Problem 25 - A First Course in Differential Equations with Modeling Applications

Question

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

Step-by-Step Solution

Answer

The solution is \( y(x) = 2 - \frac{2}{x^2} \).

Step 1: Understand the Problem

We have a second-order linear homogeneous differential equation: \[ x^2 y'' + 3xy' = 0 \]with initial conditions \( y(1) = 0 \) and \( y'(1) = 4 \). Our goal is to find the function \( y(x) \) that satisfies both the differential equation and the initial conditions.

Step 2: Solve the Differential Equation

To solve the differential equation, notice that it is a Cauchy-Euler type equation. We make the substitution \( y(x) = x^m \) and its derivatives \( y' = mx^{m-1} \) and \( y'' = m(m-1)x^{m-2} \). Substituting these into the equation gives:\[ x^2 (m(m-1)x^{m-2}) + 3x(mx^{m-1}) = m(m-1)x^m + 3mx^m = 0 \]Simplifying, we get the characteristic equation:\[ m(m-1) + 3m = 0 \]which simplifies to \( m^2 + 2m = 0 \). Factor to find \( m(m+2) = 0 \), giving roots \( m = 0 \) and \( m = -2 \). So, the general solution is:\[ y(x) = C_1x^0 + C_2x^{-2} = C_1 + \frac{C_2}{x^2} \].

Step 3: Apply Initial Conditions

We apply the initial conditions to find the constants. First, using \( y(1) = 0 \):\[ C_1 + \frac{C_2}{1^2} = 0 \]gives \( C_1 + C_2 = 0 \) or \( C_1 = -C_2 \). Next, find \( y'(x) = -2C_2x^{-3} \). Use \( y'(1) = 4 \):\[ -2C_2(1)^{-3} = 4 \]giving \( -2C_2 = 4 \) or \( C_2 = -2 \). Substitute \( C_2 = -2 \) back into \( C_1 = -C_2 \) to find \( C_1 = 2 \). Thus, the specific solution is:\[ y(x) = 2 - \frac{2}{x^2} \].

Step 4: Graph the Solution

Using a graphing utility, plot the function \( y(x) = 2 - \frac{2}{x^2} \). Observe its behavior as \( x \) moves from close to zero to larger values, keeping in mind the initial directional behavior around \( x = 1 \) and how it changes as \( y(x) \) asymptotically approaches 2 as \( x \) increases.

Key Concepts

Second-Order Differential Equations

A second-order differential equation involves derivatives of a function up to the second degree. It's a differential equation in which the highest derivative is the second derivative. In our given problem, the equation \[x^2 y'' + 3xy' = 0\] is second-order because of \(y''\), the second derivative of \(y\) with respect to \(x\). These types of equations are fundamental in understanding how systems evolve over time, as they can model physical situations such as motion, electric circuits, or fluid dynamics.The complexity in solving these equations usually arises from the need to simplify and manipulate expressions with multiple variables and derivatives. In particular, finding a general form for these solutions often involves characteristic equations or specific techniques depending on the equation's form.

Initial-Value Problem

An initial-value problem provides additional conditions to a differential equation, often helping us determine a unique solution. In essence, they specify the state of the system at a starting point. For our problem, the initial conditions are:

• \( y(1) = 0 \)
• \( y'(1) = 4 \)

These conditions tell us the value of the function \(y\) and its first derivative \(y'\) at a specific point, \(x = 1\).The purpose of initial conditions is to constrain the solution that fits the given differential equation so that rather than an infinite number of possible solutions, we find one specific solution that suits the conditions provided. Once we solve the homogeneous equation, we can use these initial values to find constants that give us the particular solution.

Characteristic Equation

Characterizing second-order differential equations often requires simplifying them into something more manageable. This is where the concept of the characteristic equation comes in, particularly with Cauchy-Euler type equations.This equation is derived by substituting a solution form \(y(x) = x^m\) into the differential equation. Doing this substitution in our problem led to:\[ m(m-1)x^m + 3mx^m = 0 \] which further simplifies to the characteristic equation:\[ m^2 + 2m = 0 \] By factoring the characteristic equation, we find the roots \(m = 0\) and \(m = -2\). These roots are crucial because they help us determine the structure of the general solution for the differential equation.

Homogeneous Differential Equation

A homogeneous differential equation is one where every term is dependent on the unknown function or its derivatives, without any standalone constant or function. In other words, if you replace the dependent variable and its derivatives with zero, the equation holds true. The problem statement:\[ x^2 y'' + 3xy' = 0 \]is homogeneous, meaning the structure allows us to use standard techniques such as finding the characteristic equation and general solution form easily.The simplicity brought by homogeneous equations is in their linearity and dependence on the same function throughout. Solutions include combinations of terms dependent solely upon the roots of their characteristic equation. In our solved equation, this leads to a general solution \(y(x) = C_1 + \frac{C_2}{x^2}\) once the roots are inserted into its general form.