= Solution
{c}
Set $g=f_1'$. The <boundary value problem> becomes
$$
\boxed{g''+\zeta g'-2g=0,\qquad g'(0)=1,\qquad
g(\zeta)\longrightarrow0\quad(\zeta\to-\infty).}
$$
For $y_1=1+\zeta^2$, direct substitution gives $y_1''+\zeta y_1'-2y_1=2+2\zeta^2-2(1+\zeta^2)=0$. This growing <polynomial> solution is useful for <reduction of order>.
Write $g=y_1v$. Substitution into the <linear ordinary differential equation> cancels the terms proportional to $v$ and gives
$$
y_1v''+(2y_1'+\zeta y_1)v'=0.
$$
An <integrating factor> for this first-order equation for $v'$ yields
$$
v'(\zeta)=C\frac{e^{-\zeta^2/2}}{(1+\zeta^2)^2}.
$$
The far-field <boundary condition> selects the convergent <integral> from $-\infty$:
$$
g(\zeta)=C(1+\zeta^2)\int_{-\infty}^{\zeta}
\frac{e^{-s^2/2}}{(1+s^2)^2}\,ds.
$$
Any additional multiple of $1+\zeta^2$ would grow at $-\infty$. The chosen expression tends to zero: for $r=-\zeta\geq1$,
$$
0\leq\int_{-\infty}^{\zeta}\frac{e^{-s^2/2}}{(1+s^2)^2}\,ds
=\int_r^\infty\frac{e^{-t^2/2}}{(1+t^2)^2}\,dt
\leq\frac{e^{-r^2/2}}{r(1+r^2)^2}.
$$
Here $(1+t^2)^{-2}\leq(1+r^2)^{-2}$ and $\int_r^\infty e^{-t^2/2}\,dt\leq r^{-1}\int_r^\infty t e^{-t^2/2}\,dt$. Thus the far-field <boundary condition> is verified without a cancellation of large terms.
At $\zeta=0$, differentiating gives $g'(0)=C$, so $C=1$. Using the antiderivative supplied in the paper gives the explicit answer
$$
\boxed{g(\zeta)=\frac{\zeta}{2}e^{-\zeta^2/2}
+\frac{\sqrt{2\pi}}4(1+\zeta^2)
\left[1+\operatorname{erf}\!\left(\frac{\zeta}{\sqrt2}\right)\right],
\qquad \zeta\leq0.}
$$
Here $\operatorname{erf}$ is the <error function>. Finally,
$$
f_1(\zeta)=\int_0^{\zeta}g(s)\,ds
$$
recovers the unique correction satisfying $f_1(0)=0$. The decay bound also shows that $f_1$ approaches a finite constant at $-\infty$, as allowed by the normal <velocity> matching. This is the <weak surface-shear perturbation of stagnation flow>.
If the printed $+\infty$ condition is instead imposed literally on a positive half-line, <reduction of order> gives
$$
g_+(\zeta)=-(1+\zeta^2)\int_{\zeta}^{\infty}
\frac{e^{-s^2/2}}{(1+s^2)^2}\,ds,
\qquad g_+(0)=-\frac{\sqrt{2\pi}}4.
$$
It satisfies $g_+'(0)=1$ but grows as $\zeta\to-\infty$, so it cannot describe the fluid domain specified in this paper. This makes the consequence of the printed sign error explicit.
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