Hi pals , possible to check where did i went wrong ?

2 / s(1 + s^2) = A/s + B/(1 + s^2)

then , 2 = A(1 + s^2) + Bs

when s = 0 , A = 2

when s = 1 and A = 2 , B = -2

however i could not get B = -2s

(Crying)

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- Dec 2nd 2009, 05:32 AMxcludedpartial fraction did wrongly...
Hi pals , possible to check where did i went wrong ?

2 / s(1 + s^2) = A/s + B/(1 + s^2)

then , 2 = A(1 + s^2) + Bs

when s = 0 , A = 2

when s = 1 and A = 2 , B = -2

however i could not get B = -2s

(Crying) - Dec 2nd 2009, 05:50 AMJester
- Dec 2nd 2009, 05:54 AMxcluded
- Dec 2nd 2009, 06:03 AMHallsofIvy
Because the denominator is an "irreducible" (unfactorable in terms of real numbers) quadratic. If x- a is a factor of the denominator, the partial fractions expansion will include $\displaystyle \frac{A}{x- a}$. If $\displaystyle x^2+ cx+ d= (x-a)^2+ b$, so that it is irreducible, then the partial fractions expansion will include $\displaystyle \frac{Ax+ B}{(x-a)^2+ b}$. If the denominator has $\displaystyle (x-a)^n$ as a factor, the partial fraction expansion will include $\displaystyle \frac{A_1}{x-a}+ \frac{A_2}{(x-a)^2}+ \cdot\cdot\cdot+ \frac{A_n}{(x-a)^n}$. Finally, if the denominator has $\displaystyle (x^2+ a)^n$ as a factor, the partial fraction expansion will include $\displaystyle \frac{A_1x+ B_1}{x^2+ a}+ \frac{A_2x+ B_2}{(x^2+a)^2}+$$\displaystyle \cdot\cdot\cdot+$$\displaystyle \frac{A_nx+ B_n}{(x^2+a)^n}$.

Since any polynomial, with real coefficients, can be factored (in the real numbers) to powers of linear or powers of quadratic terms, those are all the terms you need.