solve the initial value problem dy/dx=2yx-x^2y and determine where the solution attains its minimal value

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solve the initial value problem dy/dx=2yx-x^2y and determine where the solution attains its minimal value

Mathematics
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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
but how to get the value of c
C is a constant - it can be anything.

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Other answers:

Because if you differentiate a constant, you get 0
the value of "C" will depend on the point that is provided in the initial question. Otherwise, "C" is any constant....
but i got to solve it by separable equation
yes y(0)=3
Right, right - plug in that initial to find C.
i will type the question again
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...
I saw the question. You need your initial value (x = 0 y = __)
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
im sending the solution now
solution; separating variables gives; y^-2dy=(2+x)dx
Get out of here if you're not willing to listen to us.
-y^-1=2x+x^2/2+C
OK GO ON
so explain me plz

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