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dy/dx = 2y^2 and if y = -1 when x = 1. then when x = 2, y=?
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dy/y^2 = 2*dx, now integratin both sides -1/y = 2x + C when x = 1, y=-1 -1/-1 = 2*1 + C 1 = 2 + C C = -1, so... -1/y = 2x -1 When x=2: -1/y = 4 -1 -1/y = 3 y = -1/3
but there is no x in the derivative....
oh, integration on both sides, didn't think of that
yeah, I will make sure to include this type of problems as well, ty
:)
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I thought it was one of those, when the differential dx=... and f(x)=... find dy,. Looked at it wrong.....
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