Question Number 86294 by jagoll last updated on 28/Mar/20
$$\mathrm{y}\:=\:\mathrm{2x}\:+\:\left(\mathrm{y}'\right)^{\mathrm{2}} −\mathrm{4y}' \\ $$
Answered by mr W last updated on 28/Mar/20
$${y}'=\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{2}{x}+{y}} \\ $$$${let}\:{u}=\mathrm{4}−\mathrm{2}{x}+{y} \\ $$$$\frac{{du}}{{dx}}=−\mathrm{2}+\frac{{dy}}{{dx}} \\ $$$$\mathrm{2}+\frac{{du}}{{dx}}=\mathrm{2}\pm\sqrt{{u}} \\ $$$$\frac{{du}}{{dx}}=\pm\sqrt{{u}} \\ $$$$\int\frac{{du}}{\:\sqrt{{u}}}=\pm\int{dx} \\ $$$$\mathrm{2}\sqrt{{u}}=\pm{x}+{C} \\ $$$$\Rightarrow\mathrm{2}\sqrt{\mathrm{4}−\mathrm{2}{x}+{y}}\pm{x}={C} \\ $$$$\Rightarrow{y}=\frac{\left({C}\pm{x}\right)^{\mathrm{2}} }{\mathrm{4}}+\mathrm{2}{x}−\mathrm{4} \\ $$