Question

The Euler equation \(x^{2} y^{\prime \prime}+\) \(\alpha x y^{\prime}+\beta y=0\) can be reduced to an equation with constant coefficients by a change of the independent variable. Let \(x=e^{z},\) or \(z=\ln x,\) and consider only the interval \(x>0 .\) (a) Show that $$ \frac{d y}{d x}=\frac{1}{x} \frac{d y}{d z} \quad \text { and } \quad \frac{d^{2} y}{d x^{2}}=\frac{1}{x^{2}} \frac{d^{2} y}{d z^{2}}-\frac{1}{x^{2}} \frac{d y}{d z} $$ (b) Show that the Euler equation becomes $$ \frac{d^{2} y}{d z^{2}}+(\alpha-1) \frac{d y}{d z}+\beta y=0 $$ Letting \(r_{1}\) and \(r_{2}\) denote the roots of \(r^{2}+(\alpha-1) r+\beta=0\), show that (c) If \(r_{1}\) and \(r_{2}\) are real and different, then $$ y=c_{1} e^{r_{1} z}+c_{2} e^{r_{2} z}=c_{1} x^{r_{1}}+c_{2} x^{r_{2}} $$ (d) If \(r_{1}\) and \(r_{2}\) are real and equal, then $$ y=\left(c_{1}+c_{2} z\right) e^{r_{1} z}=\left(c_{1}+c_{2} \ln x\right) x^{r_{1}} $$ (e) If \(r_{1}\) and \(r_{2}\) are complex conjugates, \(r_{1}=\lambda+i \mu,\) then $$ y=e^{\lambda z}\left[c_{1} \cos (\mu z)+c_{2} \sin (\mu z)\right]=x^{\lambda}\left[c_{1} \cos (\mu \ln x)+c_{2} \sin (\mu \ln x)\right] $$

Short answer

Answer: The general solution of the Euler equation when the roots of the characteristic equation are complex conjugates is given by: $$y = x^{\lambda}(c_1\cos(\mu \ln x) + c_2\sin(\mu \ln x))$$ where \(r_1 = \lambda + i\mu\) and \(r_2 = \lambda - i\mu\) are the complex conjugate roots of the characteristic equation, and \(c_1\) and \(c_2\) are constants.

Step-by-step solution

Substitute the new variable and find the derivatives

First, we need to find expressions for \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) in terms of \(z\). Since \(x = e^z\), we have \(dx = e^z dz\) or \(dz = \frac{1}{x}dx\). Now we can use the chain rule to find the derivatives. $$\frac{dy}{dx} = \frac{dy}{dz}\frac{dz}{dx} = \frac{1}{x}\frac{dy}{dz}$$ Now, we differentiate again with respect to \(x\): $$\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{1}{x}\frac{dy}{dz}\right) = \frac{1}{x^2}\frac{d^2y}{dz^2} - \frac{1}{x^2}\frac{dy}{dz}$$

Transform the Euler Equation

Now, we'll substitute the expressions for the derivatives from part (a) into the given Euler equation to transform it: $$x^{2}\left(\frac{1}{x^2}\frac{d^{2}y}{dz^{2}}-\frac{1}{x^2}\frac{dy}{dz}\right)+\alpha x\left(\frac{1}{x}\frac{dy}{dz}\right)+\beta y=0$$ This simplifies to: $$\frac{d^2y}{dz^2}+(\alpha-1)\frac{dy}{dz}+\beta y=0$$

Real and Different Roots

Given that \(r_1\) and \(r_2\) are real and different, the roots of the characteristic equation \(r^2+(\alpha-1)r+\beta=0\), the general solution of the equation will be of the form: $$y = c_1 e^{r_1 z} + c_2 e^{r_2 z}$$ Since \(x = e^z\), we substitute to get the final solution: $$y = c_1 x^{r_1} + c_2 x^{r_2}$$

