Differential Equation Questions

Question Number 105283 by bemath last updated on 27/Jul/20

$$\left(\mathrm{1}\right)\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}{xy}}{\mathrm{4}{x}^{\mathrm{2}} −{y}^{\mathrm{3}} } \\$$$$\left(\mathrm{2}\right)\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{{y}\left(\mathrm{2ln}\:{y}\:+\:\mathrm{1}\right)} \\$$

Answered by bobhans last updated on 27/Jul/20

$$\left(\mathrm{2}\right)\:{y}\left(\mathrm{2ln}\:{y}+\mathrm{1}\right)\:{dy}\:=\:\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)\:{dx} \\$$$$\int\:\left(\mathrm{2}{y}\:\mathrm{ln}\:{y}\:+{y}\right){dy}\:=\:\int\:\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right){dx} \\$$$$\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} +{y}^{\mathrm{2}} \mathrm{ln}\:{y}\:−\int{y}\:{dy}\:=\:\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\:+{C} \\$$$$\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} +{y}^{\mathrm{2}} \mathrm{ln}\:{y}−\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} =\mathrm{sin}\:{x}−\mathrm{cos}\:{x}+{C} \\$$$$\rightarrow{y}^{\mathrm{2}} \:\mathrm{ln}\:{y}\:=\:\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\:+\:{C}\:\triangleright \\$$

Answered by Dwaipayan Shikari last updated on 27/Jul/20

$$\frac{{dy}}{{dx}}=\frac{{sinx}+{cosx}}{{y}\left(\mathrm{2}{logy}+\mathrm{1}\right)} \\$$$$\int{y}\left(\mathrm{2}{logy}+\mathrm{1}\right){dy}=\int{sinx}+{cosx}\:{dx} \\$$$${y}^{\mathrm{2}} {logy}−\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+\frac{{y}^{\mathrm{2}} }{\mathrm{2}}=−{cosx}+{sinx}+{C} \\$$$${y}^{\mathrm{2}} {logy}=−{cosx}+{sinx}+{C} \\$$$${y}={e}^{\frac{−{cosx}+{sinx}}{{y}^{\mathrm{2}} }} \\$$

Answered by bobhans last updated on 27/Jul/20

$$\left(\mathrm{1}\right)\:\left({y}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} \right)\:{dy}\:+\:\mathrm{2}{xy}\:{dx}\:=\:\mathrm{0} \\$$$$\frac{\partial{N}}{\partial{x}}\:=\:−\mathrm{8}{x}\:;\:\frac{\partial{M}}{\partial{y}}\:=\:\mathrm{2}{x}\:\Leftrightarrow\:\frac{\partial{N}}{\partial{x}}\:\neq\:\frac{\partial{M}}{\partial{y}} \\$$$${non}\:{exact}\:. \\$$$${by}\:{observe}\:\frac{\frac{\partial{N}}{\partial{x}}−\frac{\partial{M}}{\partial{y}}}{{M}}\:=\:\frac{−\mathrm{10}{x}}{\mathrm{2}{xy}}\:=\:−\frac{\mathrm{5}}{{y}} \\$$$${integrating}\:{factor}\:{u}\left({y}\right)={e}^{\int\:−\frac{\mathrm{5}}{{y}}\:{dy}} \\$$$${u}\left({y}\right)\:=\:{e}^{−\mathrm{5ln}\:{y}} \:=\:{y}^{−\mathrm{5}} \\$$$${we}\:{can}\:{multiply}\:{original}\:{equation}\:{by}\:{y}^{−\mathrm{5}} \\$$$$\Rightarrow\left({y}^{−\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} {y}^{−\mathrm{5}} \right){dy}\:+\mathrm{2}{xy}^{−\mathrm{4}} \:{dx}\:=\:\mathrm{0} \\$$$$\begin{cases}{\frac{\partial{F}}{\partial{x}}\:=\:{M}\left({x},{y}\right)\:=\:\mathrm{2}{xy}^{−\mathrm{4}} }\\{\frac{\partial{F}}{\partial{x}}\:=\:{N}\left({x},{y}\right)\:=\:{y}^{−\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} {y}^{−\mathrm{5}} }\end{cases} \\$$$${F}\left({x},{y}\right)\:=\:\int\left(\mathrm{2}{xy}^{−\mathrm{4}} \right){dx}\:+\:{g}\left({y}\right) \\$$$${F}\left({x},{y}\right)\:=\:{x}^{\mathrm{2}} {y}^{−\mathrm{4}} \:+\:{g}\left({y}\right) \\$$$$\Leftrightarrow{g}'\left({y}\right)\:=\:{N}\left({x},{y}\right)−\frac{\partial}{\partial{y}}\:\left({x}^{\mathrm{2}} {y}^{−\mathrm{4}} \right) \\$$$${g}'\left({y}\right)\:=\:{y}^{−\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} {y}^{−\mathrm{5}} +\mathrm{4}{x}^{\mathrm{2}} {y}^{−\mathrm{5}} \\$$$${g}'\left({y}\right)\:=\:{y}^{−\mathrm{2}} \:\rightarrow{g}\left({y}\right)\:=\:\int{y}^{−\mathrm{2}} {dy} \\$$$${g}\left({y}\right)\:=\:−{y}^{−\mathrm{1}} .\:{Therefore}\:{we}\:{get}\:{solution} \\$$$$\therefore\:{x}^{\mathrm{2}} {y}^{−\mathrm{4}} −{y}^{−\mathrm{1}} \:=\:{C}\:.\:\triangleright \\$$$$\\$$

Commented by bemath last updated on 27/Jul/20

$${very}...{cooll}\:{thanks}\: \\$$