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The equation (2xy+3) dx+(x2−1) dy=0(2xy + 3)\,dx + (x^{2} - 1)\,dy = 0 is exact. Find the potential function F(x,y)F(x, y) with Fx=2xy+3F_x = 2xy + 3, Fy=x2−1F_y = x^{2} - 1 and F(0,0)=0F(0, 0) = 0; the solutions are then the curves F(x,y)=CF(x, y) = C.
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