Question Number 87059 by jagoll last updated on 02/Apr/20
$$\left(\mathrm{y}\:'\right)^{\mathrm{2}} −\mathrm{xy}'\:+\mathrm{y}\:=\:\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution} \\ $$
Answered by mr W last updated on 03/Apr/20
$${y}'=\frac{\mathrm{1}}{\mathrm{2}}\left({x}\pm\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{y}}\right) \\ $$$${let}\:{u}={x}^{\mathrm{2}} −\mathrm{4}{y} \\ $$$$ \\ $$$${case}\:\mathrm{1}:\:{u}={constant}={C} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{4}{y}=−\mathrm{4}{C} \\ $$$$\Rightarrow{y}=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+{C} \\ $$$${y}'=\frac{{x}}{\mathrm{2}} \\ $$$$\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} −{x}\left(\frac{{x}}{\mathrm{2}}\right)+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+{C}=\mathrm{0}\:\Rightarrow{C}=\mathrm{0} \\ $$$$\Rightarrow{y}=\frac{{x}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:{u}\neq{constant} \\ $$$$\frac{{du}}{{dx}}=\mathrm{2}{x}−\mathrm{4}{y}' \\ $$$$\mathrm{2}{x}−\frac{{du}}{{dx}}=\mathrm{2}{x}\pm\mathrm{2}\sqrt{{u}} \\ $$$$\frac{{du}}{{dx}}=\pm\mathrm{2}\sqrt{{u}} \\ $$$$\int\frac{{du}}{\mathrm{2}\sqrt{{u}}}=\pm\int{dx} \\ $$$$\sqrt{{u}}={C}\pm{x} \\ $$$${u}=\left({C}\pm{x}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{C}\left({C}\pm\mathrm{2}{x}\right) \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{y}={x}^{\mathrm{2}} +{C}\left({C}\pm\mathrm{2}{x}\right) \\ $$$$\Rightarrow{y}=−\frac{{C}\left({C}\pm\mathrm{2}{x}\right)}{\mathrm{4}} \\ $$$$ \\ $$$${check}: \\ $$$${y}'=−\frac{\pm{C}}{\mathrm{2}} \\ $$$${xy}'=−\frac{\pm{Cx}}{\mathrm{2}} \\ $$$${xy}'−{y}=−\frac{\pm\mathrm{2}{Cx}}{\mathrm{4}}+\frac{{C}\left({C}\pm\mathrm{2}{x}\right)}{\mathrm{4}}=\frac{{C}^{\mathrm{2}} }{\mathrm{4}}=\left({y}'\right)^{\mathrm{2}} \\ $$
Commented by jagoll last updated on 03/Apr/20
$$\mathrm{sir}\:\mathrm{y}\:=\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}\:?\: \\ $$
Commented by mr W last updated on 03/Apr/20
$${yes}.\:{y}=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:{is}\:{also}\:{a}\:{solution},\:{see} \\ $$$${above}. \\ $$