Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/3/b/ii/solution

Now is continuous, so has no delta contribution. Integrate its normalized differential equation over . The bounded lower-order terms have integrals tending to zero, while the integral of is the difference of the two one-sided first derivatives. Hence the jump condition is
The denominator is evaluated at the source point; it is not a variable left over after taking the limits. The positivity or negativity of this jump follows the sign convention used for the leading coefficient of .

New to topics? Read the docs here!