can anybody help solve this?
x= 1/(s^2(s+2))
i need to find the partial fractions of the RHS of the equation.
(No partial fractions foundation.)
$\displaystyle \begin{aligned}
\frac{1}
{{s^2 (s + 2)}} &= \frac{4}
{{4s^2 (s + 2)}} \hfill \\
&= \frac{{4 - s^2 + s^2 }}
{{4s^2 (s + 2)}} \hfill \\
&= \frac{{(2 - s)(s + 2) + s^2 }}
{{4s^2 (s + 2)}} \hfill \\
&= \frac{1}
{4}\left( {\frac{{2 - s}}
{{s^2 }} + \frac{1}
{{s + 2}}} \right).
\end{aligned}$
just to finish this method up, since Krizalid showed you another.
$\displaystyle \Rightarrow 1 = As(s + 2) + B(s + 2) + Cs^2$ .............(1)
plug in $\displaystyle s = 0$ in (1), we get:
$\displaystyle 1 = 2B \implies \boxed{B = \frac 12}$
plug in $\displaystyle s = -2$ in (1), we get:
$\displaystyle 1 = 4C \implies \boxed{C = \frac 14}$
can we plug anything else in to eliminate B and C but not A? Nah, we will have to do A the hard way, but it will be easier considering we know B and C
expand (1):
$\displaystyle 1 = As^2 + 2As + Bs + 2B + Cs^2$
group like powers of $\displaystyle s$:
$\displaystyle 1 = (A + C)s^2 + (2A + B)s + 2B$
now the coefficients of each power of $\displaystyle s$ on the right must be the same as the coefficient of that same power on the left.
thus we have the system:
$\displaystyle A + C = 0$
$\displaystyle 2A + B = 0$
$\displaystyle 2B = 1$
and we could have easily found each variable from this system to begin with (but i wanted to show you the two methods, you usually need only one). but now let's just find A. we can find A by plugging in the value we know for C in the $\displaystyle A + C = 0$ equation