Problem 27 - Advanced Engineering Mathematics

Question

In Problems 13-28, use the procedures developed in this chapter to find the general solution of each differential equation. $$ x^{2} y^{\prime \prime}-4 x y^{\prime}+6 y=2 x^{4}+x^{2} $$

Step-by-Step Solution

Answer

The general solution is \( y = C_1x^2 + C_2x^3 + \frac{1}{8}x^4 + \frac{1}{6}x^2 \).

Step 1: Identify the Type of Equation

The differential equation given is \( x^2y'' - 4xy' + 6y = 2x^4 + x^2 \). This is a second-order linear differential equation. Since the coefficients of \( y'' \), \( y' \), and \( y \) depend on \( x \), it is also a Cauchy-Euler (or Euler-Cauchy) equation.

Step 2: Solve the Homogeneous Equation

The homogeneous part of the differential equation is \( x^2y'' - 4xy' + 6y = 0 \). For a Cauchy-Euler equation like this, we assume a solution of the form \( y = x^m \). Substituting \( y = x^m \) and its derivatives into the homogeneous equation gives the characteristic equation \( m(m-1) - 4m + 6 = 0 \).

Step 3: Solve the Characteristic Equation

The characteristic equation derived is \( m(m-1) - 4m + 6 = 0 \). Simplifying gives \( m^2 - 5m + 6 = 0 \). Solving this quadratic equation results in roots \( m = 2 \) and \( m = 3 \).

Step 4: Write the General Solution of the Homogeneous Equation

Based on the solutions to the characteristic equation, the general solution of the homogeneous equation is \( y_h = C_1x^2 + C_2x^3 \), where \( C_1 \) and \( C_2 \) are arbitrary constants.

Step 5: Find a Particular Solution

The non-homogeneous part of the equation is \( 2x^4 + x^2 \). We use the method of undetermined coefficients, assuming a particular solution of the form \( y_p = Ax^4 + Bx^2 \).

Step 6: Substitute and Solve for Coefficients

Substitute \( y_p = Ax^4 + Bx^2 \) into the original differential equation to find \( A \) and \( B \). This involves taking the derivatives and equating coefficients after substitution. This yields \( A = \frac{1}{8} \) and \( B = \frac{1}{6} \).

Step 7: Write the General Solution of the Differential Equation

The general solution is the sum of the homogenous solution and the particular solution: \( y = y_h + y_p = C_1x^2 + C_2x^3 + \frac{1}{8}x^4 + \frac{1}{6}x^2 \). This can be simplified or rearranged as needed.

Key Concepts

General Solution

When solving differential equations, finding the general solution is a crucial step. The general solution consists of two parts: the homogeneous solution and a particular solution. For the given Cauchy-Euler equation, the homogeneous solution accounts for the inherent structure of the equation when there is no external forcing function. The particular solution addresses any nonhomogeneous components, like external inputs or particular functions. In essence, the general solution provides a comprehensive description of all possible behaviors of the differential equation. To find the general solution for a Cauchy-Euler equation, we typically start with the homogeneous equation and find its solution. Then, we identify a particular solution that satisfies the non-homogeneous part. Finally, by summing up both solutions, we obtain the complete general solution. For our specific problem, the homogeneous solution was found first, yielding terms with arbitrary constants based on the roots of the characteristic equation. Afterward, we derived a particular solution using a specific method to capture the non-homogeneous part's impact. This combination results in what is known as the general solution.

Homogeneous Equation

A homogeneous equation in differential equations is one where all terms are dependent on the function and its derivatives, without any added external terms. In the context of our problem, the homogeneous equation is expressed as \(x^2y'' - 4xy' + 6y = 0\). This simplification ignores the right-hand side of the original differential equation, focusing on the structure inherent to the operators and function itself.Cauchy-Euler equations are particularly well-suited for solving using their characteristic forms. We assume a solution of the form \(y = x^m\), which exploits the power-law nature of the coefficients in the differential equation.By substituting the assumed solution and its derivatives into the homogeneous equation, we derive a characteristic equation. For this exercise, the characteristic equation \(m^2 - 5m + 6 = 0\), when solved, gives roots that guide the formation of the homogeneous solution, \(y_h = C_1x^2 + C_2x^3\), where \(C_1\) and \(C_2\) are constants determined by boundary or initial conditions when provided.

Method of Undetermined Coefficients

The method of undetermined coefficients is a clever technique used to find particular solutions to linear non-homogeneous differential equations. This method assumes a particular solution with a form similar to the non-homogeneous term of the equation itself.For our equation, the non-homogeneous part is \(2x^4 + x^2\). We assumed a particular solution \(y_p = Ax^4 + Bx^2\), where \(A\) and \(B\) are constants to be determined. This assumption is based on the observation that polynomial terms in the solution will match the form of non-homogeneous components.By substituting the assumed particular solution into the original differential equation and matching coefficients, we can solve for these constants. In this case, after substitution and simplification, we derive that \(A = \frac{1}{8}\) and \(B = \frac{1}{6}\). This allows us to write the particular solution and add it to the homogeneous solution, thus forming the general solution of the differential equation.