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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
 one year ago
 one year 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
 one year ago
 one year ago

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mahmit2012Best ResponseYou've already chosen the best response.1
dw:1337540828547:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
1 <= y <= infinity ??
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
sorry that is more compact than that you find.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
is it between 1.5 and 2 ??
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
both negative and positive?
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
f(0)=1 so 1 belonging to interval.
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
y'>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
 one year ago

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

mahmit2012Best ResponseYou've already chosen the best response.1
yes 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.
 one year ago
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