# Seperable equations

1. Apr 28, 2006

### davegillmour

Ok, I'm having trouble with the last part of this problem. After solving for the constant C, I get:
y^2 - 2y = x^3 + 2x^2 +2x + 3

My question is, how do I solve this in terms of Y? The only instruction my book gives me is "To obtain the solution explicitly we must solve for y in terms of x. This is a simple matter in this case since the equation is quadratic in y" and then they jump to the solution:
y = 1 +/- sqrt(x^3 + 2x^2 +2x + 4)

2. Apr 28, 2006

### dav2008

Well you said it yourself: it's a quadratic in terms of y.

Do you know how to solve quadratic equations?

3. Apr 28, 2006

### davegillmour

So would the expression x^3 + 2x^2 +2x + 3 be treated as the 'c' value in the quadratic eq?

4. Apr 28, 2006

### dav2008

Yep.

Remember that the quadratic equation lets you solve for a certain variable (we'll call it w here) when you have an equation in the form of $aw^2+bw+c=0$

5. Apr 28, 2006

### davegillmour

Alright I got it, thanks a lot

6. Apr 29, 2006

### HallsofIvy

Staff Emeritus
Actually, no. Since your equation is y^2 - 2y = x^3 + 2x^2 +2x + 3
and the quadratic is normally written ay^2+ by+ c= 0,
c= -x^3- 2x^2- 2x- 3.