Question Number 30188 by abdo imad last updated on 17/Feb/18
$$\:{solve}\:{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} \:+\mathrm{1}\right){y}^{'} \:−\mathrm{2}{xy}\:=\mathrm{0} \\ $$
Commented by mrW2 last updated on 17/Feb/18
$${x}^{\mathrm{3}} \left({x}^{\mathrm{2}} \:+\mathrm{1}\right){y}^{'} \:−\mathrm{2}{xy}\:=\mathrm{0} \\ $$$${x}^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+\mathrm{1}\right){y}^{'} \:=\mathrm{2}{y}\: \\ $$$$\frac{{dy}}{{y}}=\frac{\mathrm{2}{dx}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\int\frac{{dy}}{{y}}=\mathrm{2}\int\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right){dx} \\ $$$$\mathrm{ln}\:{y}=−\mathrm{2}\left(\frac{\mathrm{1}}{{x}}+\mathrm{tan}^{−\mathrm{1}} {x}\right)+{C}_{\mathrm{1}} \\ $$$$\Rightarrow{y}={ce}^{−\mathrm{2}\left(\frac{\mathrm{1}}{{x}}+\mathrm{tan}^{−\mathrm{1}} {x}\right)} \\ $$