Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

gorica

  • 3 years ago

I have initial problem y'=y^2 y(0)=1. I solved it and I got that it's solution is y=1/(1-t). In book, there's written that the solution exists only in the interval −∞ < t < 1. Why? Why not in 1< t < ∞ too?

  • This Question is Closed
  1. amistre64
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    dy/dx = y^2 y^-2 dy = dx ; y!= 0 -y^-1 = x + C ; y!= 0 -y^-1 = x + C ; C = -1 y = 1/(-x+1) since there is a split in the domain; and x=0 is only in the left side of the domain; then the initial value does not exist in the right side of the domain

  2. amistre64
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    something to do with continuity we have to split the function into 2 parts; each one being continuous in its own right i believe

  3. mark_o.
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i think its because the initial value y(0)=1is on the left side of the graph, since you are dealing only on the half left side of the graph,,,try to graph this one and you will see it

  4. gorica
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so, if it was, for example, y(2)=-1, solution would exist only in interval 1< t < ∞ ?

  5. amistre64
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    correct, if the initial value wanted a t value from the other part of the domain, then to maintain continuity, you would opt for the other side :)

  6. gorica
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks :)

  7. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy