Problem 28 - Advanced Engineering Mathematics

Question

In Problems 25-30, solve the given initial-value problem. Use a graphing utility to graph the solution curve. $$ x^{2} y^{\prime \prime}-3 x y^{\prime}+4 y=0, y(1)=5, y^{\prime}(1)=3 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = 5x^2 - 7x^2 \ln x \).

Step 1: Identify the Differential Equation

The given problem is a second-order linear homogeneous differential equation: \( x^{2} y'' - 3xy' + 4y = 0 \).

Step 2: Identify the Type of Differential Equation

Recognize that this is a Cauchy-Euler equation, which has the general form \( x^2 y'' + axy' + by = 0 \).

Step 3: Substitute the Solution Form

Assume a solution of the form \( y = x^m \), where \( m \) is a constant to be determined.

Step 4: Find Derivatives of Assumed Solution

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

Step 5: Substitute into Differential Equation

Substitute \( y = x^m \), \( y' = mx^{m-1} \), and \( y'' = m(m-1)x^{m-2} \) into the differential equation: \[ x^2(m(m-1)x^{m-2}) - 3x(mx^{m-1}) + 4x^m = 0 \] Simplify to \[ m(m-1)x^m - 3mx^m + 4x^m = 0 \].

Step 6: Simplify and Factor

Factor out \( x^m \):\[ x^m (m^2 - 4m + 4) = 0 \].

Step 7: Solve Characteristic Equation

Solve the quadratic equation \( m^2 - 4m + 4 = 0 \). This can be rewritten as \[ (m-2)^2 = 0 \], giving the double root \( m = 2 \).

Step 8: Write General Solution of ODE

Since we have a double root \( m = 2 \), the general solution is \[ y(x) = C_1 x^2 + C_2 x^2 \ln x \].

Step 9: Apply Initial Conditions to Solve for Constants

Use initial conditions to determine the constants. First, for \( y(1) = 5 \), substitute into \[ 5 = C_1 (1)^2 + C_2 (1)^2 \ln(1) \Rightarrow 5 = C_1 \].Then use \( y'(x) = 2C_1 x + C_2 (2x \ln x + x) \) and \( y'(1) = 3 \):\[ 3 = 2(5) + C_2 \Rightarrow C_2 = -7 \].

Step 10: Write Specific Solution

Substitute the values of \( C_1 \) and \( C_2 \) back into the general solution: \[ y(x) = 5x^2 - 7x^2 \ln x \].

Step 11: Graph the Solution Curve

Use a graphing utility to plot the solution curve for \( y(x) = 5x^2 - 7x^2 \ln x \). The graph will show the behavior of the function over a range of \( x \) values.

Key Concepts

second-order differential equations

Second-order differential equations are equations that involve the second derivative of a function. In these equations, the unknown function is typically a function of a single independent variable. The primary focus is on how the function's rate of change (derivatives) relates to the function itself.

These equations can be more complex than first-order differential equations because they involve higher rates of change. They often describe phenomena like oscillations or dynamic systems such as mechanical vibrations or electrical circuits. A second-order differential equation like the Cauchy-Euler differential equation in the problem given, for example, usually takes the form:

• In general: \( a(x)y'' + b(x)y' + c(x)y = 0 \).
• For Cauchy-Euler: \( ax^2y'' + bxy' + cy = 0 \).

The solution method often involves assuming a specific form for the solution, as was demonstrated in the problem where a power of \( x \) was assumed. This allows us to solve for characteristic roots—key steps to arrive at the equation's general solution.

initial-value problem

An initial-value problem specifies values of the solution and possibly its derivatives at a particular point. This provides a unique solution to differential equations, especially useful in many practical situations where initial conditions are known.

In the context of the exercise, an initial-value problem is specified by providing the function value and its derivative at a specific point, known as the initial conditions.

• The initial condition for the function: \( y(1) = 5 \)
• The initial condition for its derivative: \( y'(1) = 3 \)

The goal is to find a particular solution that satisfies both the differential equation and these conditions. By substituting the given initial conditions into the general solution, you can calculate the constants in the general solution formula. This narrows down the infinite solutions to a single specific solution that meets the initial conditions, which in this case led to \( y(x) = 5x^2 - 7x^2 \ln x \).

homogeneous differential equations

Homogeneous differential equations are a specific type of differential equations where all terms are dependent on the function (and/or its derivatives). For a homogeneous differential equation like the one given in the exercise, every term can be expressed as a multiple of the dependent variable or its derivatives.

The general form is:

• \( a(x)y'' + b(x)y' + c(x)y = 0 \)

For homogeneous equations, we assume solutions which allow transformations that simplify the differential equation. Methods often involve guessing a particular form for the solution and finding necessary constants to satisfy the equation.

In our exercise, the solution was assumed to be \( x^m \) and derivatives were plugged back into the original differential equation. This lead to a characteristic equation that was solved to find roots. For the given Cauchy-Euler form, the solution could be expressed in terms of powers and logarithms, which revealed solutions for constant coefficients due to its unique property of having terms multiplied by powers of \( x \). This approach is crucial in determining the nature of solutions that depend solely on the structural form of the equation.