Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 13 / b / Solution

The differential operator on the left is $(D-3{)}^2$, where $D=d/dt$. Its repeated characteristic root suggests writing $y=e^{3t}v$. By the product rule,so the ordinary differential equation reduces to $v^{\prime \prime }=3(t+1)$. Two antiderivatives giveThe boundary conditions imply $v(0)=1$ and $v(1)=0$, hence $B=1$ and $A=-3$. ThereforeThe calculation of $v^{\prime \prime }$ verifies the ordinary differential equation, and the bracket takes the values one and zero at the two endpoints. These boundary conditions determine both constants uniquely.