The first-order Bogomolny equation is
$$
\frac{d\phi}{dx}=\phi^2(1-\phi^2).
$$
Partial fractions and one integration give
$$
x-A=\int\frac{d\phi}{\phi^2(1-\phi^2)}
=-\frac1\phi+\frac12\log\frac{1+\phi}{1-\phi}.
$$
Thus
$$
\boxed{\frac12\log\frac{1+\phi}{1-\phi}-\frac1\phi=x-A}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 313 / 1 / iii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-313/1/iii/solution