= Solution
At these parameters the equation becomes $g''+g^2-4=0$, where $g=f'$. Multiply by $g'$ and integrate:
$$
\frac12(g')^2+\frac13g^3-4g=C.
$$
Outer matching gives $g\to-2$ and $g'\to0$, so $C=16/3$. No slip gives $g(0)=0$. At that point the <first integral> fixes $g'(0)=\pm4\sqrt{2/3}$, so there are exactly two smooth initial-value solutions of the second-order equation satisfying the matching condition. Put $\eta_0=\operatorname{arctanh}\sqrt{2/3}$. Translation invariance of the supplied profile gives them explicitly:
$$
\boxed{g_-(\eta)=4-6\tanh^2(\eta+\eta_0),\qquad
g_+(\eta)=4-6\tanh^2(\eta-\eta_0).}
$$
The first decreases monotonically from zero to $-2$. The second first rises to $4$ at $\eta_0$, crosses zero again at $2\eta_0$, and then tends to $-2$. Integrating either profile from zero recovers $f$ with $f(0)=0$.
\Image[/past-exam-of-the-theoretical-mechanics-course-of-the-university-of-oxford/2010-boundary-layer-branches.png]
{title=Two translated similarity profiles: monotone outer-flow matching and a branch with an internal reversed-flow region}
One would \b[not expect both branches to be equally stable]. The rising branch contains flow opposite to its negative outer stream and interior <inflection points>, typical ingredients for <shear instability>. The monotone branch has $g''=4-g^2>0$ at finite positive $\eta$ and no interior <inflection point>, so the classical inviscid inflection-point mechanism is absent there. This motivates expecting the reversed-flow branch to be unstable and the monotone branch to be the stable candidate. A definitive viscous temporal-stability result requires perturbing the time-dependent equations; the steady <first integral> alone is not a stability proof.
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