Problem 28 - Advanced Engineering Mathematics

Question

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 \( y = 5x^2 - \frac{7}{3}x^2 \ln{x} \), determined using initial values.

Step 1: Identify the Differential Equation Type

The given differential equation is \( x^2 y'' - 3x y' + 4y = 0 \). This is a second-order linear homogeneous differential equation with variable coefficients.

Step 2: Assume a Solution Form

To solve this type of equation, assume a solution of the form \( y = x^m \). This allows us to find roots and solutions based on the method of Frobenius.

Step 3: Substitute into the Differential Equation

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

Step 4: Solve the Characteristic Equation

Solve \( m(m-1) - 3m + 4 = 0 \) to find the roots. This simplifies to \( m^2 - 4m + 4 = 0 \), which factors to \((m-2)^2 = 0 \). The roots are thus \( m = 2 \) (a repeated root).

Step 5: Write the General Solution for Repeated Roots

Since we have repeated roots \( m = 2 \), the form of the solution is given by \( y = c_1 x^2 + c_2 x^2 \ln{x} \).

Step 6: Apply Initial Conditions to Determine Constants

Use the initial conditions \( y(1) = 5 \) and \( y'(1) = 3 \) to find \( c_1 \) and \( c_2 \).- First, for \( y(1) = 5 \): \[ c_1 (1)^2 + c_2 (1)^2 \ln{1} = 5 \Rightarrow c_1 = 5 \]- Then find \( y' \):\[ y' = 2c_1 x + c_2(2x \ln{x} + x) \]- Apply \( y'(1) = 3 \):\[ 2(5)(1) + c_2(2 + 1) = 3 \Rightarrow 10 + 3c_2 = 3 \Rightarrow c_2 = -rac{7}{3} \]

Step 7: Write the Particular Solution

The particular solution is:\[ y = 5x^2 - \frac{7}{3}x^2 \ln{x} \].

Step 8: Graph the Solution

Use a graphing utility to plot the function \( y = 5x^2 - \frac{7}{3}x^2 \ln{x} \). The important features are the values at given points and the overall curve shape.

Key Concepts

Linear Homogeneous Differential Equation

A linear homogeneous differential equation is one where every term is a derivative of the unknown function or the function itself, all added together to yield zero. This means there are no constant terms or external inputs in the equation.
For example, in the equation \( x^2 y'' - 3x y' + 4y = 0 \), each term involves either the function \( y \) or its derivatives. These terms are arranged to equal zero.
Such equations are significant in studying systems that naturally return to equilibrium without external input, such as mechanical or electrical systems' natural resonances.

Variable Coefficients

Variable coefficients are coefficients that are functions of the independent variable, rather than constants.
In the equation \( x^2 y'' - 3x y' + 4y = 0 \), the coefficients of \( y'' \) and \( y' \) depend on \( x \). In contrast, constant coefficients would be numbers like 2, -3, or 4 – not linked to \( x \).
Equations with variable coefficients are often more complex, as their behavior changes with the value of the independent variable, making them more realistic in modeling scenarios where conditions are not uniform across situations.

Method of Frobenius

The Method of Frobenius is a technique used to solve linear differential equations with variable coefficients, particularly around a singular point.
It involves assuming solutions in the form of a power series, which helps to solve equations where simple methods might fail.
Here, we assumed a solution of the form \( y = x^m \), simplifying the equation to find the characteristic roots. This step can transform seemingly complicated equations into algebraic ones, making them easier to solve.

Characteristic Equation

The characteristic equation results from substituting an assumed solution into the differential equation and simplifying.
It is an algebraic equation whose solutions provide us with the exponents or factors necessary in the solution of the differential equation.
For example, solving \( m(m-1) - 3m + 4 = 0 \) gives the root \( m = 2 \), leading us to a specific form of the solution. When roots are repeated, the solution involves additional terms like \( c_2 x^2 \ln{x} \), adding complexity to the standard power solutions.