91Ó°ÊÓ

In the realm of differential equations, understanding when a unique solution exists is crucial. When we consider a first order linear differential equation like the one presented, the theorem that defines the existence of a unique solution, often known as the Existence and Uniqueness Theorem, asserts that if the functions representing the coefficients and the non-homogeneous term are continuous over an interval, then within that interval, for any initial condition, there should be one and only one solution. What does this mean in practice? For a given initial condition, such as one of the points mentioned in the exercise - like \[\begin{equation}(a) y(5)=2, (b) y\bigg(-\frac{3}{2}\bigg)=1,(c) y(0)=0,(d) y(-5)=4,(e) y\bigg(\frac{3}{2}\bigg)=3,\bigg),\bigg(\end{equation}\] a unique solution can only be guaranteed within the bounds that the functions are continuous.

Initial conditions are an anchor for solving differential equations, defining where the solution begins. In the context of the provided exercise, an initial condition like \( y(5)=2 \) instructs us to start the solution at the point where \( t=5 \) and \( y=2 \). The importance of these conditions cannot be understated; they directly influence the interval in which a unique solution exists. Initial conditions are like stating your starting point on a map - giving a specific location from which all the other points on the solution's path must align.

For a first order linear differential equation, the coefficients of \( y \) and the terms on the other side of the equation must be continuous for the Existence and Uniqueness Theorem to hold true. In our exercise, we look at the functions \( p(t) = \frac{t}{t^{2}-4} \) and \( q(t)=\frac{e^{t}}{t-3} \), noting where they are not continuous. Discontinuities are like breaks in the road; we cannot get a full path (solution) if there is an interruption. Therefore, ensuring function continuity over an interval informs us where solutions can smoothly exist. For instance, the discontinuities at \( t = 2, -2 \) and \( t = 3 \) for the functions \( p(t) \) and \( q(t) \) respectively, essentially carve out intervals where solutions can be reliably found.

Applying theorems in differential equations is akin to utilizing a set of rules to predict where and when we can expect solutions to exist. The application of Theorem 2.1 involves verifying the conditions for continuity as outlined above and linking those with the given initial conditions. By examining the functions at those initial conditions and the adjacent points of discontinuity, we can delineate the largest interval where the theorem assures us of a unique solution. For example, in Step 7 of our solution for the initial condition \( y\bigg(\frac{3}{2}\bigg)=3 \), the existence of a unique solution is guaranteed between the discontinuities at \( t = 2 \) and \( t = 3 \), yielding the interval \( a = 2, b = 3 \). Thus, theorem application is not just about the mechanical use of mathematical rules but also about interpreting the continuity of our functions and the interaction with initial conditions to define the domain of our solutions.