Solve $y^{v} - y^{iv} = 0$

Since the equation is homogeneous linear with constant coefficients, we guess $y = e^{rt}$. Plugging this in yields:

After factoring, we obtain:

Thus the solutions for $r$ are

Since $r = 1$ is a root, one solution is

Since $r = 0$ is a root, another solution is

Since $r = 0$ is repeated 4 times, there are 4 linearly independent solutions corresponding to this root. The 4 solutions are obtained by multiplying a solution by $t$. Thus these 4 solutions, starting with $y = 1$ are

To summarize:

Since we have a fifth order differential equation, $~~~y^{v} - y^{iv} = 0$

the characteristice polynomial with be of degree 5: $~~~r^5 - r^4 = r^4(r - 1)$

Since the characteristic polynomial has degree 5, it will have 5 roots including repeats: $~~~r = 0, 0, 0, 0, 1$.

Thus we have the following 5 linearly independent solutions:

Hence the general solution is

Note we can think of this equation as quadratice in the variable $r^2$: $(r^2)^2 + 8(r^2) + 16 = 0$ Thus we can factor this in the same way we factor $x^2 + 8x + 16 = 0$, where $x = r^2$. After factoring over the real numbers, we obtain:

Thus the solutions to the characteristic equation is

Note these solutions are repeated roots

To summarize:

Since we have a fourth order differential equation, $~~~y^{iv} + 8y'' + 16y = 0$,

the characteristice polynomial with be of degree 4: $~~~r^4 + 8r^2 + 16$

Factored over the real numbers: $~~~(r^2 + 4)^2$

Factored over the complex numbers: $~~~[(r - 2i)(r + 2i)]^2 $

Since the characteristic polynomial has degree 4, it will have 4 roots including repeats: $~~~2i, 2i, -2i, -2i$.

Since $y = \pm 2i$ are purely imaginary solutions, then 2 linearly independent solutions are

Since these roots are repeated, we have the following 4 linearly independent solutions (where we multiply be $t$ for repeated roots):

Thus the general solution is