Here's how a person might be able to solve this, even without ever being taught anything about differential equations:
First, such a person might know that constant functions have derivative zero, and might then realize that if $y$ is constant, then $y'$ and $y'''$ are both identically zero, so $8y'''+8y'=0$. Thus, every constant function is a solution.
Second, such a person may know that the derivative of $\sin x$ is $\cos x$, and the derivative of $\cos x$ is $-\sin x$. It follows immediately that the second derivative of $\sin x$ is $-\sin x$, so, if $y=\sin x$, then $y''=-y$, whence $y''+y=0$, whence $8y''+8y=0$. Now differentiating gives $8y'''+8y'=0$, so $y=\sin x$ is a solution.
But then the same argument applies to $B\sin x$ for any constant $B$, so $y=B\sin x$ is a solution. And also, the second derivative of $\cos x$ is $-\cos x$, so, for any constant $C$, $y=C\cos x$ is a solution.
Then, knowing that $(f+g)'=f'+g'$, our hypothetical person may come to realize that $$y=A+B\sin x+C\cos x$$ is a solution for all constants $A,B,C$.
This is where I get stuck – I'm not sure how the person who has never been taught anything about solving differential equations could conclude that every solution is of this form, so that we have found the general solution.
A 2nd order DE specifies the second derivative of the solution. This leaves the value at some time $t_0$ and the first derivative at that same time unspecified. Those are your two degrees of freedom. One example might be a ball on a spring. Wherever and however it starts, the equation of motion tells you that the ball will follow a sinusoid motion. But it doesn't tell you where the initial location of the ball has to be, or the initial kick you can give it in its motion.
ADDED: This is not a complete explanation, but might give you an idea of how it all works. Let's say we have a DE
$\displaystyle
\frac{d^2x}{dt^2} = f(t)$
and let's say we find out that it is satisfied by some $x_0(t)$. Now, what if we randomly pick two numbers $a$ and $b$ and add the following to $x_0(t)$ to make $x_1(t)$:
$\displaystyle
x_1(t) = x_0(t) +a +bt $
Is $x_1(t)$ also a solution of the equation, or not? Yes, it is, because
$\displaystyle
\frac{d^2x_1}{dt^2} = \frac{d^2x_0}{dt^2} = f(t)$
That is, the second derivative "killed" both the $a$ and $bt$ terms. Note that this is going to happen regardless of what values you choose for $a$ and $b$. So you have two "degrees of freedom" in your solution.
Best Answer
The $D^n$ just refers to the $n'th$ derivative of $y$ with respect to $x$.
So the first equation is just: $$y''''+8y''+16y=0$$
Which is a linear homogenous ordinary differential equation. To solve this let $y=e^{\gamma x}$ and you'll be left with a quadratic characteristic equation if you can't solve that drop a comment or do some research on 'linear homogenous ODE's'.
Similarly, the second equation is going to be: $$y''-3y'+2y=\frac{1}{1+e^{-x}}$$
which is a linear nonhomogeneous ordinary differential equation. Solved almost exactly the same way however in this case, you will also have to find a particular solution (due to the inhomogeneity). Without having solved it yet myself, I would say to use the variation of parameters method to solve the particular solution.
Hope that answers your question!