anonymous
  • anonymous
solve the initial value problem dy/dx=2yx-x^2y and determine where the solution attains its minimal value
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
dy = (2yx - x^2y) dx dy = y(2x - x^2) dx dy / y = (2x - x^2) dx ln(y) = x^2 - x^3 / 3 y = e^((x^2 - x^3) / 3) + C
anonymous
  • anonymous
but how to get the value of c
anonymous
  • anonymous
C is a constant - it can be anything.

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anonymous
  • anonymous
Because if you differentiate a constant, you get 0
amistre64
  • amistre64
the value of "C" will depend on the point that is provided in the initial question. Otherwise, "C" is any constant....
anonymous
  • anonymous
but i got to solve it by separable equation
anonymous
  • anonymous
yes y(0)=3
anonymous
  • anonymous
Right, right - plug in that initial to find C.
anonymous
  • anonymous
i will type the question again
amistre64
  • amistre64
there are an infinite number of "graphs" that have the same derivitive. They all differ by the value of their constant (C). the only way to determine the value of (C) is to have information regarding some point onthe graph that contains (C) to anchor it in...
anonymous
  • anonymous
I saw the question. You need your initial value (x = 0 y = __)
anonymous
  • anonymous
solve the initial value problem dy/dx=2y^2+xy^2, y(0)=1 and determine where the solution attain its minimal value i am sending the same example and solution i got to do the same with the above one
anonymous
  • anonymous
im sending the solution now
anonymous
  • anonymous
solution; separating variables gives; y^-2dy=(2+x)dx
anonymous
  • anonymous
Get out of here if you're not willing to listen to us.
anonymous
  • anonymous
-y^-1=2x+x^2/2+C
anonymous
  • anonymous
OK GO ON
anonymous
  • anonymous
so explain me plz

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