# matching the terms in partial fractions

• Jan 5th 2013, 02:03 PM
cxz7410123
matching the terms in partial fractions
does anybody know where this "1" comes from?
• Jan 5th 2013, 02:20 PM
skeeter
Re: matching the terms in partial fractions
expanding the RHS yields the quadratic terms \$\displaystyle 2Au^2 + Cu^2 = (2A+C)u^2\$

note the coefficient of \$\displaystyle u^2\$ on the LHS is 1 , hence by matching coefficients \$\displaystyle 2A+C = 1\$
• Jan 5th 2013, 02:20 PM
Deveno
Re: matching the terms in partial fractions
u2 = (1)(u2), that is, in the polynomial u2, the coefficient of u2 is 1 (this is usually suppressed when we write polynomials).
• Jan 5th 2013, 02:56 PM
cxz7410123
Re: matching the terms in partial fractions
Thank you very much.
Your answer rises me an other problem, why in the second scan there is this "0" and not 3?.

Attachment 26483
• Jan 5th 2013, 03:13 PM
skeeter
Re: matching the terms in partial fractions
Quote:

Originally Posted by cxz7410123
Thank you very much.
Your answer rises me an other problem, why in the second scan there is this "0" and not 3?.

Attachment 26483

LHS is \$\displaystyle {\color{red}0}x^2 + 0x + 3\$

you're matching coefficients, remember?