Problem 42 - Advanced Engineering Mathematics

Question

Use the substitution \(y=\left(x-x_{0}\right)^{m}\) to solve the given equation. $$ (x-4)^{2} y^{\prime \prime}-5(x-4) y^{\prime}+9 y=0 $$

Step-by-Step Solution

Answer

Substituting \( y = t^3 \) solves the differential equation with \( t = x-4 \).

Step 1: Introduce the The Substitution

Let \( t = x - 4 \) so that \( y = t^m \). This substitution transforms the differential equation in terms of \( t \). Substitute these into the given equation.

Step 2: Apply Chain Rule for Derivatives

Calculate the first and second derivatives using the chain rule: \( y' = m t^{m-1} \) and \( y'' = m(m-1) t^{m-2} \). Substitute these into the equation replacing \( y, y', y'' \).

Step 3: Substitute Derivatives into Original Equation

Plug in \( y = t^m \), \( y' = m t^{m-1} \), and \( y'' = m(m-1) t^{m-2} \) into \( (x-4)^2 y'' - 5(x-4) y' + 9y = 0 \). This simplifies to: \[ t^2 m(m-1) t^{m-2} - 5t m t^{m-1} + 9t^m = 0. \]

Step 4: Simplify the Equation

Simplify the equation by multiplying out all terms: \[ m(m-1) t^m - 5m t^m + 9t^m = 0. \] Factor out \( t^m \) from the equation: \[ (m(m-1) - 5m + 9) t^m = 0. \]

Step 5: Solve for \( m \)

Since \( t^m eq 0 \) for \( t eq 0 \), solve \( m(m-1) - 5m + 9 = 0 \). Simplify to \( m^2 - 6m + 9 = 0 \), which factors as \( (m-3)^2 = 0 \). Thus, \( m = 3 \).

Step 6: Find the General Solution

Since \( m = 3 \) is a repeated root, the general solution for \( y \) given \( t = x - 4 \) is \( y = c_1 t^3 + c_2 t^3 \ln t \), where \( c_1 \) and \( c_2 \) are constants.

Key Concepts

Change of Variables

A powerful technique in solving differential equations is using a change of variables. It involves substituting one set of variables for another to simplify the problem. In this exercise, we use the substitution \( y = (x-4)^m \). This is particularly useful because it transforms the original differential equation, which is in terms of \( x \), into an equation in terms of \( t = x - 4 \).

Why do we do this?

• It can turn a complex differential equation into a simpler form that is easier to solve.
• By choosing a substitution that matches the structure of the equation, we can often spot patterns and solutions more easily.

In this exercise, the change of variables helped streamline calculations by reducing the complexity of the problem into a more manageable task. This is a common practice particularly in differential equations where the goal is to simplify derivatives and coefficients.

Characteristic Equation

In solving differential equations, particularly linear ones, the characteristic equation arises naturally when simplifying using substitutions and factoring. The characteristic equation is a polynomial equation derived from the differential equation that helps find specific values, which in this case is \( m \) from our substitution.

Upon transforming the differential equation with the substitution \( t = x - 4 \), we reach an expression of \((m(m-1) - 5m + 9) t^m = 0.\) Solving for \( m \) involves setting the polynomial part equal to zero and solving \( m(m-1) - 5m + 9 = 0 \).

Key aspects of the characteristic equation include:

• It transforms a differential problem into an algebraic one.
• Solving the characteristic equation gives the possible behaviors of the solution \( y \).
• In this exercise, the roots indicate the type of solution we construct, such as repeated roots affecting the form of the general solution.

Understanding the characteristic equation is crucial as it breaks down the complexity of differential equations into simpler algebraic steps.

General Solution

After solving the characteristic equation, the next step is constructing the general solution. This is a form that includes all possible solutions of the differential equation. It reflects both particular solutions and a basis of solutions for the differential equation.

In our example, solving \( m^2 - 6m + 9 = 0 \) (the characteristic polynomial) gives a repeated root \( m = 3 \). When there is a repeated root, the solutions take a specific form:

• \( y = c_1 t^3 \), a typical power solution.
• \( y = c_2 t^3 \ln t \), introduced due to the repeated root.

This gives us the general solution:\[ y = c_1 t^3 + c_2 t^3 \ln t \]where \( c_1 \) and \( c_2 \) are arbitrary constants determined by initial conditions or boundary values.

The general solution is comprehensive because it incorporates all possible variations determined by the constants.

Second-Order Differential Equation

Second-order differential equations involve the second derivative, such as the equation given in this exercise. They are more complex than first-order differential equations due to the presence of the \( y'' \) term.

The general form is:\[ a(x) y'' + b(x) y' + c(x) y = 0 \]where \( a(x) \), \( b(x) \), and \( c(x) \) are functions of \( x \). In our exercise, the terms are structured specifically around \( x - 4 \), which contributes directly to the need for substitution.

Characteristics of second-order differential equations include:

• They model a vast array of physical systems, such as oscillations and waves.
• They often require techniques such as change of variables or characteristic equation to simplify.
• Solutions can be more complex, including terms like logarithmic functions, especially when dealing with repeated roots.

Understanding how to approach these equations, particularly with substitution and simplification techniques, is fundamental in mastering their solutions.