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mahmit2012 Group Title

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

  • 2 years ago
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    |dw:1337540828547:dw|

    • 2 years ago
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    1 <= y <= infinity ??

    • 2 years ago
  3. mahmit2012 Group Title
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    sorry that is more compact than that you find.

    • 2 years ago
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    is it between 1.5 and 2 ??

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    both negative and positive?

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  6. mahmit2012 Group Title
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    f(0)=1 so 1 belonging to interval.

    • 2 years ago
  7. mahmit2012 Group Title
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    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|

    • 2 years ago
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    I didn't have a clue http://www.hostsrv.com/webmaa/app1/MSP/webm1001/IntegralCurves

    • 2 years ago
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  10. mahmit2012 Group Title
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    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.

    • 2 years ago
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