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.
The meaning is the same: a particular solution is just that, one solution,
corresponding to one choice of initial conditions. If you have a formula for the general solution, in a second order equation it will have two arbitrary parameters and each choice
of values for those parameters gives you a particular solution.
Linear equations (for any order) have the property that if you add a solution of the homogeneous equation and a solution of the non-homogeneous equation, you get another solution of the same non-homogeneous equation. If you take the general solution of the
homogeneous equation (involving, for a second-order equation, two parameters), and add
a particular solution of the non-homogeneous equation, you get the general solution of the non-homogeneous equation.
Best Answer
To say that you have found the general homogeneous solution means that this function solves the homogeneous equation for every choice of the constant $C_1$ and every solution of the homogeneous equation is of this form for some choice of $C_1$.
You can actually have more than one particular solution to a DEQ. The difference between any two particular solutions is always a homogeneous solution.
Example:
$$y' + \left(\frac{a}{t}\right)y = t^3$$
The homogeneous solution is:
$$\displaystyle y_H = c_1 t^{-a}$$
Here are two particular solutions:
$$\displaystyle y_{1P} = \frac{t^4}{4+ a}$$
$$y_{2P} = \displaystyle \frac{t^4}{4+ a} + c_1 t^{-a}$$
What is the difference between these two particular solutions?
For this, we obviously need to be given an initial condition.