partial fraction

**al2308** · March 2nd 2008, 08:04 AM

can anybody help solve this?

x= 1/(s^2(s+2))

i need to find the partial fractions of the RHS of the equation.

**Jhevon** · March 2nd 2008, 08:10 AM

Originally Posted by al2308

can anybody help solve this?

x= 1/(s^2(s+2))

i need to find the partial fractions of the RHS of the equation.

$\frac 1{s^2(s + 2)} = \frac As + \frac B{s^2} + \frac C{s + 2}$

can you continue?

**al2308** · March 2nd 2008, 08:12 AM

I got that far and then i got A=1/2, thats where i'm stuck from

**Krizalid** · March 2nd 2008, 08:15 AM

(No partial fractions foundation.)

$\begin{aligned} \frac{1} {{s^2 (s + 2)}} &= \frac{4} {{4s^2 (s + 2)}} \hfill \\ &= \frac{{4 - s^2 + s^2 }} {{4s^2 (s + 2)}} \hfill \\ &= \frac{{(2 - s)(s + 2) + s^2 }} {{4s^2 (s + 2)}} \hfill \\ &= \frac{1} {4}\left( {\frac{{2 - s}} {{s^2 }} + \frac{1} {{s + 2}}} \right). \end{aligned}$

**Jhevon** · March 2nd 2008, 08:30 AM

Originally Posted by Jhevon

$\frac 1{s^2(s + 2)} = \frac As + \frac B{s^2} + \frac C{s + 2}$

just to finish this method up, since Krizalid showed you another.

$\Rightarrow 1 = As(s + 2) + B(s + 2) + Cs^2$ .............(1)

plug in $s = 0$ in (1), we get:

$1 = 2B \implies \boxed{B = \frac 12}$

plug in $s = -2$ in (1), we get:

$1 = 4C \implies \boxed{C = \frac 14}$

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):

$1 = As^2 + 2As + Bs + 2B + Cs^2$

group like powers of $s$ :

$1 = (A + C)s^2 + (2A + B)s + 2B$

now the coefficients of each power of $s$ on the right must be the same as the coefficient of that same power on the left.

thus we have the system:

$A + C = 0$

$2A + B = 0$

$2B = 1$

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 $A + C = 0$ equation

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