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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 $u{u}_x+v{u}_y=\nu u_{yy}$. In conservative form ${\partial }_x(u^2)+{\partial }_y(uv)=\nu u_{yy}$. Integrating across the jet, with vanishing end contributions, proves $d[\rho \int u^{2}dy]/dx=0$. 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 $M=F$.

For width $b$ and velocity scale $U$, conservation gives $\rho U^{2}b\sim F$, while inertia-viscosity balance gives $U^2/x\sim \nu U/b^2$. Thus $b\sim (\rho {\nu }^2{x}^2/F{)}^{1/3}$, $U\sim [F^2/({\rho }^2\nu x){]}^{1/3}$ and stream function scale $Ub=(F\nu x/\rho {)}^{1/3}$. Put $\psi =C{x}^{1/3}f(\eta )$, $\eta =Dy{x}^{-2/3}$, with the constants in the question. ThenDirect differentiation gives $u{u}_x+v{u}_y=-(C^2{D}^2/3)x^{-5/3}[(f\prime {)}^2+ff\prime \prime ]$ and $\nu u_{yy}=\nu C{D}^3{x}^{-5/3}f\prime \prime \prime$. Since $C=\nu D$, their equality cancels the terms involving $\eta$ and givesFor $f=A\tanh (A\eta /6)$, differentiation verifies this equation directly. Moreoverfor $A>0$. Therefore momentum normalization fixes $A=(9/2{)}^{1/3}$. Independence of $x$ alone, as the last wording suggests, would not fix $A$; equality to the injected force does. FinallyEntrainment 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.