# Solving an IC

Its funny.. because I got this right on a test.. but.. I'm looking back at it.. and it doesn't make sense how I figured it out...

I had to find the IC of $$d(\frac{x^2}{y})$$

this was my work...

$$d(\frac{x^2}{y}) = 2xy - x^2y^2 - y^2$$
$$d(\frac{x^2}{y}) = \frac{2xy - x^2y}{y^2}$$

I'm just confused because I would of thought that it would of been something like this...

$$d(\frac{x^2}{y}) = 2xy^-^1 + x^2y^-^2$$

hmm...?

HallsofIvy
Homework Helper
Its funny.. because I got this right on a test.. but.. I'm looking back at it.. and it doesn't make sense how I figured it out...

I had to find the IC of $$d(\frac{x^2}{y})$$

this was my work...

$$d(\frac{x^2}{y}) = 2xy - x^2y^2 - y^2$$
$$d(\frac{x^2}{y}) = \frac{2xy - x^2y}{y^2}$$

I'm just confused because I would of thought that it would of been something like this...

$$d(\frac{x^2}{y}) = 2xy^-^1 + x^2y^-^2$$

hmm...?
What do you mean by "IC"? When I saw "Solving an IC" in the differential equation forum, I thought you mean an "initial condition" problem but I see no differential equation at all here!

Certainly, the derivative of x2/y is, by the quotient rule,
$$\frac{2xy- x^2}{y^2}$$
There is no "y" in the second term in the numerator. (I surely don't see where you get your first line
$$d(\frac{x^2}{y}) = 2xy - x^2y^2 - y^2$$
that makes no sense to me.)

Of course, you could also write x2/y as x2y-1 and use the product rule. In that case, the dervative is
$$2xy^{-1}+ (-1)x^2y^{-2}$$
Just rewrite that as
$$\frac{2x}{y}- \frac{x^2}{y^2}$$
multiply the numerator and denominator of the first fraction by y and combine and you get the same answer as with the quotient rule.

Since this has nothing to do with differential equations, I moving it to calculus.

sorry, I didn't know the difference, this was on a DE test

but thanks for explaining it