Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 2 / 5A / a / Solution

Let $D=d/dx$ and write the constant-coefficient linear ordinary differential equation as $p(D)y=0$. Since $D^k{e}^{\lambda x}={\lambda }^k{e}^{\lambda x}$,The exponential never vanishes, so $\boxed{e^{\lambda x}\text{ solves the equation }⟺p(\lambda )=0}$.

For the second claim, the product rule givesfor $k\ge 1$, with the $k=0$ term treated separately. Summing the coefficients yieldsA root of multiplicity at least two has $p(\mu )=p^{\prime }(\mu )=0$, proving $x{e}^{\mu x}$ is also a solution. This directly establishes the repeated-root contribution without assuming a solution formula. If the displayed coefficient list has leading zeros, the exponential calculation still applies to its actual polynomial degree.