +4)}$
As Captain Black pointed out, it can be factored: .$\displaystyle (s - 1 - i)(s - 1 + i)$
. . but I doubt that you want to introduce complex numbers.
If this is an integration problem:
. . $\displaystyle \int\frac{ds}{s^2 - 2x + 2} \;=\;\int\frac{ds}{(s-1)^2+1} \;=\;\tan^{-1}(s-1) + C $