# inverse laplace transform - with symbolic variables

Transform: $$F(s) = \frac{2s^2 + (a-6b)s + a^2 - 4ab}{(s^2-a^2)(s-2b)}$$

My steps: $$F(s) = \frac{2s^2 + (a-6b)s + a^2 - 4ab}{(s+a)(s-a)(s-2b)}$$ $$= \frac{A}{s+a} + \frac{B}{s-a} + \frac{C}{s-2b} + K$$ $$K = 0$$

$$A = F(s) * (s+a)$$ at s = -a $$A = \frac{2a^2 + (a-6b)(-a) +a^2 - 4ab}{4ab+2a^2}$$ $$A = \frac{a+b}{2b+a}$$

I problems like this (that I've seen) at the step above the fraction would reduce into just a number. In this case, it doesn't and its surprising because I've never seen it happened before. Is there something I am doing wrong here?

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The routine way is to use the way we get in partial fractions and you correctly noted that. If we set:

$$\frac{2s^2 + (a-6b)s + a^2 - 4ab}{(s^2-a^2)(s-2b)}=F= \frac{A}{s+a} + \frac{B}{s-a} + \frac{C}{s-2b}$$ then by doing boring :-) handy calculations we can find $A,B$ and $C$. We have then:

$$\frac{-Asa+2Aab+As^2-2Asb+Bsa-2Bab+Bs^2-2Bsb+Cs^2-Ca^2}{(-s+2b)(-s^2+a^2)}$$ Now if we put $s=-a$ in the numerators, we get $A=\frac{a+b}{a+2b}$ and this is what you already got. With the similar approach $s=+a$ for $B$ and $s=2b$ for $C$, we get: $$B=\frac{2a-5b}{a-2b},~~C=\frac{4b^2+2ab-a^2}{a^2-4b^2}$$

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Needs a TU!! +1 – amWhy Aug 29 '13 at 12:46
What is a TU ?. – 40Plot Aug 29 '13 at 15:36
I'm glad to see you. – user63181 Jan 15 '14 at 20:30