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

(2xy−sec^2 x) dx + (x^2 +2y)dy = 0

$$\left(\mathrm{2}{xy}−\mathrm{sec}\:^{\mathrm{2}} {x}\right)\:{dx}\:+\:\left({x}^{\mathrm{2}} +\mathrm{2}{y}\right){dy}\:=\:\mathrm{0} \\ $$

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

$${This}\:{is}\:{exact}\:{diff}\:{eq}\: \\ $$$${Here}\:{M}\left({x},{y}\right)=\:\mathrm{2}{xy}−\mathrm{sec}\:^{\mathrm{2}} {x}\: \\ $$$$\frac{\partial{M}}{\partial{y}}\:=\:\mathrm{2}{x}\:=\:\frac{\partial{N}}{\partial{x}}\: \\ $$$$\mathcal{F}\left({x},{y}\right)\:=\:\int\:\left(\mathrm{2}{xy}−\mathrm{sec}\:^{\mathrm{2}} {x}\right){dx}+{g}\left({y}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{x}^{\mathrm{2}} {y}−\mathrm{tan}\:{x}\:+{g}\left({y}\right) \\ $$$$\left(\rightarrow\right)\frac{\partial\mathcal{F}}{\partial{y}}\:=\:{N}\left({x},{y}\right) \\ $$$${x}^{\mathrm{2}} +\:{g}'\left({y}\right)\:=\:{x}^{\mathrm{2}} +\mathrm{2}{y}\: \\ $$$${g}'\left({y}\right)\:=\:\mathrm{2}{y}\:\Rightarrow{g}\left({y}\right)\:=\int\mathrm{2}{y}\:{dy}\:=\:{y}^{\mathrm{2}} \\ $$$$\therefore\:{we}\:{have}\:\mathcal{F}\left({x},{y}\right)\:=\:{x}^{\mathrm{2}} {y}\:−\mathrm{tan}\:{x}+{y}^{\mathrm{2}} ={C} \\ $$

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