Partial fractions question

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October 22nd 2012, 02:19 PM
Ragnarok

Partial fractions question

Hello, I am in a differential equations class need to find the inverse Laplace of
$\frac{1}{s(s^2+5)}$ , but have forgotten how to do partial fractions with quadratic factors. What I got so far:

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

$A(s^2+5)+(Bs+C)s=1$

Letting s=0 I get $A=\frac{1}{5}$ , but where do I go from there?
October 22nd 2012, 02:24 PM
MaxJasper

Re: Partial fractions question

$\frac{1}{s\left(s^2+5\right)} = \frac{1}{5 s}-\frac{s}{5 \left(5+s^2\right)}$
October 22nd 2012, 02:58 PM
Ragnarok

Re: Partial fractions question

Thanks! Could you explain how you got that?
October 22nd 2012, 03:10 PM
BobP

Re: Partial fractions question

Equate coefficients of the various powers of $s$ across the identity.

The number of $s^{2}$ on the LHS have to equal the number of $s^{2}$ on the RHS.

That gives you $A+B=0.$

Then equate coefficients of $s$ to get the value of $C.$

Alternatively you could substitute two other values for $s$ and solve the resulting simultaneous equations.
October 22nd 2012, 03:25 PM
HallsofIvy

Re: Partial fractions question

You have $A(s^2+ 5)+ (Bs+ C)s= 1$ which is to be true for all s. So take s to be some simple numbers.

If s= 0, then $A(0+ 5)+ (B(0)+ C)0= 5A= 1$ . Putting that value into the equation, $(1/5)(s^2+ 5)+ (Bs+ C)s= (1/5+ B)s^2+ Cs+ 1= 1$ or $(1/5+ B)s^2+ Cs= 0$ .
Taking s= 1, that becomes 1/5+ B+ C= 0 or B+ C= -1/5. Taking s= -1, it is 1/5+ B- C= 0 or B- C= 1/5. Adding those, 2B= 0 so B= 0. Subtracting, 2C= -2/5 so C= -1/5.

Another way to get that is to multiply out $A(s^2+ 5)+ (Bs+ C)s=(A+ B)s^2+ Cs+ 5A= 0s^2+ 0s+ 1$ . Now, because that is true for all s, we must have "corresponding coefficients" equal: A+ B= 0, C= 0, 5A= 1. That gives A= 1/5 and then (1/5)+ B= 0 so B= -1/5.
October 22nd 2012, 03:55 PM
Ragnarok

Re: Partial fractions question

Aah! Thanks a lot!