I have a question about the solution to the following homogeneous differential equation: $ydx=(x+\sqrt{xy})dy$. Specifically, I am struggling to understand the textbook's solution to it, which is: $$ydx=(x+\sqrt{xy})dy$$ $$y=vx, dy=xdv+vdx$$ $$vxdx=(x+x\sqrt{v})(vdx+xdv)$$ I understand the substitution part of the solution, but in the last line they are writing $\sqrt{vx^2}$ as $x\sqrt{v}$, without putting the absolute value bars around $x$. Why is the absolute value neglected? Shouldn't it be $|x|\sqrt{v}$? In general, my textbook seems to be quite loosey with dropping absolute values in differential equations, is it correct to do so? This is the rest of the solution: $$vxdx=(x+x\sqrt{v})(vdx+xdv) \implies -xv\sqrt{v}dx=x^2(1+\sqrt{v})dv$$ $$\implies -\frac{dx}{x}=\left(v^{-\frac{3}{2}}+\frac{1}{v} \right)dv$$ $$\int -\frac{dx}{x}=\int \left(v^{-\frac{3}{2}}+\frac{1}{v} \right)dv$$ $$-\ln|x|+c=-2\sqrt{\frac{x}{y}}+\ln\left|\frac{y}{x}\right|\implies 2\sqrt{\frac{x}{y}}=\ln|y|-c$$ $$\implies 4x=y(\ln|y|-c)^2$$
Thank you in advance!