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6x-y-2-dx-y-2xy-3y-2-dy-0-




Question Number 124050 by john_santu last updated on 30/Nov/20
(6x+y^2 )dx +y(2xy−3y^2 )dy = 0
$$\left(\mathrm{6}{x}+{y}^{\mathrm{2}} \right){dx}\:+{y}\left(\mathrm{2}{xy}−\mathrm{3}{y}^{\mathrm{2}} \right){dy}\:=\:\mathrm{0}\: \\ $$
Answered by mohammad17 last updated on 30/Nov/20
M=6x+y^2 →M_y =2y→(∗)  N=2xy−3y^2 →N_x =2y→(∗∗)  from (∗)and(∗∗) the equation is exact    ∫Mdx+∫Ndy=∫d(c)    ∫(6x+y^2 )dx+∫(2xy−3y^2 )dy=∫d(c)    3x^2 +2xy^2 +xy−y^3 =C    by:⟨M.O⟩
$${M}=\mathrm{6}{x}+{y}^{\mathrm{2}} \rightarrow{M}_{{y}} =\mathrm{2}{y}\rightarrow\left(\ast\right) \\ $$$${N}=\mathrm{2}{xy}−\mathrm{3}{y}^{\mathrm{2}} \rightarrow{N}_{{x}} =\mathrm{2}{y}\rightarrow\left(\ast\ast\right) \\ $$$${from}\:\left(\ast\right){and}\left(\ast\ast\right)\:{the}\:{equation}\:{is}\:{exact} \\ $$$$ \\ $$$$\int{Mdx}+\int{Ndy}=\int{d}\left({c}\right) \\ $$$$ \\ $$$$\int\left(\mathrm{6}{x}+{y}^{\mathrm{2}} \right){dx}+\int\left(\mathrm{2}{xy}−\mathrm{3}{y}^{\mathrm{2}} \right){dy}=\int{d}\left({c}\right) \\ $$$$ \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{xy}^{\mathrm{2}} +{xy}−{y}^{\mathrm{3}} ={C} \\ $$$$ \\ $$$${by}:\langle{M}.{O}\rangle \\ $$

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