Rewriting $1/(v^2-(mg/k))$ as two fractions

Question

I'm looking at the solution someone gave to me and I'm having a bit of trouble following one of the steps. This step in particular,

$\frac{1}{k}\frac{1}{v^2-\frac{mg}{k}}=\frac{1}{2k\sqrt{mg/k}}(\frac{1}{v-\sqrt{mg/k}}- \frac{1}{v+\sqrt{mg/k}})$. All letters are constants except for $v$. Any help is appreciated.

Answer 1

Note that $$v^2 - \frac{mg}{k} = (v-\sqrt{mg/k})(v+\sqrt{mg/k})$$ So we can use partial fraction decomposition, which will result in the formula you posted.

---

$$\frac{1}{k}\frac{1}{v^2-\frac{mg}{k}}=\frac{1}{k}\left(\frac{A}{v-\sqrt{mg/k}}+ \frac{B}{v+\sqrt{mg/k}}\right)$$

If you solve for $A, B$ as one does in partial fraction decomposition, you'll find that $$A =\frac 1{2\sqrt{(mg)/k}},\;B = -A$$

This gives us, as desired,

$$\frac{1}{k}\frac{1}{v^2-\frac{mg}{k}}=\frac{1}{2k\sqrt{\frac{mg}{k}}}\left(\frac{1}{v-\sqrt{\frac{mg}{k}}}-\frac{1}{v+\sqrt{\frac{mg}{k}}}\right)$$

Answer 2

$$\frac{1}{k}\frac{1}{v^2-\frac{mg}{k}}=\frac{1}{k}\frac{1}{v+\sqrt\frac{mg}{k}}\frac{1}{v-\sqrt\frac{mg}{k}}=$$ $$\ =\frac{1}{k}(\frac{A}{v+\sqrt\frac{mg}{k}}+\frac{B}{v-\sqrt\frac{mg}{k}})$$ $$\text{now you bring every term in the brackets under a common denominator}$$ $$\text{so you get:}\begin{cases} A+B=0\\ -A+B= \sqrt\frac{k}{mg} \end{cases} \iff \begin{cases} A=-\frac{1}{2\sqrt\frac{mg}{k}}\\ B= \frac{1}{2\sqrt\frac{mg}{k}} \end{cases} $$

Answer 3

consider the right hand side we know that in general

$$\frac1a-\frac1b = \frac{b-a}{ab}$$

here $$a=v-\sqrt{mg/k} , b=v+\sqrt{mg/k}$$

therefore $$b-a = 2\sqrt{mg/k}$$

and since $$x^2-y^2=(x+y)(x-y)$$

therefore $$a*b = v^2-\frac{mg}k$$

and again $$\frac{b-a}{ab}=\frac{2\sqrt{mg/k}}{v^2-\frac{mg}k}$$ This value multiplied by $$\frac1{2k\sqrt{\frac{mg}k}}$$

provides you the left hand side.