Question Number 103931 by bemath last updated on 18/Jul/20
$$\left({y}^{\mathrm{2}} +\mathrm{2}\right)\:{dx}\:=\:\left({xy}+\mathrm{2}{y}+{y}^{\mathrm{3}} \right)\:{dy} \\ $$
Answered by bobhans last updated on 18/Jul/20
$$\frac{{dy}}{{dx}}\:=\:\frac{{y}^{\mathrm{2}} +\mathrm{2}}{{y}\left({x}+\mathrm{2}+{y}^{\mathrm{2}} \right)} \\ $$$${set}\::\:{q}\:=\:{x}+\mathrm{2}+{y}^{\mathrm{2}} \:\Rightarrow\frac{{dq}}{{dx}}\:=\:\mathrm{1}+\mathrm{2}{y}\:\frac{{dy}}{{dx}} \\ $$$$\mathrm{2}{y}\:\frac{{dy}}{{dx}}\:=\:\frac{{dq}}{{dx}}−\mathrm{1}\Rightarrow\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{2}{y}}\:\frac{{dq}}{{dx}}−\frac{\mathrm{1}}{\mathrm{2}{y}} \\ $$$$\Leftrightarrow\:\frac{\mathrm{1}}{\mathrm{2}{y}}\:\frac{{dq}}{{dx}}−\frac{\mathrm{1}}{\mathrm{2}{y}}\:=\:\frac{{q}−{x}}{{qy}}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{dq}}{{dx}}−\mathrm{1}\right)\:=\mathrm{1}−\frac{{x}}{{q}}\: \\ $$$$\frac{{dq}}{{dx}}\:−\mathrm{1}\:=\:\mathrm{2}−\frac{\mathrm{2}{x}}{{q}}\:\Rightarrow\frac{{dq}}{{dx}}\:=\:\mathrm{3}−\frac{\mathrm{2}{x}}{{q}} \\ $$$$\frac{{dq}}{{dx}}\:+\:\frac{\mathrm{2}{x}}{{q}}\:=\:\mathrm{3}\: \\ $$
Answered by bramlex last updated on 18/Jul/20
$$\frac{\partial{M}}{\partial{y}}\:=\:\mathrm{2}{y}\:;\:\frac{\partial{N}}{\partial{x}}\:=\:−{y}\:\Rightarrow{non}\:{exact}\: \\ $$$$\frac{\partial{N}}{\partial{x}}−\frac{\partial{M}}{\partial{y}}\:=\:−\mathrm{3}{y}\:{is}\:{a}\:{function}\:{only}\:{in}\:{y}\:.\: \\ $$$${integrating}\:{factor}\:{u}\left({y}\right)={e}^{\int\left(\frac{−\mathrm{3}{y}}{{y}^{\mathrm{2}} +\mathrm{2}}\right)\:{dy}} \\ $$$${u}\left({y}\right)=\:{e}^{−\frac{\mathrm{3}}{\mathrm{2}}\int\:\frac{{d}\left({y}^{\mathrm{2}} +\mathrm{2}\right)}{{y}^{\mathrm{2}} +\mathrm{2}}} =\:\frac{\mathrm{1}}{\left({y}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{3}/\mathrm{2}} } \\ $$$$\Rightarrow\left(\frac{{y}^{\mathrm{2}} +\mathrm{2}}{\:\sqrt{\left({y}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{3}} }}\right)\:{dx}\:−\:\left(\frac{{xy}+\mathrm{2}{y}+{y}^{\mathrm{3}} }{\:\sqrt{\left({y}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{3}} }}\right)\:{dy}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{2}}}\:{dx}\:−\:\frac{{x}+\mathrm{2}{y}+{y}^{\mathrm{3}} }{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{2}}}\:{dy}=\mathrm{0} \\ $$$$\begin{cases}{\frac{\partial{f}}{\partial{x}}\:=\:{M}\left({x},{y}\right)=\frac{\mathrm{1}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{2}}}}\\{\frac{\partial{f}}{\partial{y}}={N}\left({x},{y}\right)=\frac{{x}+\mathrm{2}{y}+{y}^{\mathrm{3}} }{\:\sqrt{\left({y}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{3}} }}}\end{cases} \\ $$$${F}\left({x},{y}\right)=\int{M}\left({x},{y}\right){dx}+{g}\left({y}\right) \\ $$$${g}'\left({y}\right)\:=\frac{−{y}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{2}}} \\ $$$${g}\left({y}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{d}\left({y}^{\mathrm{2}} +\mathrm{2}\right)}{\left({y}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{1}/\mathrm{2}} }\: \\ $$$${g}\left({y}\right)\:=\:\sqrt{{y}^{\mathrm{2}} +\mathrm{2}} \\ $$$${F}\left({x},{y}\right)=\:\int\:\frac{{dx}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{2}}}\:+\:\sqrt{{y}^{\mathrm{2}} +\mathrm{2}} \\ $$$${F}\left({x},{y}\right)\:=\:\frac{{x}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{2}}}\:+\sqrt{{y}^{\mathrm{2}} +\mathrm{2}}\: \\ $$$${F}\left({x},{y}\right)\:=\:\frac{{x}+{y}^{\mathrm{2}} +\mathrm{2}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{2}}}\:.\bigstar\: \\ $$