Question Number 104592 by bemath last updated on 22/Jul/20

(2xy−sec ^2 x) dx + (x^2 +2y)dy = 0

Answered by john santu last updated on 28/Jul/20

This is exact diff eq   Here M(x,y)= 2xy−sec ^2 x   (∂M/∂y) = 2x = (∂N/∂x)   F(x,y) = ∫ (2xy−sec ^2 x)dx+g(y)                   = x^2 y−tan x +g(y)  (→)(∂F/∂y) = N(x,y)  x^2 + g′(y) = x^2 +2y   g′(y) = 2y ⇒g(y) =∫2y dy = y^2   ∴ we have F(x,y) = x^2 y −tan x+y^2 =C