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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 (1+x^2 )y^′ −2xy =(1+x^2 )^2   (he) ⇒(1+x^2 )y^′ −2xy =0 ⇒  (1+x^2 )y^′  =2xy ⇒(y^′ /y)  =((2x)/(1+x^2 )) ⇒  ln∣y∣=ln(1+x^2 ) +c ⇒  y = k(1+x^2 )   mvc mrthod give  y^′ =k^, (1+x^2 ) +2xk   (e) ⇒(1+x^2 )k′(1+x^2 ) +2xk(1+x^2 )  −2xk(1+x^2 ) =(1+x^2 )^2  ⇒  k^′ =1 ⇒k(x)=x +λ ⇒  y(x) =(1+x^2 )(x+λ)    ⇒  y(x) =x^3  +x  +λ(1+x^2 )  we see that y is a polynom..

$${the}\:{equation}\:{is}\:{not}\:{clear}\:{but}\:{i}\:{suppose} \\ $$$${that}\:{is}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} −\mathrm{2}{xy}\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\left({he}\right)\:\Rightarrow\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} −\mathrm{2}{xy}\:=\mathrm{0}\:\Rightarrow \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} \:=\mathrm{2}{xy}\:\Rightarrow\frac{{y}^{'} }{{y}}\:\:=\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${ln}\mid{y}\mid={ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:+{c}\:\Rightarrow \\ $$$${y}\:=\:{k}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\:\:{mvc}\:{mrthod}\:{give} \\ $$$${y}^{'} ={k}^{,} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:+\mathrm{2}{xk}\: \\ $$$$\left({e}\right)\:\Rightarrow\left(\mathrm{1}+{x}^{\mathrm{2}} \right){k}'\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:+\mathrm{2}{xk}\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$$−\mathrm{2}{xk}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:\Rightarrow \\ $$$${k}^{'} =\mathrm{1}\:\Rightarrow{k}\left({x}\right)={x}\:+\lambda\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({x}+\lambda\right)\:\:\:\:\Rightarrow \\ $$$${y}\left({x}\right)\:={x}^{\mathrm{3}} \:+{x}\:\:+\lambda\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$${we}\:{see}\:{that}\:{y}\:{is}\:{a}\:{polynom}.. \\ $$

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