A community for students.
Here's the question you clicked on:
 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/(1t).
In book, there's written that the solution exists only in the interval −∞ < t < 1. Why? Why not in
1< t < ∞ too?
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/(1t). 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2dy/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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2something to do with continuity we have to split the function into 2 parts; each one being continuous in its own right i believe

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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

gorica
 3 years ago
Best ResponseYou've already chosen the best response.0so, if it was, for example, y(2)=1, solution would exist only in interval 1< t < ∞ ?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2correct, 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 :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.