91Ó°ÊÓ

A second-order differential equation involves the second derivative of a function. It is called "second-order" because the highest derivative is the second one. These equations often appear in physics and engineering when describing systems with acceleration or curvature. Euler's equation, a type seen in this exercise, is a linear homogeneous second-order differential equation of the form \[ ax^2 y'' + bx y' + c y = 0 \] In the given problem, \[-x^2 y'' + x y' - y = 0 \] functions as such an equation where the coefficients are based on powers of \( x \). This specific equation becomes manageable by identifying patterns and employing specific mathematical techniques, like substitution.

The general solution to a differential equation provides a relationship or formula that encompasses all possible solutions. For second-order differential equations, it often includes constants that can be determined with initial conditions.

In the context of Euler's equation, solving for a general solution usually requires determining the form of the solution first. Here, we use the substitution \( y = x^m \) after recognizing the equation's homogeneous nature and special characteristics. This choice allows the transformation of the equation into a simpler algebraic form. The final general solution is then expressed in terms of arbitrary constants, \( C_1 \) and \( C_2 \), multiplied respectively by functions representing separate solution families.

The method of change of variables is a powerful technique to simplify differential equations, making them more approachable.

In this problem, the original equation involves terms tied with powers of \( x \). By substituting \( y = x^m \), we effectively transform a second-order differential equation into an easier algebraic equation. This new equation relates \( m \), the power, to the constants \( a \) and \( b \) of Euler's equation. Here, derivatives become simpler:- \( y' = m x^{m-1} \)- \( y'' = m(m-1)x^{m-2} \)
Thus, this algebraic transformation results in \[-m^2 + m + 1 = 0 \], a quadratic form that is straightforward to solve.

The quadratic formula is a staple in solving equations where the variable is squared. For any equation written as \( ax^2 + bx + c = 0 \), the solutions can be found using:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In Euler's equation problem, we reach a point where solving for \( m \) in \(-m^2 + m + 1 = 0\) is necessary. The quadratic formula is applied here with \( a = 1 \), \( b = 1 \), and \( c = 1 \). Using these values:- First, compute the discriminant: \( 1^2 - 4 \cdot 1 \cdot 1 = -3 + 4 = 1 \)- Then solve: \[ m = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]
Thus, we uncover two distinct solutions for \( m \), which when substituted back provide the two-parameter general solution.