How to solve this equation by integration

Solution 1:

$$\frac{x’’}{x}-\frac{x’}{x^2} - \frac{x’^2}{x^2} =0,$$ $$\frac{xx’'-x'^2}{x^2}=\frac{x’}{x^2} $$ $$\left(\frac{x'}{x}\right)'=-\left(\frac{1}{x}\right)' $$ Reduce the order of the DE by integration.

Edit $$\left(\frac{x'}{x}\right)'=-\left(\frac{1}{x}\right)' $$ $$\frac{x'}{x}=-\frac{1}{x} +C $$ $$x'=Cx-1 $$ This DE is separable. $$\frac{dx}{Cx-1}=dy$$

Solution 2:

There is another way to solve this equation : switch variables and write it as $$-\frac 1x \frac{ y''}{[y']^3}-\frac 1{x^2}\frac 1{ y'}-\frac 1{x^2}\frac 1{ [y']^2}=0$$ that is to say $$x y''+y'+[y']^2=0$$ and the reduction of order becomes obvious. With $p=y'$ $$x p'+p+p ^2=0$$ $$p=\frac 1 q \implies x q'=1+q\implies q=c_1x-1\implies p=\frac 1{c_1 x-1}$$ Just finish and inverse.