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

$$(2xy-\sec {}^{2}x)dx+(x^2+2y)dy=0$$

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

$$Thisisexactdiffeq$$$$HereM(x,y)=2xy-\sec {}^{2}x$$$$\frac{\partial M}{\partial y}=2x=\frac{\partial N}{\partial x}$$$$\mathcal{F}(x,y)=\int (2xy-\sec {}^{2}x)dx+g\left(y\right)$$$$=x^{2}y-\tan x+g\left(y\right)$$$$(\to )\frac{\partial \mathcal{F}}{\partial y}=N(x,y)$$$$x^2+g^{\prime }\left(y\right)=x^2+2y$$$$g^{\prime }\left(y\right)=2y\Rightarrow g\left(y\right)=\int 2ydy=y^2$$$$\therefore wehave\mathcal{F}(x,y)=x^{2}y-\tan x+y^2=C$$

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