= Solution
For the proposed form,
$$
u=x^{\alpha-\beta}f',\quad
V=-x^{\alpha-1}(\alpha f-\beta\eta f'),\quad
u_Y=x^{\alpha-2\beta}f''.
$$
Matching the outer <velocity> requires $\alpha-\beta=m$. The convective and outer-<pressure> terms have order $x^{2m-1}$, while $u_{YY}$ has order $x^{\alpha-3\beta}$. Equality of exponents gives
$$
\boxed{\alpha=\frac{1+m}{2},\qquad\beta=\frac{1-m}{2}.}
$$
The two terms containing $\eta f'f''$ cancel in $uu_x+Vu_Y$, leaving $x^{2m-1}[m(f')^2-\alpha ff'']$. Since $U_sU_s'=mA^2x^{2m-1}$, the reduced equation is
$$
\boxed{f^{(3)}+\frac{1+m}{2}ff''+m\{A^2-(f')^2\}=0.}
$$
The wall and matching conditions become $\boxed{f(0)=f'(0)=0,\ f'(\infty)=A}$. This establishes a consistent similarity reduction; existence of an actual profile is a separate boundary-value question for a particular choice of parameters.
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