Differential Equation Questions

Question Number 105710 by john santu last updated on 31/Jul/20

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

Answered by Smail last updated on 31/Jul/20

$${let}\:{M}=\mathrm{2}{xy}+{cosy}\:{and}\:\:{N}={x}^{\mathrm{2}} −{xsiny}−\mathrm{2}{y} \\$$$$\frac{{dM}}{{dy}}=\mathrm{2}{x}−{siny} \\$$$$\frac{{dN}}{{dx}}=\mathrm{2}{x}−{siny} \\$$$${so}\:\frac{{dN}}{{dx}}=\frac{{dM}}{{y}} \\$$$${This}\:{means}\:{that}\:{the}\:{equation}\:{is}\:{exact}\:{and} \\$$$${there}\:{a}\:{function}\:{f}\left({x},{y}\right)={c}\:{such}\:{that}\:\frac{{df}}{{dx}}={M}\:{and}\:\frac{{df}}{{dy}}={N} \\$$$$\frac{{df}}{{dx}}={M}=\mathrm{2}{xy}+{cosy} \\$$$${f}\left({x},{y}\right)={x}^{\mathrm{2}} {y}+{xcosy}+{q}\left({y}\right) \\$$$$\frac{{df}}{{dy}}={x}^{\mathrm{2}} −{xsiny}+{q}'\left({y}\right)={x}^{\mathrm{2}} −{xsiny}−\mathrm{2}{y} \\$$$${q}'\left({y}\right)=−\mathrm{2}{y}\Rightarrow{q}\left({y}\right)=−{y}^{\mathrm{2}} \\$$$${thus} \\$$$${f}\left({x},{y}\right)={x}^{\mathrm{2}} {y}+{xcosy}−{y}^{\mathrm{2}} \\$$

Answered by bobhans last updated on 31/Jul/20

$$\frac{\partial{M}}{\partial{y}}\:=\:\mathrm{2}{x}−\mathrm{sin}\:{y}\:;\:\frac{\partial{N}}{\partial{x}}\:=\:\mathrm{2}{x}−\mathrm{sin}\:{y} \\$$$$\Leftrightarrow\:\frac{\partial{M}}{\partial{y}}\:=\:\frac{\partial{N}}{\partial{x}}\:\Leftrightarrow\:{exact}\:{diff}\:{equation} \\$$$${F}\left({x},{y}\right)=\:\int\:\left(\mathrm{2}{xy}+\mathrm{cos}\:{y}\right)\:{dx}\:+\:{g}\left({y}\right) \\$$$${F}\left({x},{y}\right)=\:{x}^{\mathrm{2}} {y}+{x}\mathrm{cos}\:{y}\:+\:{g}\left({y}\right) \\$$$${g}\:'\left({y}\right)=\:\left({x}^{\mathrm{2}} −{x}\mathrm{sin}\:{y}−\mathrm{2}{y}\right)−\frac{\partial}{\partial{y}}\left[\:{x}^{\mathrm{2}} {y}+{x}\mathrm{cos}\:{y}\:\right] \\$$$${g}\left({y}\right)\:=\:\int\:\left({x}^{\mathrm{2}} −{x}\mathrm{sin}\:{y}−\mathrm{2}{y}−\left({x}^{\mathrm{2}} −{x}\mathrm{sin}\:{y}\right)\right){dy} \\$$$${g}\left({y}\right)\:=\:\int\left(−\mathrm{2}{y}\right){dy}\:=\:−{y}^{\mathrm{2}} \\$$$$\mathcal{S}{olution}\: \\$$$${x}^{\mathrm{2}} {y}\:+\:{x}\mathrm{cos}\:{y}−{y}^{\mathrm{2}} \:=\:{C}\:\bigstar \\$$

Answered by 1549442205PVT last updated on 31/Jul/20

$$\mathrm{This}\:\mathrm{is}\:\mathrm{an}\:\mathrm{full}\:\mathrm{differential}\:\mathrm{equation} \\$$$$\mathrm{since}\:\mathrm{P}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{2xy}+\mathrm{cosy},\mathrm{Q}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{xsiny}−\mathrm{2y} \\$$$$\mathrm{and}\:\frac{\partial\mathrm{P}}{\partial\mathrm{y}}=\frac{\partial\mathrm{Q}}{\partial\mathrm{x}}=\mathrm{2x}−\mathrm{siny} \\$$$$\mathrm{We}\:\mathrm{find}\:\mathrm{general}\:\mathrm{integration}\:\mathrm{byformular} \\$$$$\int_{\mathrm{x}_{\mathrm{0}} } ^{\mathrm{x}} \mathrm{P}\left(\mathrm{x},\mathrm{y}\right)\mathrm{dx}+\int_{\mathrm{y}_{\mathrm{0}} } ^{\:\mathrm{y}} \mathrm{Q}\left(\mathrm{x},\mathrm{y}\right)\mathrm{dy}=\mathrm{C} \\$$$$\mathrm{Take}\:\mathrm{x}_{\mathrm{0}} =\mathrm{0},\mathrm{y}_{\mathrm{0}} =\mathrm{0},\mathrm{we}\:\mathrm{obtain}: \\$$$$\int_{\mathrm{0}} ^{\:\mathrm{x}} \left(\mathrm{2xy}+\mathrm{cosy}\right)\mathrm{dx}+\int_{\mathrm{0}} ^{\:\mathrm{y}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{xsiny}−\mathrm{2y}\right)\mathrm{dy} \\$$$$\Leftrightarrow\left(\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{xcosy}\right)\mid_{\mathrm{0}} ^{\mathrm{x}} +\left(\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{xcosy}−\mathrm{y}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\mathrm{y}} =\mathrm{C}_{\mathrm{1}} \\$$$$\Leftrightarrow\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{xcosy}+\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{xcosy}−\mathrm{y}^{\mathrm{2}} −\mathrm{x}=\mathrm{C}_{\mathrm{1}} \\$$$$\Leftrightarrow\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{xcosy}\right)−\mathrm{x}−\mathrm{y}^{\mathrm{2}} =\mathrm{C} \\$$