Question Number 84477 by jagoll last updated on 13/Mar/20
$$\left(\mathrm{ycos}\:\mathrm{x}+\mathrm{2xe}^{\mathrm{y}} \right)\mathrm{dx}+\left(\mathrm{sin}\:\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\mathrm{y}} −\mathrm{1}\right)\mathrm{dy}=\mathrm{0} \\ $$
Answered by john santu last updated on 13/Mar/20
![(∂M/∂y) = cos x+2xe^y (∂N/∂x) = cos x+2xe^y ⇒ (∂M/∂y) =(∂N/∂x) diff equation exact F(x,y) = ∫ (ycos x+2xe^y )dx + g(y) = ysin x+x^2 e^y +g(y) (∗) g′(y)= N(x,y)−(∂/∂y)[ ysin x+x^2 e^y ] g′(y)= sin x+x^2 e^y −1−[ sin x+x^2 e^y ] g(y)=∫ −dy = −y ∴ F(x,y) = C ∴ ysin x+x^2 e^y −y = C](https://www.tinkutara.com/question/Q84482.png)
$$\frac{\partial\mathrm{M}}{\partial\mathrm{y}}\:=\:\mathrm{cos}\:\mathrm{x}+\mathrm{2xe}^{\mathrm{y}} \\ $$$$\frac{\partial\mathrm{N}}{\partial\mathrm{x}}\:=\:\mathrm{cos}\:\mathrm{x}+\mathrm{2xe}^{\mathrm{y}} \:\Rightarrow\:\frac{\partial\mathrm{M}}{\partial\mathrm{y}}\:=\frac{\partial\mathrm{N}}{\partial\mathrm{x}} \\ $$$$\mathrm{diff}\:\mathrm{equation}\:\mathrm{exact} \\ $$$$\mathrm{F}\left(\mathrm{x},\mathrm{y}\right)\:=\:\int\:\left(\mathrm{ycos}\:\mathrm{x}+\mathrm{2xe}^{\mathrm{y}} \right)\mathrm{dx}\:+\:\mathrm{g}\left(\mathrm{y}\right) \\ $$$$=\:\mathrm{ysin}\:\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\mathrm{y}} \:+\mathrm{g}\left(\mathrm{y}\right) \\ $$$$\left(\ast\right)\:\mathrm{g}'\left(\mathrm{y}\right)=\:\mathrm{N}\left(\mathrm{x},\mathrm{y}\right)−\frac{\partial}{\partial\mathrm{y}}\left[\:\mathrm{ysin}\:\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\mathrm{y}} \:\right] \\ $$$$\mathrm{g}'\left(\mathrm{y}\right)=\:\mathrm{sin}\:\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\mathrm{y}} −\mathrm{1}−\left[\:\mathrm{sin}\:\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\mathrm{y}} \:\right] \\ $$$$\mathrm{g}\left(\mathrm{y}\right)=\int\:−\mathrm{dy}\:=\:−\mathrm{y}\: \\ $$$$\therefore\:\mathrm{F}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{C} \\ $$$$\therefore\:\mathrm{ysin}\:\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\mathrm{y}} −\mathrm{y}\:=\:\mathrm{C} \\ $$
Answered by TANMAY PANACEA last updated on 13/Mar/20
$${ycosxdx}+{sinxdy}+\mathrm{2}{xe}^{{y}} {dx}+{x}^{\mathrm{2}} {e}^{{y}} {dy}−{dy}=\mathrm{0} \\ $$$${yd}\left({sinx}\right)+{sinxd}\left({y}\right)+{e}^{{y}} {d}\left({x}^{\mathrm{2}} \right)+{x}^{\mathrm{2}} {d}\left({e}^{{y}} \right)−{dy}=\mathrm{0} \\ $$$${d}\left({y}.{sinx}\right)+{d}\left({x}^{\mathrm{2}} .{e}^{{y}} \right)−{dy}={dC} \\ $$$${ysinx}+{x}^{\mathrm{2}} {e}^{{y}} −{y}={C} \\ $$