1/((u^2)-4) = A/(u-2) + B/(u+2)
how do u solve for A and B?
i am told that A and B can be easily solved to be A=.25 B=-.25 but i dont see how to get that. Please help
Ahh, I'm going to assume that this is partial fractions decomposition? If so you can do this. Multiplying through by $\displaystyle u^2-4$ gives
$\displaystyle 1=A(u+2)+B(u-2)\quad{\color{red}\star}$
Now we use this little trick, if we let $\displaystyle u=2$ we get
$\displaystyle 1=4A+0B\implies{A=\frac{1}{4}}$
And letting $\displaystyle u=-2$ gives $\displaystyle 1=0A-4B\implies{B=\frac{-1}{4}}$
The other way is to compare coefficents. Expanding the starred step gives
$\displaystyle 1=\left(A+B\right)u+2A-2B$
Now we know that for this expression to be true that each of the coefficients must match. So $\displaystyle A+B=0$ and $\displaystyle 2A-2B=1$. Solving those gives the correct values.
Because think about it if you have something like
$\displaystyle C=A(x+d)+B(x+e)$
And you want to isolate either $\displaystyle A$ or $\displaystyle B$ you want the other one to dissapear, and how do you do that? You make its coefficient zero, so you basically need to use the value of $\displaystyle x$ that makes either $\displaystyle A$ or $\displaystyle B$ zero. In this case it would be $\displaystyle x=-d$ to Isolate $\displaystyle B$ and $\displaystyle x=-e$ for $\displaystyle A$
Consider:
$\displaystyle \frac{A}{u-2}+\frac{B}{u+2}=\frac{A(u+2)+B(u-2)}{(u-2)(u+2)}=\frac{u(A+B)+2(A-B)}{u^2-4}$
Now if:
$\displaystyle \frac{A}{u-2}+\frac{B}{u+2}=\frac{u(A+B)+2(A-B)}{u^2-4}=\frac{1}{u^2-4}$
we must have:
$\displaystyle u(A+B)+2(A-B)=1$
so:
$\displaystyle A+B=0$
and
$\displaystyle A-B=\frac{1}{2}$
Solve these and you will get the given answers.
CB