Problem 13 - Advanced Engineering Mathematics

Question

In Problems 1-18, solve the given differential equation. $$ 3 x^{2} y^{\prime \prime}+6 x y^{\prime}+y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y = x^{-\frac{1}{2}} (C_1 \cos(\frac{\sqrt{3}}{6} \ln x) + C_2 \sin(\frac{\sqrt{3}}{6} \ln x)) \).

Step 1: Recognize the type of differential equation

The given equation is a linear second-order homogeneous differential equation with variable coefficients: \(3x^2y'' + 6xy' + y = 0\). This identifies it as an equation that can potentially be solved using special methods such as the method of Frobenius.

Step 2: Transform the equation

Divide the entire equation by \(3x^2\) to simplify it: \(y'' + \frac{2}{x}y' + \frac{1}{3x^2}y = 0\). This is a Cauchy-Euler equation.

Step 3: Suggest a substitution

Use the substitution \(y = x^m\) because Cauchy-Euler equations have solutions of this form. So, \(y' = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\).

Step 4: Substitute and simplify

Substitute the expressions for \(y\), \(y'\), and \(y''\) back into the differential equation: \[ m(m-1)x^{m-2} + \frac{2}{x}mx^{m-1} + \frac{1}{3x^2}x^m = 0 \]Simplify to get:\[ m(m-1) + 2m + \frac{1}{3} = 0 \].

Step 5: Solve the characteristic equation

This gives us a quadratic: \[ m^2 + m + \frac{1}{3} = 0 \].Use the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = 1\), and \(c = \frac{1}{3}\).

Step 6: Evaluate the roots

Calculate the discriminant: \(b^2 - 4ac = 1^2 - 4 \times 1 \times \frac{1}{3} = 1 - \frac{4}{3} = -\frac{1}{3}\), which is negative, indicating complex roots: \[ m = \frac{-1 \pm i\sqrt{\frac{1}{3}}}{2} \].

Step 7: Write the general solution

With complex roots \(m = \alpha \pm \beta i\), the solution to the differential equation is: \[ y = x^{\alpha}(C_1 \cos(\beta \ln x) + C_2 \sin(\beta \ln x)) \].Here, \(\alpha = -\frac{1}{2}\) and \(\beta = \frac{\sqrt{3}}{6}\), so: \[ y = x^{-\frac{1}{2}} \left(C_1 \cos\left(\frac{\sqrt{3}}{6} \ln x\right) + C_2 \sin\left(\frac{\sqrt{3}}{6} \ln x\right)\right) \].

Key Concepts

method of Frobenius

The method of Frobenius is a powerful technique often used to solve second-order linear differential equations, especially those with variable coefficients like the Cauchy-Euler equations. This method becomes particularly useful when the point of interest, often called the regular singular point, causes trouble for other solving techniques.
The core idea of the Frobenius method is to assume a solution in the form of a power series:

• Start with the form: \( y = \sum_{n=0}^{\infty} a_n x^{n+c} \)
• Here, the constant \( c \) is determined such that a valid solution exists.

Then, by substituting this series into the differential equation, you can find values for the coefficients \( a_n \), giving you a series solution. This method is especially robust in handling complex roots, allowing for solutions to be expressed even when traditional methods might falter.

second-order differential equation

A second-order differential equation involves derivatives up to the second order, meaning it includes terms such as \( y'' \). These types of equations describe a wide range of physical phenomena such as oscillations, mechanical vibrations, and electrical circuits.
When dealing with a second-order differential equation, it can be simplified into one of three main forms:

• Homogeneous: All terms are dependent on the function and its derivatives.
• Non-homogeneous: Includes additional terms or functions independent of the function and its derivatives.
• Variable coefficients: Coefficients of the terms may not be constants, but functions of the variable \( x \).

Recognizing the order and type of a differential equation sets the stage for selecting an appropriate method for finding its solution.

homogeneous differential equation

A homogeneous differential equation is one in which every term is a multiple of the function itself or its derivatives. In other words, there are no independent or constant terms.
Formally, a homogeneous differential equation might look like:

• \( a_2(x)y'' + a_1(x)y' + a_0(x)y = 0 \)

This means if you substitute \( y = 0 \) into the equation, it holds true because all terms include \( y \) or its derivatives.
Homogeneous equations often allow us to utilize techniques like the method of Frobenius, exploiting their inherent symmetry to find general solutions that describe the behavior of the system accurately.

complex roots

Complex roots arise in the characteristic equation when its discriminant is negative. For a quadratic like \( m^2 + m + \frac{1}{3} = 0 \), the discriminant \( (b^2 - 4ac) \) being negative implies that the solutions for \( m \) involve complex numbers with an imaginary component.
Such complex roots take the form \( m = \alpha \pm \beta i \), where \( \alpha \) and \( \beta \) are real numbers. The presence of complex roots in the solution indicates oscillatory behavior in the differential equation's solutions.
For second-order differential equations, complex roots lead to a beautifully structured general solution that involves trigonometric functions:

• \( y = x^{\alpha}(C_1 \cos(\beta \ln x) + C_2 \sin(\beta \ln x)) \)

This reflects vibrational modes or wave-like solutions common in many physical systems.

general solution

The general solution of a differential equation encompasses all possible solutions. For a second-order differential equation, especially with complex roots, the general solution would cover both trigonometric and exponential specific solutions.
The beauty of the general solution lies in its completeness, often expressed with arbitrary constants like \( C_1 \) and \( C_2 \). These constants are valuable as they can be determined when initial or boundary conditions are applied to a problem.

• In our given problem, the general solution was expressed as: \[ y = x^{-\frac{1}{2}} \left(C_1 \cos\left(\frac{\sqrt{3}}{6} \ln x\right) + C_2 \sin\left(\frac{\sqrt{3}}{6} \ln x\right)\right) \]

This form encapsulates all potential solutions to the differential equation, allowing for versatile applications in different contexts.