= Solution
For $U(y)=c^2-y^2$, <completing the square> gives the <Bogomolny bound>
$$
\begin{aligned}
I[y]
&=\frac12\int_{-\infty}^{\infty}\left[(y'-U(y))^2+2U(y)y'\right]dx\\
&=\frac12\int_{-\infty}^{\infty}(y'-U(y))^2dx
+\int_{-c}^{c}(c^2-y^2)\,dy\\
&=\frac12\int_{-\infty}^{\infty}(y'-U(y))^2dx+\frac{4c^3}{3}.
\end{aligned}
$$
Consequently
$$
\boxed{I[y]\geq\frac{4c^3}{3}},
$$
with equality exactly when the <Bogomolny equation>
$$
y'=c^2-y^2
$$
holds everywhere. Separating variables, or differentiating the proposed form directly, gives all solutions with the required limits:
$$
\boxed{y(x)=c\tanh\{c(x-x_0)\}},
\qquad x_0\in\mathbb R.
$$
These are the translated <Phi-four kink>[phi-four kinks].
Differentiating the first-order equation yields
$$
y''=-2yy'=(-2y)(c^2-y^2)=U'(y)U(y),
$$
which is precisely the Euler-Lagrange equation from part (a). Thus every configuration saturating the first-order bound is automatically a stationary solution of the second-order variational equation.
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