# partial fraction problem

• Mar 26th 2009, 03:15 AM
ben.mahoney@tesco.net
partial fraction problem
i amstruggling with the partial fraction:
8/s^2*(s^2+4)) = A/s^2 + B/s + Cs+D/(s^2+4)

i get A=2 but cannot find a decent s value to give B,C and D
• Mar 26th 2009, 04:08 AM
HallsofIvy
Quote:

Originally Posted by ben.mahoney@tesco.net
i amstruggling with the partial fraction:
8/s^2*(s^2+4)) = A/s^2 + B/s + Cs+D/(s^2+4)

i get A=2 but cannot find a decent s value to give B,C and D

Multiplying on both sides by $s^2(s^2+ 4)$ gives
$8= A(s^2+4)+ Bs(s^2+ 4)+ (Cs+D)s^2$. Setting s= 0 gives 8= 4A so A= 2. I presume that is what you have already done.

Although s= 0 is the only value that makes the equation that simple, any value of s give an equation.

If s= 1, 8= 5A+ 5B+ C+ D and you already know that A= 2: 5B+ C+ D= -2.

If s= -1, 8= 5A- 5B- C+ D so 5B+ C- D= 2.
Adding those two equations eliminates D immediately.

If s= 2, 8= 8A+ 16B+ 8C+ 4D so 16B+ 8C+ 4D= -8 or 4B+ 2C+ D= -2.
Adding that to 5B+ C+ D= -2 again eliminates D leaving two equations to solve for B and C.

Another method for problems like this is to combine like powers of s.

From $8= A(s^2+4)+ Bs(s^2+ 4)+ (Cs+D)s^2$, $8= As^2+ 4A+ Bs^3+ 4Bs+ Cs^3+ Ds^2$ or, combining like powers, $8= (B+ C)s^3+ (A+ D)s^2+ (4B)s+ 4A$. For that to be true for all x, we must have coefficients of the same powers equal on both sides: B+ C= 0, A+ D= 0, 4B= 0, 4A= 8.

This method is often harder that just choosing value for s but here is much easier.