# Thread: Elimination of Arbitrary Constants

1. ## 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?

2. 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 ?

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

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

3. 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

Your workings have brought me closer to understanding this.

Why do we have to multiply both sides by 2?

5. Only to remove the "inconvenient" fraction (a half) on the right, nothing more!
Even without going that far, you were still correct where you finished up.

6. Ok great, thanks a lot Archie Meade and TheEmptySet