Problem 26 - Advanced Engineering Mathematics

Question

Solve the given initial-value problem. Use a graphing utility to graph the solution curve. $$ x^{2} y^{\prime \prime}-5 x y^{\prime}+8 y=0, y(2)=32, y^{\prime}(2)=0 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = 8x^2 - x^4 \).

Step 1: Identify the Type of Equation

The given differential equation is \( x^2 y'' - 5x y' + 8y = 0 \), which is a second-order linear homogeneous differential equation with variable coefficients.

Step 2: Assume a Solution Form

Assume a solution of the form \( y = x^m \). Substituting in the given differential equation gives \( x^2 m(m-1)x^{m-2} - 5x m x^{m-1} + 8x^m = 0 \). Simplify to get the characteristic equation.

Step 3: Derive the Characteristic Equation

Simplifying the assumed solution results in: \( m(m-1) - 5m + 8 = 0 \). This yields \( m^2 - 6m + 8 = 0 \).

Step 4: Solve the Characteristic Equation

Factor the quadratic equation: \( m^2 - 6m + 8 = 0 \). The roots are \( m = 2 \) and \( m = 4 \).

Step 5: Write the General Solution

The general solution for this differential equation is \( y(x) = C_1 x^2 + C_2 x^4 \), where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.

Step 6: Apply the Initial Condition for \( y(2)=32 \)

Substitute into the solution: \( 32 = C_1 (2)^2 + C_2 (2)^4 \) gives us the first equation, \( 32 = 4C_1 + 16C_2 \).

Step 7: Apply the Initial Condition for \( y'(2)=0 \)

First, find \( y'(x) = 2C_1 x + 4C_2 x^3 \). Then, substitute \( x=2 \) into \( y'(x) \) to get \( 0 = 4C_1 + 32C_2 \), which is the second equation.

Step 8: Solve the System of Equations

We have two equations: \( 4C_1 + 16C_2 = 32 \) and \( 4C_1 + 32C_2 = 0 \). Solving these simultaneously, we find \( C_1 = 8 \) and \( C_2 = -1 \).

Step 9: Write the Particular Solution

Thus, the solution to the initial value problem is \( y(x) = 8x^2 - x^4 \).

Step 10: Graph the Solution

Use a graphing utility to plot the function \( y(x) = 8x^2 - x^4 \) to visualize the solution curve.

Key Concepts

Initial Value Problem

An initial value problem in differential equations is essential for predicting specific behaviors of dynamic systems. Here, we seek a solution to a differential equation that satisfies given initial conditions. In our exercise, we have the differential equation \( x^2 y'' - 5x y' + 8y = 0 \) with initial conditions \( y(2) = 32 \) and \( y'(2) = 0 \). These conditions mean that at \( x = 2 \), the value of the function \( y \) is 32, and the derivative \( y' \) is 0.
This setup allows us to find a particular solution that not only fits the general structure of solutions but also passes through the specific points described by the initial conditions. By solving the corresponding system of equations formed from these conditions, we can determine the constants in the general solution to find the unique solution for this problem.

Variable Coefficients

Variable coefficients occur in differential equations when coefficients of the derivatives are functions of the independent variable instead of constants. In this problem, the given equation \( x^2 y'' - 5x y' + 8y = 0 \) clearly demonstrates variable coefficients, as the terms \( x^2 \) and \(-5x \) depend on \( x \).

• Such equations are typically more complex and challenging to solve compared to constant coefficient equations.
• They often require special methods or substitutions, such as assuming a solution like \( y = x^m \).

Handling these coefficients demands careful manipulation, as shown when deriving the characteristic equation. This process involves transforming the equation into a simpler form while maintaining equivalence, which aids in identifying a solvable structure.

Characteristic Equation

The characteristic equation is a crucial step in solving linear differential equations with variable coefficients. It arises from assuming a solution form and substituting it back into the original equation. In this case, we assume \( y = x^m \), leading us to derive the characteristic equation from \( m(m-1) - 5m + 8 = 0 \).
This polynomial equation transforms our differential equation into an algebraic one, greatly simplifying the process of finding solutions. The roots of the characteristic equation, here \( m = 2 \) and \( m = 4 \), represent the powers of \( x \) in our solution.

• Each root contributes a term to the general solution, forming \( y(x) = C_1 x^2 + C_2 x^4 \).

This method highlights the elegance of solving second-order linear differential equations by reducing them to finding the roots of polynomial equations.

General Solution

The general solution of a differential equation represents the set of all possible solutions before applying specific conditions. For our equation, it is given by \( y(x) = C_1 x^2 + C_2 x^4 \), with \( C_1 \) and \( C_2 \) as arbitrary constants.

• This solution encompasses all functions that satisfy the differential equation.
• It becomes a particular solution once specific initial conditions are applied.

Using initial conditions \( y(2) = 32 \) and \( y'(2) = 0 \), we find \( C_1 \) and \( C_2 \). Solving the system of equations gives \( C_1 = 8 \) and \( C_2 = -1 \).
This tailors our solution to the unique case described by the exercise, producing \( y(x) = 8x^2 - x^4 \), which satisfies both the differential equation and the initial conditions. The general solution is thus refined into a particular solution, offering a comprehensive understanding of the dynamic behavior specified.