Prove that if $y' = f(t,y)$ is an Euler homogeneous equation and $y_{1}(t)$ is a solution, then $y(t) = \frac 1k y_{1}(kt)$ is also a solution for every non-zero $k \in \mathbb{R}$.
Now a differential equation $y' = f(t,y)$ is Euler homogeneous if $f(kt, ky) = f(t, y)$ for any $k \in \mathbb{R}$.
We also know that an Euler homogeneous equation can be transformed into a separable equation of the form $\frac{v'}{F(v) - v} = \frac{1}{t}$ via the substitution $v = y/t$, where $F(v) = f(t, y)$.
That is as much as I know. Thanks.
If $y_1(t)$ is a solution, then $y_1'(t) = f(t, y_1(t))$ and for $k \in \mathbb{R}-\{0\}$, we have $$\left(\dfrac1ky_1(kt)\right)'=y_1'(kt)=f(kt, y_1(kt)) = f(kt, k\dfrac1ky_1(kt))=f(t, \dfrac1ky_1(kt))$$