Problem 4 - Advanced Engineering Mathematics

Question

In Problems, use (1) to find the general solution of the given differential equation on \((0, \infty)\). $$ 16 x^{2} y^{\prime \prime}+16 x y^{\prime}+\left(16 x^{2}-1\right) y=0 $$

Step-by-Step Solution

Answer

The general solution is \(y(x) = C_1x^{-\frac{1}{4}} + C_2x^{-1}\).

Step 1: Recognize the Type of Differential Equation

The given differential equation is \(16x^2y'' + 16xy' + (16x^2 - 1)y = 0\). This is a second-order linear differential equation in the standard form of a Cauchy-Euler equation.

Step 2: Apply the Substitution for Cauchy-Euler Equations

For Cauchy-Euler equations, we use the substitution \(y = x^m\). This results in derivatives \(y' = m x^{m-1}\) and \(y'' = m(m-1)x^{m-2}\). Substitute these into the differential equation.

Step 3: Substitute y, y', and y'' into the Equation

Substitute \(y = x^m\), \(y' = m x^{m-1}\), and \(y'' = m(m-1)x^{m-2}\) into the equation:\[16x^2 (m(m-1)x^{m-2}) + 16x(mx^{m-1}) + (16x^2 - 1)x^m = 0.\] Simplify each term.

Step 4: Simplify the Equation

After substitution, the differential equation simplifies to:\[16m(m-1)x^m + 16mx^m + (16x^2 - 1)x^m = 0.\] Combine like terms to simplify further.

Step 5: Factor and Solve for m

Combine and factor the terms: \[(16m(m-1) + 16m + (16x^2 - 1))x^m = 0.\]Since \(x^m eq 0\) for \(x\in (0, \infty)\), we focus on solving the quadratic equation for \(m\): \[16m^2 + 16m - 1 = 0.\] Use the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 6: Calculate the Roots Using the Quadratic Formula

Identify coefficients: \(a = 16\), \(b = 16\), \(c = -1\). Substitute into the quadratic formula:\[m = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 16 \cdot (-1)}}{2 \cdot 16}.\]Calculate the discriminant and find the roots.

Step 7: Determine the General Solution

After solving, the roots are \(m_1 = \frac{-1}{4}\) and \(m_2 = -1\). Therefore, the general solution of the differential equation is:\[y(x) = C_1x^{m_1} + C_2x^{m_2} = C_1x^{-\frac{1}{4}} + C_2x^{-1},\]where \(C_1\) and \(C_2\) are arbitrary constants.

Key Concepts

Second-Order Linear Differential Equations

Second-order linear differential equations are a fundamental part of differential equations, commonly encountered in various scientific fields. These equations have the standard form \(a(x) y'' + b(x) y' + c(x) y = f(x)\), where \(y''\) is the second derivative of \(y\), \(y'\) is the first derivative, and \(y\) is the function of \(x\) itself.

The equation may also have specific coefficients \(a(x)\), \(b(x)\), and \(c(x)\) that depend on \(x\), and \(f(x)\) can be any function representing sources or sinks. The solution to a second-order linear differential equation typically aims to find a function \(y(x)\) that satisfies the equation.

• In homogeneous cases, like challenging the Cauchy-Euler equation provided, \(f(x) = 0\).
• This means solving for \(y(x)\) without a source term simplifies calculations.

Solving these equations focuses understanding and applying consistent methods, enabling accurate prediction and interpretation of systems described by these equations.

Quadratic Formula

The quadratic formula is a classic mathematical tool used to find the roots of quadratic equations, given by the general form \(ax^2 + bx + c = 0\). It's applicable when the polynomial equation is set to zero, allowing the determination of variable values that satisfy the equation. The formula is:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

This equation provides both roots, accounting for \( \pm \) which depicts different scenarios dependent on the discriminant \(b^2 - 4ac\).

• If the discriminant is positive, two real and distinct roots exist.
• A zero discriminant reveals a double root, and a negative one indicates complex or imaginary roots.

In our differential equation problem, using the quadratic formula effectively assists in obtaining the values for \(m\) that lead to the general solution. Identifying these roots allows us to form the principal part of solving the second-order equation.

General Solution of Differential Equations

The concept of a general solution is central when dealing with differential equations. A general solution comprises all possible solutions that satisfy a given differential equation. It typically includes arbitrary constants representing the family of solutions.

For a second-order linear differential equation, such as a Cauchy-Euler equation, the general solution often follows the form:

\[y(x) = C_1 x^{m_1} + C_2 x^{m_2}\]

where \(C_1\) and \(C_2\) are constants, and \(m_1\) and \(m_2\) are the roots obtained via the quadratic formula.

• This structure holds for distinct real roots.
• In cases of repeated roots, or complex ones, the solution form slightly modifies to accommodate these variations.

Overall, the general solution provides a comprehensive representation of behavior described by the differential equation over its applicable domain.