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(x^2 y^2 1)dx x(x-2y)dy=0-Let's look more closely at how d dx (y 2) becomes 2y dy dx The Chain Rule says du dx = du dy dy dx Substitute in u = y 2 d dx (y 2) = d dy (y 2) dy dx And then d dx (y 2) = 2y dy dx Basically, all we did was differentiate with respect to y and multiply by dy dx 12 Orodruin said The differential equation is not well defined in (x,y) = (1,1) as you have an expression of the form 0/0 for dy/dx Oh yes
(x^2 y^2 1)dx x(x-2y)dy=0のギャラリー
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