1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

About simple first order differential equation

  1. Mar 24, 2012 #1
    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?
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?