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mahmit2012
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
|dw:1337540828547:dw|
1 <= y <= infinity ??
sorry that is more compact than that you find.
is it between 1.5 and 2 ??
both negative and positive?
f(0)=1 so 1 belonging to interval.
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|
I didn't have a clue http://www.hostsrv.com/webmaa/app1/MSP/webm1001/IntegralCurves
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.