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^2 (dy/dx)=2u(du/dx) ⇒4u(du/dx)+x=4u ⇒(du/dx)=1−(x/(4u)) u=vx (du/dx)=v+x(dv/dx) ⇒v+x(dv/dx)=1−(1/(4v)) ⇒x(dv/dx)=1−(1/(4v))−v=((4v−1−4v^2 )/(4v))=−(((2v−1)^2 )/(4v)) ⇒∫((4v)/((2v−1)^2 )) dv=−∫(dx/x) w=2v ⇒∫(w/((w−1)^2 )) dw=−∫(dx/x) ⇒∫((w−1+1)/((w−1)^2 )) dw=−∫(dx/x) ⇒∫[1+(1/((w−1)^2 ))] dw=−∫(dx/x) w−(1/(w−1))=−ln x+C w=2v=((2u)/x)=((2(√y))/x) ⇒((2(√y))/x)−(1/(((2(√y))/x)−1))=−ln x+C ⇒((2(√y))/x)−(x/(2(√y)−x))=−ln x+C](https://www.tinkutara.com/question/Q26195.png)
$${y}={u}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}{u}\frac{{du}}{{dx}} \\ $$$$\Rightarrow\mathrm{4}{u}\frac{{du}}{{dx}}+{x}=\mathrm{4}{u} \\ $$$$\Rightarrow\frac{{du}}{{dx}}=\mathrm{1}−\frac{{x}}{\mathrm{4}{u}} \\ $$$$ \\ $$$${u}={vx} \\ $$$$\frac{{du}}{{dx}}={v}+{x}\frac{{dv}}{{dx}} \\ $$$$\Rightarrow{v}+{x}\frac{{dv}}{{dx}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{v}} \\ $$$$\Rightarrow{x}\frac{{dv}}{{dx}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{v}}−{v}=\frac{\mathrm{4}{v}−\mathrm{1}−\mathrm{4}{v}^{\mathrm{2}} }{\mathrm{4}{v}}=−\frac{\left(\mathrm{2}{v}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}{v}} \\ $$$$\Rightarrow\int\frac{\mathrm{4}{v}}{\left(\mathrm{2}{v}−\mathrm{1}\right)^{\mathrm{2}} }\:{dv}=−\int\frac{{dx}}{{x}} \\ $$$$ \\ $$$${w}=\mathrm{2}{v} \\ $$$$\Rightarrow\int\frac{{w}}{\left({w}−\mathrm{1}\right)^{\mathrm{2}} }\:{dw}=−\int\frac{{dx}}{{x}} \\ $$$$\Rightarrow\int\frac{{w}−\mathrm{1}+\mathrm{1}}{\left({w}−\mathrm{1}\right)^{\mathrm{2}} }\:{dw}=−\int\frac{{dx}}{{x}} \\ $$$$\Rightarrow\int\left[\mathrm{1}+\frac{\mathrm{1}}{\left({w}−\mathrm{1}\right)^{\mathrm{2}} }\right]\:{dw}=−\int\frac{{dx}}{{x}} \\ $$$${w}−\frac{\mathrm{1}}{{w}−\mathrm{1}}=−\mathrm{ln}\:{x}+{C} \\ $$$$ \\ $$$${w}=\mathrm{2}{v}=\frac{\mathrm{2}{u}}{{x}}=\frac{\mathrm{2}\sqrt{{y}}}{{x}} \\ $$$$\Rightarrow\frac{\mathrm{2}\sqrt{{y}}}{{x}}−\frac{\mathrm{1}}{\frac{\mathrm{2}\sqrt{{y}}}{{x}}−\mathrm{1}}=−\mathrm{ln}\:{x}+{C} \\ $$$$\Rightarrow\frac{\mathrm{2}\sqrt{{y}}}{{x}}−\frac{{x}}{\mathrm{2}\sqrt{{y}}−{x}}=−\mathrm{ln}\:{x}+{C} \\ $$