SOLUTION: Consider the initial value problem y′=3y^2,y(0)=y0. For what value(s) of y0 will the solution have a vertical asymptote at t=2 and a t-interval of existence −∞<t<2? y0

Algebra ->  Expressions -> SOLUTION: Consider the initial value problem y′=3y^2,y(0)=y0. For what value(s) of y0 will the solution have a vertical asymptote at t=2 and a t-interval of existence −∞<t<2? y0      Log On


   



Question 1184040: Consider the initial value problem
y′=3y^2,y(0)=y0.
For what value(s) of y0 will the solution have a vertical asymptote at t=2 and a t-interval of existence −∞ y0= My answer is 1/21 but it's wrong.

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The particular y%5B0%5D value that will satisfy the IVP is 1%2F6.
The function that will satisfy the IVP is y%28t%29+=+%281%2F3%29%2F%282-t%29.

Note that at t = 2 there is a vertical asymptote because the denominator is zero there.
Also the maximal interval of existence is (-infinity, 2), since the initial condition can be found in this interval, and not at t > 2.