# Elimination of Arbitrary Constants

• Jan 31st 2011, 12:12 PM
sparky
Elimination of Arbitrary Constants
How do I eliminate the arbitrary constant in the equation y = ax^2 -1?

The answer at the back of the book is 2y = x(dy/dx)-2 however I don't understand how they arrived at this.

Here is my attempt:

y = ax^2 -1

dy/dx = 2ax

(dy/dx) (1/2x) = a

therefore y = (dy/dx) (1/2x)x^2 - 1

Where did I go wrong?
• Jan 31st 2011, 12:18 PM
Quote:

Originally Posted by sparky
How do I eliminate the arbitrary constant in the equation y = ax^2 -1?

The answer at the back of the book is 2y = x(dy/dx)-2 however I don't understand how they arrived at this.

Here is my attempt:

y = ax^2 -1

dy/dx = 2ax

(dy/dx) (1/2x) = a

therefore y = (dy/dx) (1/2x)x^2 - 1

Where did I go wrong?

Stopping too soon ?

$\displaystyle\ y=\frac{dy}{dx}\left(\frac{1}{2x}\right)x^2-1$

$2y=2\displaystyle\left[\frac{dy}{dx}\left(\frac{x}{2}\right)-1\right]$
• Jan 31st 2011, 12:23 PM
TheEmptySet
Quote:

Originally Posted by sparky
How do I eliminate the arbitrary constant in the equation y = ax^2 -1?

The answer at the back of the book is 2y = x(dy/dx)-2 however I don't understand how they arrived at this.

Here is my attempt:

y = ax^2 -1

dy/dx = 2ax

(dy/dx) (1/2x) = a

therefore y = (dy/dx) (1/2x)x^2 - 1

Where did I go wrong?

You didn't do anything wrong you just need to keep going. Your answer is equivalent.

$\displaystyle y=\frac{dy}{dx}\left( \frac{1}{2x}\right)x^2-1 \iff y=\left( \frac{x}{2} \right) \frac{dy}{dx}-1 \iff 2y=x\frac{dy}{dx}-2$

Edit: too slow
• Jan 31st 2011, 01:02 PM
sparky