Homogenous Differential Equation The problem is to solve for the general solution of:
$$yy' + x = (x^2 + y^2)^{1/2}$$
My teacher said to homogenize the equation as a hint during our exam, and afterwards in the solutions he gave this as the homogenized equation before solving it:
$$y' = \frac{\left[1 + \left(\frac yx\right)^2\right]^{1/2} - 1}{\frac yx}$$
I understand that this rearrangement helps us solve the equation by substituting v = (y/x) and going from there, but I still do not understand how this is "homogenizing" the equation. What does that mean exactly, in this context? I thought that a homogeneous equation was an equation with all terms raised to the same power, which is what we now have, I guess. My teacher kept emphasizing that homogeneous equations are where you try to rearrange and get (y/x) in the and substitute for that, but I still don't really understand why. Can anyone kindly explain this example in more depth? Thanks. 
 A: Well the thing is that the equation is already homogeneous. 
$\frac{dy}{dx}  = \frac{\sqrt{(x^2 + y^2)} - x}{x}$
Any function "$f(x,y)$" satisfying "$f(\lambda x,\lambda y) = \lambda ^n f(x,y)$" where $n$ is any integer, is a homogeneous function and differential equation which involve homogeneous function is called homogeneous differential equation.
The equation already satisfies the condition of homogeneity. If in doubt, put $x=\lambda x$ and $y = \lambda y$ and clear it. You need not convert it all into the form of $\frac{y}{x}$, it just simplifies calculations.
From the equation put $y=vx$ you get $\frac{dy}{dx}=v + x\frac{dv}{dx}$
putting in the equation you get,
$v+x\frac{dv}{dx} = \frac{\sqrt{(x^2 + v^2x^2)} - x}{vx}$
on taking $x^2$ common and dividing and multiplying by x you get 
$\frac{dy}{dx} = v+x\frac{dv}{dx} = \frac{\sqrt{(1 + v^2)} - 1}{v}$ 
and now if you write $\frac{y}{x} = v$ 
you will get to the same thing the teacher emphasized. He/she did it because if you already have converted it to the $\frac{y}{x}$ form it should be less cumbersome to calculate as then you could have directly substituted v as $\frac{y}{x}$.
Hope I didn't confuse you.
A: with $y=ux$ we obtain
$$u+xu'=\frac{(1+u^2)^{1/2}-1}{u}$$ and after some algebra we get
$$\frac{u}{(1+u^2)^{1/2}-1-u^2}du=\frac{dx}{x}$$
A: Let $$x^2+y^2=u^2$$  so we have that $$yy'+x=uu'$$ and from $$yy'+x=\sqrt{x^2+y^2}$$ we have $$uu'=u$$ therefore from here we see that $$u=\sqrt{x^2+y^2}=x+C$$
