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In the realm of second-order linear differential equations, the 'characteristic equation' plays a pivotal role. It's a way to convert the problem of solving a differential equation into an algebra problem. Essentially, we make an educated guess that the solution will look like an exponential function, specifically in the form of x = e^{rt}, where r is the rate of growth or decay in the function. When we plug this assumed solution back into the original differential equation, we systematically erase the function itself, leaving us with an equation solely in terms of the mysterious r. This clever manipulation yields the characteristic equation, which, for our exercise, is r^2 + 3r - 10 = 0.

Solving this algebraic equation provides us with the roots, which represent crucial constants in the general solution of the differential equation. In our exercise, the factorization method splits the quadratic into (r+5)(r-2) = 0, revealing the roots r = -5 and r = 2. These roots indicate the exponential solutions to the original differential equation.

When we say a differential equation has 'constant coefficients,' we're highlighting that the coefficients in the equation do not change; they are not functions of the variable we’re differentiating with respect to. This feature is significant because it means the differential equation behaves uniformly across the entire domain. The predictability of constant coefficients greatly simplifies the process of finding solutions and is a key reason why we can use characteristic equations so effectively.

In our example, the equation \(\frac{d^{2} x}{d t^{2}} + 3 \frac{d x}{d t} - 10 x = 0\) is a textbook model of constant coefficients with 3 and -10 serving as our steadfast constants. The simplicity lends itself to an algebraic analysis through the characteristic equation, allowing us to solve for r without worrying about the coefficients changing as t varies.

The 'general solution' of a differential equation encapsulates all possible solutions to the equation. It does this by incorporating constant terms, which can be altered to fit specific initial conditions or boundary values. What's remarkable is that the general solution accounts for different scenarios without fixing the constants, providing a flexible expression ready to conform to additional information.

In our example, finding roots r = -5 and r = 2 leads us to two particular solutions of the form e^{rt}. The general solution becomes a linear combination of these solutions: x(t) = C_1 e^{-5t} + C_2 e^{2t}. Here, C_1 and C_2 are arbitrary constants which will be determined by the context of the problem—such as initial values or other constraints. Hence, we not only have a solution that solves the differential equation in general, but also a versatile answer that can be tailored to meet specific requirements.