= Solution
The outer fluid is at rest, so its pressure is constant; <boundary layer> normal balance gives no transverse pressure change, and the streamwise <pressure gradient> is zero. The steady equations are $u_x+v_y=0$ and $uu_x+vu_y=\nu u_{yy}$. In conservative form $\partial_x(u^2)+\partial_y(uv)=\nu u_{yy}$. Integrating across the jet, with vanishing end contributions, proves $d[\rho\int u^2dy]/dx=0$. This is conserved <momentum flux>. A control volume around the force source shows that its injected momentum per unit time and span is $F$, so \b[$M=F$].
For width $b$ and velocity scale $U$, conservation gives $\rho U^2b\sim F$, while inertia-viscosity balance gives $U^2/x\sim\nu U/b^2$. Thus $b\sim(\rho\nu^2x^2/F)^{1/3}$, $U\sim[F^2/(\rho^2\nu x)]^{1/3}$ and <stream function> scale $Ub=(F\nu x/\rho)^{1/3}$. Put $\psi=Cx^{1/3}f(\eta)$, $\eta=Dyx^{-2/3}$, with the constants in the question. Then
$$
u=CDx^{-1/3}f',\qquad v=-\frac C3x^{-2/3}(f-2\eta f').
$$
Direct differentiation gives $uu_x+vu_y=-(C^2D^2/3)x^{-5/3}[(f\prime)^2+ff\prime\prime]$ and $\nu u_{yy}=\nu CD^3x^{-5/3}f\prime\prime\prime$. Since $C=\nu D$, their equality cancels the terms involving $\eta$ and gives
$$
\boxed{3f'''+(f')^2+ff''=0}.
$$
For $f=A\tanh(A\eta/6)$, differentiation verifies this equation directly. Moreover
$$
M=F\int_{-\infty}^\infty(f')^2d\eta
=F\frac{A^4}{36}\frac6A\frac43=F\frac{2A^3}{9}
$$
for $A>0$. Therefore momentum normalization fixes \b[$A=(9/2)^{1/3}$]. Independence of $x$ alone, as the last wording suggests, would not fix $A$; equality to the injected force does. Finally
$$
\boxed{Q(x)=\psi(x,\infty)-\psi(x,-\infty)=2A(F\nu x/\rho)^{1/3}}.
$$
<Entrainment> draws ambient fluid into the widening jet, so volume flux grows while <momentum flux> remains constant. These are the <laminar plane jet from a line force> scalings.
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