= Solution
Use $u=\psi_Y$, $V=-\psi_x$, and put $\eta=Y/x^\gamma$, $\psi=x^\beta f(\eta)$. Then
$$
u=x^{\beta-\gamma}f',\qquad V=-x^{\beta-1}(\beta f-\gamma\eta f').
$$
Matching $u\sim x^m$ requires $\beta-\gamma=m$. Equating the powers in streamwise <inertia> $x^{2m-1}$ and transverse diffusion $x^{m-2\gamma}$ gives
$$
\boxed{\gamma=\frac{1-m}{2},\qquad\beta=\frac{1+m}{2}.}
$$
The two convective terms combine to $x^{2m-1}[m(f')^2-\beta f f'']$, while $UU'=mx^{2m-1}$. Hence the <Falkner-Skan equation> in this scaling is
$$
\boxed{f^{(3)}+\beta f f''+m[1-(f')^2]=0,\quad f(0)=f'(0)=0,\quad f'(\infty)=1.}
$$
The additive stream-function constant is fixed by $f(0)=0$; when $\beta\ne0$ this also encodes <no-penetration boundary condition>[no penetration]. For $\beta=0$ it is a harmless choice of additive constant choice. Existence of an attached physical branch for a given $m$ is a separate issue from deriving the similarity reduction.
On an attached branch with $w=f'>0$ away from the wall, use $f$ itself as the independent variable. The <chain rule> gives $f''=w w_f$ and $f^{(3)}=w(w_f^2+w w_{ff})$. Therefore
$$
\boxed{w^2w_{ff}+w(w_f)^2+\beta f w w_f+m(1-w^2)=0,\qquad w(0)=0,\quad w(f)\to1\text{ as }f\to\infty.}
$$
If the <wall shear stress>[wall shear] $s=f''(0)>0$, then $f\sim s\eta^2/2$, so $w(f)\sim\sqrt{2sf}$. Thus $w_f$ is singular at zero while $w w_f\to s$ is finite. Imposing a finite arbitrary $w_f(0)$ would remove the usual attached branch. This is the <velocity as dependent variable in the Falkner-Skan equation> reduction.
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