# About simple first order differential equation

1. Mar 24, 2012

### jwqwerty

1. The problem statement, all variables and given/known data
dy/dx = (3x^2+4x+2)/2(y-1) , y(0)=1

2. Relevant equations

3. The attempt at a solution
I get the answer and the steps are shown:

2(y-1)dy=(3x^2+4x+2)dx and integrate both sides
y^2-2y=x^3+2x^2+2x+c
By initial condition, c=-1 and by solving for y,
y = 1±√(x^3+2x^x+2x) (x>0)

But i do not understand why x≥0 isn't the answer. If it is because of y≠1 (x=0 when y=1) by the equation above, then why does initial condition hold true?