What relation must hold between α and β so that y'=ax^α+by^β…

What relation must hold between $\alpha$ and $\beta$ so that $y'=ax^\alpha+by^\beta$ can be reduced to a homogeneous equation by the substitution $y=z^m$?

Step 1: Substitute and isolate $z'$

With $y=z^m$, $y'=mz^{m-1}z'$, so

\[z'=\frac{a}{m}x^\alpha z^{1-m}+\frac{b}{m}z^{m\beta-m+1}.\]

Step 2: Impose degree-zero homogeneity

For $z'$ to depend only on $z/x$, each term must be invariant under $(x,z)\mapsto(\lambda x,\lambda z)$, i.e. have total degree $0$:

\[\alpha+(1-m)=0 \qquad\text{and}\qquad m\beta-m+1=0.\]

Step 3: Solve for the relation

The first equation gives $m=\alpha+1$; the second gives $m=\dfrac{1}{1-\beta}$ (for $\beta\neq1$). Equating,

\[\boxed{\,\alpha+1=\frac{1}{1-\beta}\ \ \Longleftrightarrow\ \ \beta=\frac{\alpha}{\alpha+1},}\]

with $m=\alpha+1$. (As a check, $\alpha=2,\ \beta=\tfrac23$ gives $m=3$, and $y'=ax^2+by^{2/3}$ becomes, with $y=z^3$, $z'=\tfrac{a}{3}(x/z)^2+\tfrac b3$ — a function of $x/z$, confirming homogeneity.)