Please draw the slope field for y\'= x+y^2 - please explain how you get the slop

Question

Please draw the slope field for y'= x+y^2 - please explain how you get the slope at each point. Lastly, please sketch the solution curve that passes through the point (0,0)

Explanation / Answer

i checked wikipedia.org and google for the topic and what I saw " sucked " if dy / dx = f(x,y) and you desire a slope field on the x-y plane then sketch f(x,y) = constant for about 10 different choices of the constant. After you sketch each curve then draw 'tiny' line segments { at least 20 } with slope = constant through the curve. Repeat the process...tedious. You should now somewhat see the behavior of the solution curve..for a particular solution sketch a curve containing the segments { smoothly } 1 . sketch 0 = 1 - xy {hyperbola xy = 1 } and draw 'tiny ' with slope 0...1 = 1 - xy--> xy =0 , the axes with 'tiny' slopes of 1, ....-1 = 1-xy---> xy = 2 {hyperbola with 'tiny' slope of -1 } , continue...finally hold the paper about one foot away to see the behavior of the ' solutions'...the solution through the origin will be similar to an inverse tangent function the solution is e^(-x²/2) [ int over tin [0,x] of {e^(t²/2) dt} + C e^(-x²/2) , no trivial function

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