Problem 5 - Advanced Engineering Mathematics

Question

Solve the given differential equation. $$ x^{2} y^{\prime \prime}+x y^{\prime}+4 y=0 $$

Step-by-Step Solution

Answer

The general solution is \(y(x) = c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)\).

Step 1: Identify the type of differential equation

The given equation \(x^2 y'' + x y' + 4y = 0\) is a second-order linear differential equation with variable coefficients. It is a Cauchy-Euler equation.

Step 2: Apply Cauchy-Euler equation substitution

For a Cauchy-Euler equation of the form \(a x^2 y'' + b x y' + c y = 0\), we use the substitution \(y = x^m\) to find solutions. Compute \(y' = m x^{m-1}\) and \(y'' = m(m-1) x^{m-2}\).

Step 3: Substitute and simplify

Substitute \(y = x^m\), \(y'\), and \(y''\) into the equation: \(x^2(m(m-1)x^{m-2}) + x(mx^{m-1}) + 4x^m = 0\), simplifying to \(m(m-1) x^m + m x^m + 4 x^m = 0\).

Step 4: Solve the characteristic equation

Combine terms to form the algebraic equation \((m(m-1) + m + 4)x^m = 0\), leading to the characteristic equation \(m^2 + 4 = 0\). Solve it to find \(m^2 = -4\), hence \(m = \pm 2i\).

Step 5: Write the general solution

Given the roots are complex \(m = \pm 2i\), the general solution for the differential equation is \(y(x) = c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)\), where \(c_1\) and \(c_2\) are arbitrary constants.

Key Concepts

Second-order Linear Differential Equation

A second-order linear differential equation is an equation involving the second derivative of an unknown function. This type of equation can be written in the general form: \[ a(x)y'' + b(x)y' + c(x)y = f(x) \] where \(y''\) is the second derivative of \(y\), and \(a(x)\), \(b(x)\), and \(c(x)\) are functions of the variable \(x\). If \(f(x) = 0\), the equation is homogeneous, like the one we have here. To solve these equations, especially when they are of the Cauchy-Euler type, specific methods and substitutions are employed. It is not uncommon to encounter equations where the coefficients \(a(x)\), \(b(x)\), and \(c(x)\) are powers of \(x\), which leads us to the concept of variable coefficients.

Variable Coefficients

In differential equations, coefficients are numbers or functions that multiply the derivatives of the unknown function. When these coefficients are powers of the independent variable \(x\), as in a Cauchy-Euler equation, they are called variable coefficients. In our specific example, we have an equation with coefficients dependent on \(x\):

• \(a(x) = x^2\)
• \(b(x) = x\)
• \(c(x) = 4\)

These variable coefficients change the behavior and solutions to the differential equation. To solve Cauchy-Euler equations with variable coefficients, we typically use the substitution \(y = x^m\), which turns the problem into an algebraic equation involving \(m\), the characteristic equation.

Characteristic Equation

The characteristic equation is an algebraic equation obtained from a differential equation through a specific substitution. For Cauchy-Euler equations, substituting \(y = x^m\) helps convert the differential equation into such an algebraic form. After substitution, the equation becomes:\[ m(m-1)x^m + m x^m + 4 x^m = 0 \] This equation can be simplified to:\[ m^2 + 4 = 0 \]The quadratic characteristic equation determines the nature of the solutions based on the roots obtained. These roots, \(m\), dictate the general form of the solution to the differential equation.

Complex Roots

Complex roots occur in the characteristic equation when solving second-order linear differential equations, particularly when the roots of the equation are not real numbers. In our example, solving the characteristic equation \(m^2 + 4 = 0\) leads to the roots:

• \( m = 2i \)
• \( m = -2i \)

These are complex conjugates. When the roots are complex, the general solution of the differential equation has a specific form:\[ y(x) = c_1 \, \cos(2\ln x) + c_2 \, \sin(2\ln x) \] where \(c_1\) and \(c_2\) are arbitrary constants. This solution captures the oscillatory behavior introduced by the complex nature of the roots.