As part of trying to solve a differential equation using Laplace transforms, I have the fraction $\frac{-10s}{(s^2+2)(s^2+1)}$ which I am trying to perform partial fraction decomposition on so that I can do a inverse Laplace transform. When I try to work out this fraction, I get that $-10=0(A+B)$ since $s^1$ does not show up in the fraction. How does partial fraction decomposition work for a fraction like this?
2 Answers
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I think you should work on the following, instead:
$$\frac{As+B}{s^2+2}+\frac{Cs+D}{s^2+1}$$
Or use the convolution theorem.
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$\begingroup$ Thank you, I can't believe I forgot about that. It's been about a year since I last did partial fractions. $\endgroup$ Commented Mar 6, 2015 at 3:28
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$\begingroup$ @ChrisCrutchfield: Thanks! I am I could help you. :-) $\endgroup$– MikasaCommented Mar 6, 2015 at 6:24
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$\dfrac{-10s}{(s^2+2)(s^2+1)}=\dfrac{10s}{s^2+2}-\dfrac{10s}{s^2+1}$
Beacuse
$$\dfrac{-10s}{(s^2+2)(s^2+1)}=\frac{As+B}{s^2+2}+\frac{Cs+D}{s^2+1}$$
then
$$(A+C)s^3+(B+D)s^2+(A+2C)s+(B+2D)=-10s$$
$A+C=0$
$B+D=0$
$A+2C=-10$
$B+2D=0$
imply $A=10$, $B=0$, $C=-10$ and $D=0$.
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$\begingroup$ Would you be able to explain where that comes from? $\endgroup$ Commented Mar 6, 2015 at 3:15
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$\begingroup$ are irreducible quadratics in the denominator $\endgroup$– jimboCommented Mar 6, 2015 at 3:24
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$\begingroup$ math.stackexchange.com/questions/773442/… $\endgroup$– jimboCommented Mar 6, 2015 at 3:27