Question Number 26191 by sorour87 last updated on 21/Dec/17
$$\mathrm{2}\frac{{dy}}{{dx}}+{x}=\mathrm{4}\sqrt{{y}} \\ $$
Answered by mrW1 last updated on 22/Dec/17
$${y}={u}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}{u}\frac{{du}}{{dx}} \\ $$$$\Rightarrow\mathrm{4}{u}\frac{{du}}{{dx}}+{x}=\mathrm{4}{u} \\ $$$$\Rightarrow\frac{{du}}{{dx}}=\mathrm{1}−\frac{{x}}{\mathrm{4}{u}} \\ $$$$ \\ $$$${u}={vx} \\ $$$$\frac{{du}}{{dx}}={v}+{x}\frac{{dv}}{{dx}} \\ $$$$\Rightarrow{v}+{x}\frac{{dv}}{{dx}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{v}} \\ $$$$\Rightarrow{x}\frac{{dv}}{{dx}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{v}}−{v}=\frac{\mathrm{4}{v}−\mathrm{1}−\mathrm{4}{v}^{\mathrm{2}} }{\mathrm{4}{v}}=−\frac{\left(\mathrm{2}{v}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}{v}} \\ $$$$\Rightarrow\int\frac{\mathrm{4}{v}}{\left(\mathrm{2}{v}−\mathrm{1}\right)^{\mathrm{2}} }\:{dv}=−\int\frac{{dx}}{{x}} \\ $$$$ \\ $$$${w}=\mathrm{2}{v} \\ $$$$\Rightarrow\int\frac{{w}}{\left({w}−\mathrm{1}\right)^{\mathrm{2}} }\:{dw}=−\int\frac{{dx}}{{x}} \\ $$$$\Rightarrow\int\frac{{w}−\mathrm{1}+\mathrm{1}}{\left({w}−\mathrm{1}\right)^{\mathrm{2}} }\:{dw}=−\int\frac{{dx}}{{x}} \\ $$$$\Rightarrow\int\left[\mathrm{1}+\frac{\mathrm{1}}{\left({w}−\mathrm{1}\right)^{\mathrm{2}} }\right]\:{dw}=−\int\frac{{dx}}{{x}} \\ $$$${w}−\frac{\mathrm{1}}{{w}−\mathrm{1}}=−\mathrm{ln}\:{x}+{C} \\ $$$$ \\ $$$${w}=\mathrm{2}{v}=\frac{\mathrm{2}{u}}{{x}}=\frac{\mathrm{2}\sqrt{{y}}}{{x}} \\ $$$$\Rightarrow\frac{\mathrm{2}\sqrt{{y}}}{{x}}−\frac{\mathrm{1}}{\frac{\mathrm{2}\sqrt{{y}}}{{x}}−\mathrm{1}}=−\mathrm{ln}\:{x}+{C} \\ $$$$\Rightarrow\frac{\mathrm{2}\sqrt{{y}}}{{x}}−\frac{{x}}{\mathrm{2}\sqrt{{y}}−{x}}=−\mathrm{ln}\:{x}+{C} \\ $$