Final answer:
To find the homogeneous linear higher-order differential equation for y(x) = Ae^5x + Be^x cos(3x) + Ce^x sin(3x), we consider the characteristic polynomial (r - 5)(r - 1 - 3i)(r - 1 + 3i), leading to a third-order equation.
Step-by-step explanation:
The given function y(x) = Ae5x + Bex cos(3x) + Cex sin(3x) represents a combination of exponential and oscillatory solutions which suggests that the corresponding homogeneous linear higher-order differential equation with constant coefficients would need to accommodate these forms. For the exponential part Ae5x, we expect a characteristic equation to have a root at 5. For the oscillatory part Bex cos(3x) + Cex sin(3x), we anticipate a pair of complex conjugate roots 1 ± 3i, since the solutions oscillate with ex as the envelope and 3 as the frequency of oscillation.
Combining these roots, the characteristic polynomial would be (r - 5)(r - 1 - 3i)(r - 1 + 3i). Multiplying out these factors gives us a third-order polynomial with real coefficients, providing the base for our differential equation. Thus, by factoring and expanding, we would find the differential equation that has y(x) as its general solution.