## UnkleRhaukus 3 years ago Partial-Fraction Decomposition $\frac1{p^2(p^2-4)}$

1. UnkleRhaukus

\begin{align*}\frac1{p^2(p^2-4)}&=\frac1{p^2(p+2)(p-2)}\\ \\&=\frac Ap+\frac B{p^2}+\frac C{p+2}+\frac D{p-2}\\ \\\end{align*}

2. UnkleRhaukus

$1=Ap(p+2)(p-2)+B(p+2)(p-2)+Cp^2(p-2)+Dp^2(p+2)$

3. Yahoo!

Put Values for p and Find A , B , C D

4. Yahoo!

1st put P = 2 and Find D

5. RolyPoly

Covering up! Nice!

6. UnkleRhaukus

$1=Ap(p+2)(p-2)+B(p+2)(p-2)+Cp^2(p-2)+Dp^2(p+2)$ $\text{for }p=0\\\qquad1=-4B$$\qquad B=\frac{-1}4$ $\text{for }p=2$$\qquad1=16D$$\qquad D=\frac 1{16}$ $\text{for }p=-2$$\qquad1=-16C$$\qquad C=\frac {-1}{16}$ $\text{for }p=1$$\qquad1=3A+3B-C+3D$ $\qquad1=3A-\frac34+\frac 1{16}+\frac3{16}$$\qquad1=3A-\frac34+\frac 4{16}$$\qquad\frac32=3A$$\qquad A=\frac{1}{2}$ $\frac1{p^2(p^2-4)}=\frac 1{2p}-\frac 1{4p^2}-\frac 1{16(p+2)}+\frac 1{16(p-2)}$

7. UnkleRhaukus

i think $$A$$ is ment to be zero , , can you see my mistake ?

8. UnkleRhaukus

found it / (them)\begin{align*}\text{for }p=1&&1&=-3A-3B-C+3D\\ &&1&=-3A+\frac34+\frac 1{16}+\frac3{16}\\ &&1&=-3A+1\\ \\ &&A&=0\\ \end{align*}

9. UnkleRhaukus

$F(p)=\frac1{p^2(p^2-4)}=-\frac 1{4p^2}-\frac 1{16(p+2)}+\frac 1{16(p-2)}$