I was solve equation differential using Laplace transform, but I need use partial fraction of A.

Julia Chang
2022-09-26
Answered

Need find a partial-fraction expansion of $A=\frac{1}{(s+1{)}^{2}(s-1)(s+5)}.$

I was solve equation differential using Laplace transform, but I need use partial fraction of A.

I was solve equation differential using Laplace transform, but I need use partial fraction of A.

You can still ask an expert for help

berzamauw

Answered 2022-09-27
Author has **9** answers

The partial fraction expansion is:

$$\frac{1}{(s+1{)}^{2}(s-1)(s+5)}=\frac{1}{24(s-1)}-\frac{1}{32(s+1)}-\frac{1}{8(s+1{)}^{2}}-\frac{1}{96(s+5)}$$

The fraction expansion can be expressed as:

$$\frac{1}{(s+1{)}^{2}(s-1)(s+5)}=\frac{A}{s-1}+\frac{B}{s+1}+\frac{C}{(s+1{)}^{2}}+\frac{D}{s+5}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}A(s+1{)}^{2}(s+5)+B(s+1)(s-1)(s+5)+C(s-1)(s+5)+D(s-1)(s+1{)}^{2}=1$$

Using $s=1$, we get:

$$24A=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}A=\frac{1}{24}$$

Using $s=-1$, we get:

$$-8C=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=-\frac{1}{8}$$

Using $s=-5$, we get:

$$-96D=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}D=-\frac{1}{96}$$

For B, none of these will work. So we can choose something that makes life easier. Using s=0, we get

$$5A-5B-5C-D=1$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{5}{24}-5B+\frac{5}{8}+\frac{1}{96}=1$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}B=-\frac{1}{5}(1-\frac{20}{96}-\frac{60}{96}-\frac{1}{96})=-\frac{1}{32}$$

$$\frac{1}{(s+1{)}^{2}(s-1)(s+5)}=\frac{1}{24(s-1)}-\frac{1}{32(s+1)}-\frac{1}{8(s+1{)}^{2}}-\frac{1}{96(s+5)}$$

The fraction expansion can be expressed as:

$$\frac{1}{(s+1{)}^{2}(s-1)(s+5)}=\frac{A}{s-1}+\frac{B}{s+1}+\frac{C}{(s+1{)}^{2}}+\frac{D}{s+5}$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}A(s+1{)}^{2}(s+5)+B(s+1)(s-1)(s+5)+C(s-1)(s+5)+D(s-1)(s+1{)}^{2}=1$$

Using $s=1$, we get:

$$24A=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}A=\frac{1}{24}$$

Using $s=-1$, we get:

$$-8C=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=-\frac{1}{8}$$

Using $s=-5$, we get:

$$-96D=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}D=-\frac{1}{96}$$

For B, none of these will work. So we can choose something that makes life easier. Using s=0, we get

$$5A-5B-5C-D=1$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{5}{24}-5B+\frac{5}{8}+\frac{1}{96}=1$$

$$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}B=-\frac{1}{5}(1-\frac{20}{96}-\frac{60}{96}-\frac{1}{96})=-\frac{1}{32}$$

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which only uses the definition that

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I can only find a proofs which use the trig-representation of complex numbers.