Problem 40 - Advanced Engineering Mathematics

Question

Use the substitution \(y=\left(x-x_{0}\right)^{m}\) to solve the given equation. $$ (x-1)^{2} y^{\prime \prime}-(x-1) y^{\prime}+5 y=0 $$

Step-by-Step Solution

Answer

The solution is \(y(x) = c_1 (x-1)^{1+i} + c_2 (x-1)^{1-i}\).

Step 1: Substitution Setup

Given the equation \((x-1)^{2} y'' - (x-1) y' + 5y = 0\), we will use the substitution \(y = (x-1)^m\). Here, the substitution implies that the solution has a form dependent on a power of \((x-1)\).

Step 2: First Derivative

Compute the derivative of \(y\). Given \(y = (x-1)^m\), the first derivative \(y'\) is \(y' = m(x-1)^{m-1}\).

Step 3: Second Derivative

Compute the second derivative of \(y\). The second derivative \(y''\) is \(y'' = m(m-1)(x-1)^{m-2}\).

Step 4: Substitute Derivatives into Original Equation

Substitute \(y\), \(y'\), and \(y''\) back into the original equation: \((x-1)^2 \cdot m(m-1)(x-1)^{m-2} - (x-1) \cdot m(x-1)^{m-1} + 5(x-1)^m = 0\), which simplifies to: \(m(m-1)(x-1)^m - m(x-1)^m + 5(x-1)^m = 0\).

Step 5: Simplify the Equation

Factor out \((x-1)^m\) from each term: \((x-1)^m [m(m-1) - m + 5] = 0\).

Step 6: Solve for m

Since \((x-1)^m\) cannot be zero, solve the equation \(m(m-1) - m + 5 = 0\): \(m^2 - 2m + 5 = 0\). This is a quadratic equation with no real solutions (since the discriminant \(b^2 - 4ac = 4 - 20\) is negative). So \(m = 1 \pm 2i\) are the complex roots.

Step 7: Write General Solution

The solution to the differential equation, taking into account the complex roots, is: \(y(x) = c_1 (x-1)^{1+i} + c_2 (x-1)^{1-i}\), where \(c_1\) and \(c_2\) are constants.

Key Concepts

Second Order Differential Equations

A second-order differential equation involves the second derivative of a function. Here, the presence of the second derivative in our equation: \[ (x-1)^{2} y^{ ext{''}}-(x-1) y^{ ext{'}}+5 y=0 \]makes it a second-order differential equation.
These equations commonly appear in physics and engineering, modeling systems like oscillations or dynamics.
In our problem, the equation has a form that depends on an independent variable, typically denoted by \(x\), and its function \(y(x)\). Second-order differential equations often require complex methods to solve, which may include techniques such as substitution or employing characteristic equations.
Understanding the order of these equations and their characteristics is crucial in identifying the right solving technique.

Complex Roots

The quadratic equation derived in our solution steps: \[ m^2 - 2m + 5 = 0 \] results in complex roots due to the negative discriminant.
A complex root occurs when the standard quadratic formula yields a number under the square root that is negative.
This scenario implies that we cannot have real solutions, so complex roots occur.
In this case, the complex nature of roots influences the form of the general solution to the differential equation.
The roots are expressed in terms of imaginary numbers, shown as: \( m = 1 \pm 2i \).
This impacts how the solutions express periodic or oscillatory behavior, aligning well with natural phenomena such as waveforms.

Substitution Method

In solving differential equations, changing the form of the equation can be crucial.
The substitution method involves introducing a substitution to simplify the equation, often transforming into a more manageable form.
For our given equation, we used:\[ y = (x-1)^m \]This substitution converts the original differential equation into one concerning powers of \(x-1\).

• Helps reduce complexity and tackles both the derivatives and terms simultaneously.
• Enables the solution of the equation where classical methods might struggle.

By substituting directly into the equation, it facilitates identifying root behavior, whether real or complex.

Auxiliary Equations

An auxiliary equation arises from simplifying differential equations, particularly linear homogeneous ones.
It's a characteristic equation derived during the solving process.
For our problem, after substitution, the equation \[ m(m-1) - m + 5 = 0 \]represents the auxiliary equation.

• This equation is essential in identifying the nature of roots - critical when roots dictate the form of the solution.
• Once solved, provides the necessary framework to express the differential equation's solutions.

In this context, determining complex roots from the auxiliary equation allows us to express the solution in terms of exponential functions or polynomials involving complex numbers, as seen in the general solution.