The terms independent of $v$ in the substituted PDE satisfy
$$
B'(t)+\frac r2+aA(t)=r,
$$
so
$$
B'(t)=\frac r2-aA(t),
\qquad B(T)=0.
$$
Integrating backward from $T$ yields
$$
\boxed{
B(t)=-\frac12(T-t)r
+a\int_t^TA(s)ds.}
$$
Thus the requested constant is
$$
\boxed{k=a.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 211 / 6 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-211/6/c/solution