To find the complementary function of the given equation, we need to solve for the solution to the homogeneous equation D^8 + 6D^6 - 32D^2 = 0.
First, we can observe that the polynomial equation can be factored as (D^2)(D^6 + 6D^4 - 32) = 0.
Setting D^2 = 0, we find one solution, which is D = 0.
Now, let's focus on the equation D^6 + 6D^4 - 32 = 0. This is a sixth-degree polynomial equation, and it can be difficult to factor or solve directly. However, we can make a change of variable to solve it more easily.
Let u = D^2. The equation then becomes u^3 + 6u^2 - 32 = 0, which is a third-degree polynomial equation.
Using synthetic division or any other method, we find that the solutions to u^3 + 6u^2 - 32 = 0 are u = -4, u = 2, and u = -8.
Now, we can substitute back u = D^2 to find the original values of D:
For u = -4: D^2 = -4, D = ±2i
For u = 2: D^2 = 2, D = ±√2
For u = -8: D^2 = -8, D = ±2√2i
So the solutions to the homogeneous equation D^8 + 6D^6 - 32D^2 = 0 are D = 0, ±2i, ±√2, ±2√2i.
Therefore, the complementary function of the given equation is:
y(x) = c1 + c2e^(2ix) + c3e^(-2ix) + c4e^(√2x) + c5e^(-√2x) + c6e^(2√2ix) + c7e^(-2√2ix) + c8x