Problem 11 - Advanced Engineering Mathematics

Question

Solve the given differential equation. $$ x^{2} y^{\prime \prime}+5 x y^{\prime}+4 y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y = C_1 x^{-2} + C_2 x^{-2} \ln{x} \).

Step 1: Identify the Type of Differential Equation

This is a second-order linear homogeneous differential equation with variable coefficients: \( x^{2} y^{\prime \prime}+5 x y^{\prime}+4 y=0 \).

Step 2: Assume a Solution Form

Assume a solution of the form \( y = x^{m} \) where \( m \) is an unknown exponent to be determined. Substitute \( y = x^{m} \) into the equation.

Step 3: Derive the First and Second Derivatives

Calculate the first derivative, \( y' = mx^{m-1} \), and the second derivative, \( y'' = m(m-1)x^{m-2} \).

Step 4: Substitute Derivatives into the Original Equation

Substitute \( y = x^m \), \( y' = mx^{m-1} \), and \( y'' = m(m-1)x^{m-2} \) into \( x^{2} y^{\prime \prime}+5 x y^{\prime}+4 y=0 \). This gives: \[ x^2(m(m-1)x^{m-2}) + 5x(mx^{m-1}) + 4x^{m} = 0. \]

Step 5: Simplify the Equation

Simplify the terms: \[ m(m-1)x^m + 5mx^m + 4x^m = 0. \] Combine the terms with \( x^m \): \[ (m^2 + 4m) x^m = 0. \]

Step 6: Solve the Indicial Equation

For a nontrivial solution, set the coefficient of \( x^m \) to zero: \( m^2 + 4m + 4 = 0 \). Factor to find \( m \): \( (m+2)^2 = 0 \). This gives \( m = -2 \) as a double root.

Step 7: Write the General Solution

The general solution for each root \( m \) is \( y = C_1 x^{-2} + C_2 x^{-2} \ln{x} \), where \( C_1 \) and \( C_2 \) are constants. The \( \ln{x} \) term arises due to the repeated root.

Key Concepts

Linear Homogeneous Differential Equations

Linear homogeneous differential equations are a special class of differential equations. These equations can be spotted by their structure, which involves an equation set to zero. Finding the general solution involves considering terms of similar order. For example, the equation:
\[ x^2 y^{\prime \prime} + 5x y^{\prime} + 4y = 0 \]
is a second-order linear homogeneous differential equation. Here, "homogeneous" means that every term involves the function \( y \) or its derivatives. There are no constant or independent terms, just those containing \( y \), \( y^{\prime} \), or \( y^{\prime \prime} \).

• "Linear" indicates that the equation and its derivatives are linear combinations, rather than containing any terms like \( y^2 \) or \( (y^{\prime})^2 \).
• The "second-order" refers to the highest derivative, which in this case is \( y^{\prime \prime} \).

This structure makes such equations easier to approach systematically by leveraging their homogeneous, linear nature.

Variable Coefficients

Variable coefficients in differential equations present a unique challenge and distinction. Unlike constant coefficients, variable coefficients are defined as functions of the independent variable, in this case, \( x \). In the given equation:
\[ x^2 y^{\prime \prime} + 5x y^{\prime} + 4y = 0 \]
each coefficient next to the derivatives (\(x^2\) with \(y^{\prime\prime}\) and \(5x\) with \(y^{\prime}\)) varies with \( x \). This adds a layer of complexity when solving the equation.

• Variable coefficients make it difficult to directly apply constant-coefficient solution methods, such as characteristic equations.
• Special techniques, like the method of Frobenius or assuming a power series solution, may be adapted for these cases.

By "creating" the coefficients from the independent variable, they give each term a different weight, influencing the general behavior of the solution as \( x \) changes.

Indicial Equation

The indicial equation arises when seeking solutions to differential equations with variable coefficients. After assuming a tentative solution form, such as \( y = x^m \), and substituting back into the differential equation, one often derives an algebraic equation in terms of the parameter \( m \). This algebraic equation is known as the indicial equation.
For the given differential equation:\[ x^2(m(m-1)x^{m-2}) + 5x(mx^{m-1}) + 4x^{m} = 0 \]

• Simplification leads to an equation solely in terms of \( m \): \((m^2 + 4m + 4) x^m = 0\).
• Setting the coefficent of \(x^m\) to zero gives us the indicial equation: \(m^2 + 4m + 4 = 0\).
• This equation is solved to find \(m = -2\) as a double root, which guides the form of the general solution.

The roots of the indicial equation indicate the exponents in the solution, determining crucial aspects of the behavior of all possible solutions.

General Solution

The general solution to a differential equation forms the complete set of potential solutions that satisfy the equation. For linear homogeneous equations, the solution is usually expressed as a combination of specific functions derived from the roots of the indicial equation.
In our example, with a double root \( m = -2 \), the solutions take a specific form due to the repeated solutions in the indicial equation:

• The typical solution is \( y = C_1 x^{-2} \), where \( C_1 \) is a constant.
• Due to the double root, a second term including a logarithm, \( y = C_2 x^{-2} \ln{x} \), is included where \( C_2 \) is another constant.

Thus, the general solution becomes:\[ y = C_1 x^{-2} + C_2 x^{-2} \ln{x} \] This expression accounts for all possible behaviors in the system described by the differential equation, with the constants \( C_1 \) and \( C_2 \) determined by initial or boundary conditions unique to specific problem contexts.