Real and Equal Roots

If the roots are real and equal, let \(r_1 = r_2 = r\). The general solution for the equation will be of the form: $$y = (c_1 + c_2 z)e^{r z}$$ Substituting \(x = e^z\), we get the final solution: $$y = (c_1+c_2 \ln x)x^{r}$$

Complex Conjugate Roots

If the roots are complex conjugates, let \(r_1 = \lambda + i\mu\) and \(r_2 = \lambda - i\mu\). The general solution of the equation will be of the form: $$y = e^{\lambda z}(c_1\cos(\mu z) + c_2\sin(\mu z))$$ Substituting \(x = e^z\), we get the final solution: $$y = x^{\lambda}(c_1\cos(\mu \ln x) + c_2\sin(\mu \ln x))$$

Key concepts

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations that involve functions and their derivatives. In mathematics, these equations play a critical role in modeling dynamic systems where the rate of change of a variable is a function of the variable itself. For example, the simple equation \( \frac{dy}{dx} = ky \) models exponential growth or decay with the proportionality constant \( k \). ODEs can have different orders, with the order determined by the highest derivative present in the equation.

In the Euler equation exercise given, we deal with a second-order ODE, which means the highest derivative is the second derivative of the function \( y \). The equation involves both the first and second derivatives of \( y \) with respect to \( x \) and is characteristic of phenomena where the change rate itself changes over time. Understanding how to solve these equations is vital for fields such as physics, engineering, economics, and biology where such dynamic behavior is modeled.

Euler-Cauchy Equation

The Euler-Cauchy equation, also known as the Euler's homogeneous equation, is a special form of a second-order linear differential equation with variable coefficients that can be transformed into a second-order linear differential equation with constant coefficients. It has the form \( x^{2}y'' + \alpha xy' + \beta y = 0 \) where \( y' \) and \( y'' \) are the first and second derivatives of \( y \) with respect to \( x \), respectively, and \( \alpha \) and \( \beta \) are constants. The textbook exercise involves changing the independent variable from \( x \) to \( z \) by letting \( x=e^{z} \) and \( z=\ln x \) to simplify the equation into one with constant coefficients, making it easier to solve.

This approach is essential because it converts a seemingly difficult differential equation into a form where we can use established methods for constant coefficient differential equations to find the solution. This demonstrates the ingenuity in mathematical techniques to simplify and solve complex problems.

Characteristic Equation

The characteristic equation is a fundamental concept used to solve linear homogeneous differential equations with constant coefficients. It is an algebraic equation derived from the differential equation by assuming a solution of a specific form, typically an exponential function. For the Euler-Cauchy equation transformed with the substitution \( z=\ln x \), the characteristic equation becomes \( r^{2}+(\alpha-1)r+\beta=0 \).

To solve it, one typically looks for roots of this quadratic equation, which represent the powers of the exponential terms in the equation's solution. The nature of the roots—whether they are real and distinct, real and equal, or complex conjugates—determines the form of the general solution to the differential equation. Understanding how to obtain and solve the characteristic equation is crucial as it facilitates the process of finding the general solution to differential equations.

Complex Roots

When the characteristic equation yields complex roots, it indicates that the solution to the differential equation will involve trigonometric functions. If the roots are complex conjugates, expressed as \( r_{1} = \lambda + i\mu \) and \( r_{2} = \lambda - i\mu \), the general solution contains the exponential function multiplied by linear combinations of sine and cosine functions.

In our exercise example, the complex conjugate roots lead to a solution involving \( e^{\lambda z} \), representing the exponential decay or growth determined by the real part \( \lambda \), and trigonometric functions \( \cos(\mu z) \) and \( \sin(\mu z) \) with the imaginary part \( \mu \) affecting the oscillation. This form of solution is prevalent in systems that exhibit damped oscillations, such as in mechanical vibrations and electrical circuits. It's fascinating how the interplay between exponents and trigonometry can elegantly describe such dynamic and oscillatory systems.