A community for students.
Here's the question you clicked on:
 0 viewing
mahmit2012
 2 years ago
Q5
find a best bound of y for x>=0: a<=y=<b a,b=? if
y'=dy/dx=1/(x^2+y^2) x>0 ; y(0)=1
mahmit2012
 2 years ago
Q5 find a best bound of y for x>=0: a<=y=<b a,b=? if y'=dy/dx=1/(x^2+y^2) x>0 ; y(0)=1

This Question is Closed

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1337540828547:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.01 <= y <= infinity ??

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1sorry that is more compact than that you find.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0is it between 1.5 and 2 ??

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0both negative and positive?

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1f(0)=1 so 1 belonging to interval.

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1y'>0 in conclude y is a ascending function and because of y(0)=1 we can show for x>=0 ; y>=1 so a=1 for get a reasonable upper bound of y or b:dw:1337618744074:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0I didn't have a clue http://www.hostsrv.com/webmaa/app1/MSP/webm1001/IntegralCurves

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1yes there are many ways to solve y'=f(x,y) but there is no always a close function for each question. In this case or yours we can only find approximately curve with numerical calculation.
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.