Question Number 60357 by Sardor2211 last updated on 20/May/19
Commented by Mr X pcx last updated on 20/May/19
$${the}\:{equation}\:{is}\:{not}\:{clear}\:{but}\:{i}\:{suppose} \\ $$$${that}\:{is}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} −\mathrm{2}{xy}\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\left({he}\right)\:\Rightarrow\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} −\mathrm{2}{xy}\:=\mathrm{0}\:\Rightarrow \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} \:=\mathrm{2}{xy}\:\Rightarrow\frac{{y}^{'} }{{y}}\:\:=\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${ln}\mid{y}\mid={ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:+{c}\:\Rightarrow \\ $$$${y}\:=\:{k}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\:\:{mvc}\:{mrthod}\:{give} \\ $$$${y}^{'} ={k}^{,} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:+\mathrm{2}{xk}\: \\ $$$$\left({e}\right)\:\Rightarrow\left(\mathrm{1}+{x}^{\mathrm{2}} \right){k}'\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:+\mathrm{2}{xk}\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$$−\mathrm{2}{xk}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:\Rightarrow \\ $$$${k}^{'} =\mathrm{1}\:\Rightarrow{k}\left({x}\right)={x}\:+\lambda\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({x}+\lambda\right)\:\:\:\:\Rightarrow \\ $$$${y}\left({x}\right)\:={x}^{\mathrm{3}} \:+{x}\:\:+\lambda\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$${we}\:{see}\:{that}\:{y}\:{is}\:{a}\:{polynom}.. \\ $$