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!

Initial Value Problem

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve the initial value problem
    [tex]y'=y^2 , y(0)=1[/tex] and determine the interval in which the solution exists.

    3. The attempt at a solution
    So after solving the differential equation, we get
    [tex] y= 1/(1-t) [/tex] However, I don't understand why the interval is only [tex] (- \infty , 1 ) [/tex] and not [tex](- \infty, 1 ) U ( 1 , \infty)[/tex]. From the y(x) equation, we see that it is only not defined when t=1, so why don't we include the other interval as a solution? Could someone please enlighten me please.
  2. jcsd
  3. Sep 14, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    You think of the differential equation as describing how the system evolves starting from t=0 to another value of t. You can't reach t in (1,inf) without going through the singularity at t=1.
  4. Sep 14, 2008 #3
    Well, when you graph the differential equation y'=y^2, the function of course only exists above the x axis (because y^2 will always make any y value positive). Therefore, when you solve the differential equation and get y=1/(1-t), any t value you plug may or may not make the y= value positive/negative; however, the slope will always be positive, so the graph should only go from [tex] (- \infty , 1 ) [/tex]. Note the asymptotes at x=1 and y=0, anything to the right of x=1 will be negative but the differential (y^2) will only account for positive values.
  5. Sep 14, 2008 #4
    Oh, so the value t=0 which was part of the initial condition y(0)=1, must be contained in a valid interval until there is discontinuity on either ends or goes to +/-infinity?

    Thanks so much guys for your valuable inputs.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Initial Value Problem
  1. Initial value problem (Replies: 7)

  2. Initial-value problem (Replies: 2)