Differential Equations and The - Braun, Martin 7579 · Problem 2.15.8

The key insight is that if $y_1(t) = e^t \cos t$ satisfies the differential equation, then its characteristic root must be encoded in this solution's form. Since $e^t \cos t$ comes from the complex exponential $e^{(1+i)t}$, the number $1+i$ is a root of the characteristic polynomial—and so is its complex conjugate $1-i$. Once you identify these two roots, you can factor the degree-4 characteristic equation and find the remaining roots, which will give you four linearly independent solutions to build the general solution.

The key insight is that if $y_1(t) = e^t \cos t$ satisfies the differential equation, then its characteristic root must be encoded in this solution's form. Since $e^t \cos t$ comes from the complex exponential $e^{(1+i)t}$, the number $1+i$ is a root of the characteristic polynomial—and so is its complex conjugate $1-i$. Once you identify these two roots, you can factor the degree-4 characteristic equation and find the remaining roots, which will give you four linearly independent solutions to build the general solution.