Question

Partial-Fraction Decomposition
\[\frac1{p^2(p^2-4)}\]

Answer 1

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

Answer 2

\[1=Ap(p+2)(p-2)+B(p+2)(p-2)+Cp^2(p-2)+Dp^2(p+2)\]

Answer 3

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

Answer 4

1st put P = 2 and Find D

Answer 5

Covering up! Nice!

Answer 6

\[1=Ap(p+2)(p-2)+B(p+2)(p-2)+Cp^2(p-2)+Dp^2(p+2)\]
\[\text{for }p=0\\\qquad1=-4B\]\[\qquad B=\frac{-1}4\]

\[\text{for }p=2\]\[\qquad1=16D\]\[\qquad D=\frac 1{16}\]

\[\text{for }p=-2\]\[\qquad1=-16C\]\[\qquad C=\frac {-1}{16}\]

\[\text{for }p=1\]\[\qquad1=3A+3B-C+3D\]
\[\qquad1=3A-\frac34+\frac 1{16}+\frac3{16}\]\[\qquad1=3A-\frac34+\frac 4{16}\]\[\qquad\frac32=3A\]\[\qquad A=\frac{1}{2}\]

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

Answer 7

i think \(A\) is ment to be zero , ,
can you see my mistake ?

Answer 8

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

Answer 9

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