Consider the differential equation dy/dx = -1 + (y^2/ x).

Question

Let y = g(x) be the particular solution to the differential equation dy/ dx = -1 + (y^2/ x) with initial condition g(4) = 2. Does g have a relative minimum, a relative maximum, or neither at ? Justify your answer.

Answer 1

Well where is dy/dx = 0 ? (horizontal either max or min or inflection)
dy/dx = -1 + (y^2/ x) = 0
or y^2 = x
It said start at (2, 4)
well that very spot is horizontal 4 = 2^2
now is it a min or a max there?
what is d^2y/dx^2 ? at (2,4) ?
dy/dx = -1 + (y^2/ x)
d/dx(dy/x) = 0 + d/dx(y^2/x)= [x(-2y dy/dx) -y^2] /x^2
at (2,4)= [ 2*-8* (0 at that point) - 16 ] / 4 = -32/4 oh that is a maximum because headed down