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I have to find the general integral of the following ODE $$ xy'^2+2xy'-y=0 $$ The book says that, integrating w.r.t. $y'$, we get two homogeneous equations $$ y'=-1+\sqrt{1+\frac{y}{x}},\qquad y'=-1-\sqrt{1+\frac{y}{x}} $$ defined for $x(x+y)>0$. Until this point all is clear. Now it says that the general integrals of such homogeneous equations are $$ \left(\sqrt{1+\frac{y}{x}}-1\right)^2=\frac{C}{x},\qquad \left(\sqrt{1+\frac{y}{x}}+1\right)^2=\frac{C}{x}. $$ I do not understand how the book deduces such general integrals. Can someone help me?

Thank You

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  • $\begingroup$ What book are you reading? $\endgroup$ – user587192 Dec 15 '18 at 14:44
  • $\begingroup$ @user587192 Cecconi, Stampacchia - Mathematical Analysis 2 (it is an italian book) $\endgroup$ – Jeji Dec 15 '18 at 16:30
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Hint. Let $u(x):=\sqrt{1+\frac{y(x)}{x}}>0$ then $y(x)=xu^2(x)-x$ and $$y'(x)=u^2(x)+2xu'(x)u(x)-1.$$ Hence $$y'(x)=-1\pm\sqrt{1+\frac{y(x)}{x}}$$ becomes $$u^2(x)+2xu'(x)u(x)-1=-1\pm u(x)$$ that is $$u^2(x)+2xu'(x)u(x)=\pm u(x)$$ or $$u(x)+2xu'(x)=\pm 1$$ which is a linear ODE of first order. Can you take it from here?

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Substitute $$y(x)=xv(x)$$ then we get $$x\left(x\frac{dv(x)}{dx}+v(x)\right)^2+2x\left(x\frac{dv(x)}{dx}+v(x)\right)-xv(x)=0$$ and we get $$\frac{dv(x)}{dx}=\frac{-x-\sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$ $$\frac{dv(x)}{dx}=\frac{-x+\sqrt{x^2+x^2v(x)}-xv(x)}{x^2}$$ Can you finish?

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Set $y=u-x$, then $y'=u'-1$ and $$ u=y+x=x(y'^2+2y'+1)=xu'^2\tag1 $$ which has a derivative $$ u'=u'^2+2xu'u''\implies u'(2xu''+u'-1)=0 $$ Thus you get linear pieces $u=C$ and solutions of the linear ODE $2xu''+u'=1$. They can switch from one to the other where both factors are zero at the same time.

For $x>0$ the second equation solves as $$ (\sqrt xu')'=\frac{2xu''+u'}{2\sqrt x}=\frac1{2\sqrt x}\implies \sqrt xu'=\sqrt{x}+C $$ and the original equation in form (1) gives $$ u(x)=xu'^2=x+2C\sqrt x+C^2 $$

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