Getting stuck with partial fractions

**steph3824** · April 10th 2010, 09:38 AM

Ok this is driving me nuts so I hope someone can help!

$8s^2-4s+12/s(s^2+4)$

I am using partial fraction decomposition of this to find the A, B and C. So $8s^2-4s+12=A/s +[B/(s-0)^2+4] +[C/(s-0)^2+4]$ and thus $8s^2-4s+12=A[(s-0)^2+4]+Bs+Cs$

When I use the sub values in for S method, I am subbing in a 0 for the S, and this gets me 12=4A so A=3. BUT when you do it with the expand and compare method, I get A=8.

Where am I going wrong here??

**maddas** · April 10th 2010, 10:19 AM

${12\over s(s^2+4)} = {3\over s} - {3s \over s^2 + 4}$ . Or maybe you want ${8s^2-4s +12\over s(s^2+4)} = {3\over s} + {5s-4\over s^2 + 4}$ , I can't really tell, sorry.

**zzzoak** · April 10th 2010, 02:36 PM

$<br /> \frac{8s^2-4s+12}{s(s^2+4)}=\frac{A}{s}+\frac{Bs+C}{s^2+4}=\f rac{3}{s}+\frac{5s-4}{s^2+4}<br />$ .
There is no solution for
$<br /> \frac{8s^2-4s+12}{s(s^2+4)}\ne\frac{A}{s}+\frac{B}{s^2+4}=\fr ac{A(s^2+4)+Bs}{s(s^2+4)}=\frac{As^2+Bs+4A}{s(s^2+ 4)}<br />$
A=8
B=-4
4A=12 $\Rightarrow$ A=3
because the same time must be A=8 and A=3 what is impossible.

Thread: Getting stuck with partial fractions

LinkBack

Thread Tools

Display

Getting stuck with partial fractions

Similar Math Help Forum Discussions

[SOLVED] Partial derivative. Completely stuck

Partial fractions...stuck on one part

[SOLVED] Partial Fractions/algebraic fractions

Laplace transform - stuck on the partial fraction

[SOLVED] Stuck on improper fractions in integrals

Search Tags