Arnold F. answered • 04/30/16

College Professor & Expert Tutor In Statistics and Calculus

^{2}+ 36)] = A/s + (Bs+C)/(s

^{2}+ 36)

Michael P.

asked • 04/30/16HI Looking to make 72/36s+s^3 into partial fraction.

Can find the value of A ok but can't seem to cancel out A value to find B. Expanding the denominator I got s(s^2+36), then when I come to cancel out A value a squared value + 36 will not =0.

Thanks for your help in advance.

Michael

More

Arnold F. answered • 04/30/16

College Professor & Expert Tutor In Statistics and Calculus

Not sure exactly what you have but the expression should look like:

72/[s(s^{2} + 36)] = A/s + (Bs+C)/(s^{2} + 36)

then you are solving for A,B, and C.

Is this what you tried?

Philip P. answered • 04/30/16

Effective and Patient Math Tutor

72/(36s+s^{3}) = 72/s(36+s^{2}) = 72/s(s+6i)(s-6i)

We want:

72/s(s+6i)(s-6i) = A/s + B/(s+61) + C/(s-6i)

72/s(s+6i)(s-6i) = [ A(s+6i)(s-6i) + Bs(s-6i) + Cs(s+6i) ] / s(s+6i)(s-6i)

72 = A(s+6i)(s-6i) + Bs(s-6i) + Cs(s+6i)

For s=0:

72 = A*(-36i^{2})

72 = A*36

2 = A

For s = s+6i:

72 = C*6i*12i

72 = C*(-72)

-1 = C

For s = -6i:

72 = b*(-6i)*(-12i)

72 = -c*72

-1 = C

To check it, put the RHS over the common denominator.

John R. answered • 04/30/16

Financial professional with MBA/CPA looking for tutoring

A/s+ (Bs+C)/(s^2 +36)= 72/s(s^2 +36)

As^2 +36A+Bs^2 +Cs=72

A+B=0

C=0

36A=72

A=2

B=-2

So we have 2/s - 2s/(s^2+36)= 72/s(s^2+36)

Arnold F.

John, I was trying to engage Michael in solving his own question...

Report

04/30/16

Ask a question for free

Get a free answer to a quick problem.

Most questions answered within 4 hours.

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.

John R.

04/30/16