Homogenous differential equation and its solution

Question

The book Ordinary Differential Equations by Wolfgang Walter considers on page 21 the following ODE: $$y'=f\left(\frac{y}{x}\right)$$

1. How do we know that this equation is homogeneous? I.e. how can we know that $f$ is homogeneous?

Then in one relevant example one page 22 the book considers the IVP

$$y'=\frac{y}{x}-\frac{x^2}{y^2}, y(1) =1$$

and says that this IVP transforms into $$u'=-\frac{1}{xu^2}, u(1)=1$$

But I don't see how this is done. For let $u=\frac{x}{y}$, then $u'=\frac{u^4}{x}$.

Could someone please clarify how to obtain $u'=-\frac{1}{xu^2}$?

Answer 1

1. $f$ is not homogeneous, but $f(x/y)$, because $f(kx/ky) = f(x/y).$ That's the definition of "homogeneous (of degree $0$.)"

2. If $u = y/x$, then $y = ux$ and $y' = u+u'x.$ So the equation becomes

$$u+u'x = u-\frac{1}{u^2}$$

$$u'x=-\frac{1}{u^2}$$

$$u'=\frac{1}{-u^2x}.$$

which is separable.