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?