91Ó°ÊÓ

Understanding second-order linear differential equations is crucial for many fields of science and engineering. These equations are characterized by the highest derivative being the second derivative of the unknown function. In the given exercise, the form of the equation is \(2 x^{2} y^{\prime \prime}(x)+13 x y^{\prime}(x)+15 y(x)=0\), which represents a second-order linear differential equation with variable coefficients. The solution approach involves a substitution that simplifies the equation to a solvable form.

By using the substitution \(y(x)=x^r\), where \(r\) is a constant we want to determine, the equation transforms into a polynomial that can be solved for specific values of \(r\). In essence, this technique, often named the power series method or Frobenius method, reduces the complexity of the differential equation by eliminating the variable nature of the coefficients and leaning on the properties of polynomials instead.

Polynomial equation solving is a foundational skill in algebra, impacting fields as diverse as economics, physics, and computer science. The polynomial arising in this exercise, \(2 r^2 -2r +13 r+15=0\), needs to be solved for \(r\) to progress towards the differential equation's solution. Polynomial equations are composed of terms that involve a variable raised to a whole number exponent and can often be simplified and rearranged to reveal their roots.

The process of solving involves several possible techniques, including factorization, long division, synthetic division, and the use of formulas. For second-degree polynomials, like the one we have in this case, the quadratic formula becomes particularly relevant, providing a direct path to finding the roots.

The quadratic formula is a silver bullet for clearing the hurdle of quadratic equations, which are polynomial equations of the second degree. In the provided solution, we have the quadratic in the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are known coefficients. The quadratic formula: \(r = (b \pm \sqrt{b^2 - 4ac}) / (2a)\), helps to determine the roots of such equations effectively.

Using this formula bypasses the need for factoring, which can be challenging when the roots are not integers or are complex. It's a catch-all method, capable of handling a wide array of quadratic equations, and reflects the importance of formulas that generalize solutions across mathematics.

The characteristic equation is the cornerstone of solving linear differential equations with constant coefficients, and it similarly assists with variable coefficients after an appropriate substitution. In our exercise, when we factor out \(x^r\) and remove the dependency on \(x\), we're left with a polynomial in terms of \(r\). This is our characteristic equation: \(2 r(r-1) + 13 r + 15 = 0\).

This equation encapsulates the behavior of the differential equation in terms of the eigenvalue \(r\). Solving for \(r\) provides the exponents for the solution to the original differential equation. In other words, the roots of the characteristic equation are directly linked to the solution of the second-order differential equation, defining the nature of the solution, whether it's real or complex, and thus guiding us to the general solution.