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A second-order differential equation is a type of differential equation that involves the second derivative of a function. In other words, it includes terms with the function's second derivative, which is often denoted as \(y''\) or \(d^2y/dx^2\). These equations can be more complex than first-order differential equations because they involve higher-level calculus concepts. In the specific context of an Euler differential equation, we typically observe a second-order equation of the form:\[16x^2y'' + 56xy' + 25y = 0\]Euler equations are special because they involve variable coefficients, often powers of \(x\), which require unique methods for solving. When you are given a second-order Euler equation, like in the exercise, your goal is to determine the unknown function \(y(x)\) that satisfies this equation for all \(x\) within a certain range.

In the context of differential equations, roots refer to the solutions of the characteristic equation derived from the differential equation. When solving a second-order characteristic quadratic equation, you may encounter a scenario where the discriminant, which appears in the quadratic formula, is zero. This indicates a repeated root.For example, when you solve the quadratic equation:\[16m^2 + 40m + 25 = 0\]using the quadratic formula, you calculate that the discriminant (\[b^2 - 4ac\]) equals zero:\[40^2 - 4 \cdot 16 \cdot 25 = 0\]This means there's not two distinct roots, but rather one root repeated. In this problem, the repeated root \(m\) is \(-\frac{5}{4}\). Having a repeated root changes the form of the general solution of the differential equation.

The general solution to a differential equation provides an expression that encompasses all possible solutions of the equation. For second-order Euler equations with constant coefficients and a repeated root, like in this exercise, the general solution has a distinctive form.When you have a repeated root, the solution takes into account not just the root itself, but also a logarithmic term. Specifically, if the repeated root is \(m\), the general solution for this type of Euler equation is given by:\[y(x) = c_1 x^m + c_2 x^m \ln x\]Here, \(c_1\) and \(c_2\) are constants that can be determined if initial conditions are given. In the given problem, with the repeated root \(-\frac{5}{4}\), the general solution specifically comes out to:\[y(x) = c_1 x^{-\frac{5}{4}} + c_2 x^{-\frac{5}{4}} \ln x\]This formula effectively combines both the direct power term and a modification factor through the logarithm, capturing all the possible behaviors of \(y(x)\) that satisfy the original differential equation.