Question

Find all values of \(\alpha\) for which all solutions of \(x^{2} y^{\prime \prime}+\alpha x y^{\prime}+(5 / 2) y=0\) approach zero as \(x \rightarrow \infty\).

Short answer

Answer: The solutions of the given differential equation approach zero as x tends to infinity for all values of α > 0.

Step-by-step solution

Identify the given equation and substitution

We are given the second-order Euler-Cauchy linear differential equation: \(x^2y'' + \alpha xy' + \frac{5}{2}y = 0\). We will use the substitution \(y = x^m\) to find the solution of this equation.

Differentiate and substitute in the equation

Find the first and second derivatives of \(y = x^m\) with respect to x: \(y' = mx^{m-1}\) and \(y'' = m(m - 1)x^{m-2}\). Now substitute these expressions into the given differential equation: $$x^2\left[m(m - 1)x^{m-2}\right] + \alpha x\left[mx^{m-1}\right] + \frac{5}{2}\left[x^m\right] = 0.$$

Simplify the equation and find the characteristic equation

Factor out the common term \(x^m\) and simplify: $$m(m-1)x^m + \alpha mx^m + \frac{5}{2}x^m = 0.$$ Now we can divide by \(x^m\) (as it's nonzero) and obtain the characteristic equation: $$m(m - 1) + \alpha m + \frac{5}{2} = 0.$$

Solve the characteristic equation

To solve this quadratic equation in term of m, we can use the quadratic formula: $$m = \frac{-\alpha \pm \sqrt{\alpha^2 - 4\left(\frac{5}{2}\right)(1-1)}}{2(1-1)}.$$ Simplifying the equation, we get: $$m = \frac{-\alpha \pm \sqrt{\alpha^2 - 10}}{2}.$$

Find the general solution

Now we know that the general solution to this Euler-Cauchy equation is given by: $$y = C_1x^{\frac{-\alpha + \sqrt{\alpha^2-10}}{2}} + C_2x^{\frac{-\alpha - \sqrt{\alpha^2-10}}{2}}.$$

Analyze the solution for the desired property

We are looking for values of α for which the general solution approaches zero as \(x \rightarrow \infty\). For this to happen, both exponents must be negative. Let's analyze the conditions for both exponents to be negative: 1. \(\frac{-\alpha + \sqrt{\alpha^2-10}}{2} < 0 \Rightarrow \alpha > \sqrt{\alpha^2 - 10}\) 2. \(\frac{-\alpha - \sqrt{\alpha^2-10}}{2} < 0 \Rightarrow \alpha > -\sqrt{\alpha^2 - 10}\) The first condition is always true, since \(\alpha\) and \(\sqrt{\alpha^2 - 10}\) have the same sign and the latter is smaller in absolute value. The second condition implies that \(\alpha > 0\). So, the system of conditions that we found for α is: 1. \(\alpha > \sqrt{\alpha^2 - 10}\) 2. \(\alpha > 0\)

Find the set of values for α

We have already established that the first condition is always true, so we can disregard it. This leaves us with the second condition as the only requirement for the solutions to approach zero when \(x \rightarrow \infty\). We conclude that for all values of \(\alpha > 0\), the given differential equation has solutions that approach zero as \(x \rightarrow \infty\).

Key concepts

Characteristic Equation

The characteristic equation is a pivotal element when dealing with linear differential equations, especially in the case of Euler-Cauchy equations. It serves as a bridge between differential operators and algebra, allowing us to convert a calculus problem into an algebraic one. By substituting a trial solution of the form y = x^m into the differential equation, we derive a polynomial where the exponent m becomes our variable of interest.

For the equation m(m - 1) + \text{\(\text{\alpha}\)} m + \(\text{\frac{5}{2}}\) = 0, a quadratic in terms of m, we are essentially seeking roots that will correspond to the power m in the trial solution x^m, which defines the behavior of the differential equation's solution. These roots are instrumental in determining the nature and asymptotic behavior of the solution as x approaches infinity. Understanding and solving the characteristic equation is hence a crucial step towards the comprehensive analysis of differential equations.

Differential Equations

Differential equations are powerful mathematical tools used to describe various phenomena in the fields of physics, engineering, economics, and beyond. They relate functions to their derivatives, representing rates of change and enabling the modeling of dynamic systems.

In our exercise, we encountered the Euler-Cauchy differential equation, which is a type of second-order linear differential equation characterized by varying coefficients that are powers of the independent variable x. To solve such equations, we look for solutions where the function y and its derivatives y' and y'' are proportional to x raised to some power m. This method simplifies the problem significantly, transforming a complicated differential equation into a more manageable characteristic equation, as shown in the provided step-by-step solution.

It's also vital for students to appreciate the importance of boundary conditions and initial values to determine the constants in these solutions, which in turn define the precise behavior of the system being studied.

Asymptotic Behavior

Understanding the asymptotic behavior of solutions to differential equations is crucial when predicting long-term trends in dynamic systems. Asymptotic behavior refers to the tendencies of functions as the input approaches a particular value, often infinity. In our exercise scenario, we're interested in the behavior as x approaches infinity.

The values of \(\alpha\) determine the powers of x in the solution to the Euler-Cauchy equation, influencing whether the solution grows, decays, or oscillates as x increases. We look for conditions leading to decay to ensure the function approaches zero. The constraints derived from the characteristic equation's roots result in requiring that both roots be negative for the solution to approach zero as x goes to infinity.

Thus, analyzing the asymptotic behavior is essential as it helps us assert the stability of solutions over time, which can be particularly important in physical systems where unbounded growth could be an indication of system failure or instability.