91Ó°ÊÓ

A first-order differential equation involves derivatives of the first degree, which means it contains the first derivative of a function but no higher derivatives. A fundamental example of such an equation is \( \frac{dy}{dt} = ky \), where \( y \) is the function of \( t \) that we need to determine and \( k \) is a constant coefficient.

To solve these equations, we often use separation of variables or integrate factors methods. Typically, the solution involves an exponential function because the rate of change of the function is proportional to the function itself. For instance, the solution to \( \frac{dy}{dt} = 2y \) is \( y = Ce^{2t} \), where \( C \) is an arbitrary constant determined by initial conditions. This solution captures the essence of growth or decay processes frequently occurring in natural phenomena.

Homogeneous equations have a characteristic that they can be written in the form \( \frac{dy}{dt} = f(y/t) \). For the differential equation \( \frac{dy}{dt} = ky \), it is homogeneous and linear because it depends on the function \( y \) but not directly on \( t \).

In this context, homogeneous doesn't mean the same thing as in chemistry. It indicates the absence of a standalone term, meaning no independent or forcing function affects \( y \) and \( t \) directly. These equations exhibit solutions that are self-similar. The solution is multiplied by any constant \( C \), leading to another valid solution. This property highlights why solutions using forms like \( Ce^{kt} \) can be generalized to encompass a family of solutions rather than a single one.

Verification of solutions in differential equations involves ensuring that the proposed solution satisfies the given differential equation. For the equation \( \frac{dy}{dt} = 2y \), any valid solution must align with this relationship.

To verify, you substitute the proposed solution into the differential equation. Consider a general solution \( y = Ce^{2t} \). Derive \( y \) with respect to \( t \) to get \( \frac{dy}{dt} = 2Ce^{2t} \). This matches \( 2y \), thus confirming it is a solution. The exercise illustrates the importance of verifying by comparing different forms, such as \( y = C(3^{kt}) \). By substitution and comparison, if the forms don't align, as with the exponents not matching, then \( C(3^{kt}) \) is not a solution for the given equation, hence the statement is false.