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Question Number 160903 by blackmamba last updated on 08/Dec/21

(1+x^2 )y ′= 2xy +(1+x^2 )^2

$$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}\:'=\:\mathrm{2}{xy}\:+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \: \\ $$

Answered by GuruBelakangPadang last updated on 09/Dec/21

(1+x^2 )y ′= (1+x^2 )′y +(1+x^2 )^2 (1+x^2 )^(-1) y ′=(1+x^2 )^(-2) (1+x^2 )′y (1+x^2 )^(-1) y ′−(1+x^2 )^(-2) (1+x^2 )′y=0 (1+x^2 )^(-1) y ′+((1+x^2 )^(-1) )^′ y=0 (1+x^2 )^(-1) y=(y/(1+x^2 ))=C

$$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}\:'=\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)'{y}\:+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \: \\ $$$$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{-\mathrm{1}} {y}\:'=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{-\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)'{y} \\ $$$$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{-\mathrm{1}} {y}\:'−\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{-\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)'{y}=\mathrm{0} \\ $$$$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{-\mathrm{1}} {y}\:'+\left(\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{-\mathrm{1}} \right)^{'} {y}=\mathrm{0} \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{-\mathrm{1}} {y}=\frac{{y}}{\mathrm{1}+{x}^{\mathrm{2}} }={C} \\ $$

Answered by yeti123 last updated on 09/Dec/21

(1 + x^2 )(dy/dx) = 2xy + (1 + x^2 )^2 (dy/dx) − ((2xy)/(1 + x^2 )) = 1 + x^2 (1 + x^2 )^(−1) (dy/dx) − ((2xy)/(1 + x^2 ))(1 + x^2 )^(−1) = (1 + x^2 )(1 + x^2 )^(−1) (d/dx){(1 + x^2 )^(−1) y} = 1 (1 + x^2 )^(−1) y = x + c y = (x + c)(1 + x^2 )

$$\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}\:=\:\mathrm{2}{xy}\:+\:\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}\:−\:\frac{\mathrm{2}{xy}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:=\:\mathrm{1}\:+\:{x}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)^{−\mathrm{1}} \frac{{dy}}{{dx}}\:−\:\frac{\mathrm{2}{xy}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)^{−\mathrm{1}} \:=\:\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)^{−\mathrm{1}} \\ $$$$\frac{{d}}{{dx}}\left\{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {y}\right\}\:=\:\mathrm{1} \\ $$$$\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {y}\:=\:{x}\:+\:{c} \\ $$$${y}\:=\:\left({x}\:+\:{c}\right)\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right) \\ $$

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