Question

I am trying to seperate 1/(s^2+1)(s^2+4) using partial fractions

Answer 1

Answer:

$\frac{1/3}{s^2+1}- \frac{1/3}{s^2+4}$

Step-by-step explanation:

You already have the bottom in factors, write the partial fraction in the form As+b for each factor

$\frac{1}{(s^2+1)(s^2+4)} =\frac{As+B}{s^2+1}+ \frac{Cs+D}{s^2+4}$

Multiply both sides by the bottom to stop using fractions:

$\frac{1}{(s^2+1)(s^2+4)}(s^2+1)(s^2+4)=(\frac{As+B}{s^2+1}+ \frac{Cs+D}{s^2+4})(s^2+1)(s^2+4)$

$1=(As+B)(s^2+4)+(Cs+D)(s^2+1)}$

To find the value of the constants A, B, C, and D, separate the equation by the grade of the s, for example, all that multiplies s^3 in one equation like this:

$0=As^3+Cs^3$

Because in the left part of the equation doesn't have constants multiplying the s^3 we put a zero in that side of the equation, do the same with s^2,s^1, and s^0

$0=4As+Cs\\0=Bs^2+Ds^2\\1=4B+D$

In this case, you can group the equation by constant to solve. First A and D, then B and C.

$As^3+Cs^3=0\\4As+Cs=0$

$Bs^2+Ds^2=0\\4B+D=1$

Solving for A and C:

$(A+C)(s^3)=0\\(4A+C)s=0$  

$A+C=0\\4A+C=0$

This system only has one solution A=0 and C=0.

Solving for B and D:

$(B+D)(s^2)=0\\4B+D=1$

$\\B+D=0\\4B+D=1\\\\D=-B\\4B-B=1\\3B=1\\B=1/3\\D=-1/3$

Then the solution is:

$\frac{1/3}{s^2+1}- \frac{1/3}{s^2+4}$