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Question Number 68855 by ramirez105 last updated on 16/Sep/19

Commented by kaivan.ahmadi last updated on 16/Sep/19

set M=6x+y^2   and N=y(2x−3y)  (∂M/∂y)=2y =(∂N/∂x)⇒the equation is exact  (∂u/∂x)=6x+y^2    and  (∂u/∂y)=y(2x−3y)=2xy−3y^2   ⇒u(x,y)=∫(6x+y^2 )dx=3x^2 +xy^2 +h(y)  ⇒(∂u/∂y)=2xy−3y^2 =2xy+((d(h))/dy)⇒((d(h))/dy)=−3y^2 ⇒  h=−y^3 +c′  u(x,y)=c⇒3x^2 +xy^2 −y^3 +c′=c⇒  u(x,y)=3x^2 +xy^2 −y^3 =C

$${set}\:{M}=\mathrm{6}{x}+{y}^{\mathrm{2}} \:\:{and}\:{N}={y}\left(\mathrm{2}{x}−\mathrm{3}{y}\right) \\ $$$$\frac{\partial{M}}{\partial{y}}=\mathrm{2}{y}\:=\frac{\partial{N}}{\partial{x}}\Rightarrow{the}\:{equation}\:{is}\:{exact} \\ $$$$\frac{\partial{u}}{\partial{x}}=\mathrm{6}{x}+{y}^{\mathrm{2}} \:\:\:{and}\:\:\frac{\partial{u}}{\partial{y}}={y}\left(\mathrm{2}{x}−\mathrm{3}{y}\right)=\mathrm{2}{xy}−\mathrm{3}{y}^{\mathrm{2}} \\ $$$$\Rightarrow{u}\left({x},{y}\right)=\int\left(\mathrm{6}{x}+{y}^{\mathrm{2}} \right){dx}=\mathrm{3}{x}^{\mathrm{2}} +{xy}^{\mathrm{2}} +{h}\left({y}\right) \\ $$$$\Rightarrow\frac{\partial{u}}{\partial{y}}=\mathrm{2}{xy}−\mathrm{3}{y}^{\mathrm{2}} =\mathrm{2}{xy}+\frac{{d}\left({h}\right)}{{dy}}\Rightarrow\frac{{d}\left({h}\right)}{{dy}}=−\mathrm{3}{y}^{\mathrm{2}} \Rightarrow \\ $$$${h}=−{y}^{\mathrm{3}} +{c}' \\ $$$${u}\left({x},{y}\right)={c}\Rightarrow\mathrm{3}{x}^{\mathrm{2}} +{xy}^{\mathrm{2}} −{y}^{\mathrm{3}} +{c}'={c}\Rightarrow \\ $$$${u}\left({x},{y}\right)=\mathrm{3}{x}^{\mathrm{2}} +{xy}^{\mathrm{2}} −{y}^{\mathrm{3}} ={C} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by ramirez105 last updated on 17/Sep/19

thank you sir!

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}! \\ $$$$ \\ $$

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