= Solution
Put
$$
\delta=\sqrt{\frac{\nu}{\alpha}},
\qquad
\eta=\frac y\delta,
\qquad
\psi=\alpha x\delta f(\eta).
$$
Then
$$
u=\alpha x f'(\eta),
\qquad
v=-\alpha\delta f(\eta).
$$
Substitution in the boundary-layer momentum equation, using $\nu/\delta^2=\alpha$, gives
$$
\boxed{f'''+ff''-(f')^2=0}.
$$
The wall and far-field conditions become
$$
\boxed{f(0)=0,\qquad f'(0)=1,\qquad f'(\infty)=0}.
$$
These are the similarity equations for the <boundary layer over a linearly stretching sheet>.
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