Final answer:
The general solution to the differential equation y''' + 8y'' + 25y' = 0 includes a constant term and two exponential terms multiplied by trigonometric functions. These are as a result of the real and complex roots of the characteristic equation associated with the differential equation. The final solution is y(x) = c₁ + c₂*e^(-4x)cos(3x) + c₃*e^(-4x)sin(3x).
Step-by-step explanation:
To solve the differential equation y''' + 8y'' + 25y' = 0, we first need to find the characteristic equation which is derived by replacing every y''' with r^3, y'' with r^2, and y' with r. Doing this, we get the characteristic equation r^3 + 8r^2 + 25r = 0. We can factor out an r to get r(r^2 + 8r + 25) = 0. The roots of this equation are r = 0 and the roots of the quadratic r^2 + 8r + 25, which has complex roots since its discriminant (b^2 - 4ac) is negative (8^2 - 4(1)(25) = 64 - 100 = -36).
The complex roots can be found using the quadratic formula: r = (-8 ± √(-36))/2. Simplifying, we get two complex roots: -4 + 3i and -4 - 3i. For the differential equation, the general solution comprises terms that correspond to these roots. For the real root r = 0, the solution is c₁. For the complex roots, the solutions will be in the form of e^(at) times a linear combination of cos(bt) and sin(bt), where a = -4 and b = 3, leading to the solution c₂*e^(-4x)cos(3x) + c₃*e^(-4x)sin(3x).
Therefore, the general solution to the differential equation y''' + 8y'' + 25y' = 0 is y(x) = c₁ + c₂*e^(-4x)cos(3x) + c₃*e^(-4x)sin(3x), where c₁, c₂, and c₃ are arbitrary constants.