= Solution
Let
$$
\epsilon=\sqrt{\frac\nu Q},
\qquad
\eta=\frac{y}{\epsilon x},
\qquad
\psi=\sqrt{\nu Q}\,f(\eta)=Q\epsilon f(\eta).
$$
Using the <Cartesian streamfunction> convention $u=\psi_y$, $v=-\psi_x$ gives
$$
u=\frac Qx f'(\eta),
\qquad
v=\frac{Q\epsilon}{x}\eta f'(\eta).
$$
At fixed $y$,
$$
u_x=-\frac Q{x^2}[f'+\eta f''],
\qquad
u_y=\frac Q{\epsilon x^2}f'',
\qquad
u_{yy}=\frac Q{\epsilon^2x^3}f'''.
$$
The terms proportional to $\eta f'f''$ cancel, leaving
$$
u u_x+v u_y=-\frac{Q^2}{x^3}(f')^2.
$$
Since $U=-Q/x$, $UU'=-Q^2/x^3$, while $\nu u_{yy}=Q^2f'''/x^3$. The boundary-layer equation therefore becomes
$$
\boxed{f'''=1-(f')^2}.
$$
Choosing the wall value of the streamfunction as zero, imposing no slip, and matching to $U=-Q/x$ give
$$
\boxed{f(0)=0,
\qquad f'(0)=0,
\qquad f'(\eta)\to-1\quad(\eta\to\infty)}.
$$
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