Problem 25 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Differential Equation Type

The given differential equation is \(x^2 y'' + 3x y' = 0\). This is a second-order, linear, homogeneous differential equation. Notice that it does not contain the variable \(y\) explicitly, so it forms a Cauchy-Euler type equation.

Step 2: Solve the Homogeneous Differential Equation

For a Cauchy-Euler equation of the form \(a x^2 y'' + b x y' + c y = 0\), we solve using the characteristic equation \(ar(r-1) + br + c = 0\). Substituting \(a = 1, b = 3, c = 0\), we get the characteristic equation \(r(r-1) + 3r = 0\). Simplifying, we obtain \(r^2 + 2r = 0\), which factors to \(r(r + 2) = 0\). The solutions are \(r_1 = 0\) and \(r_2 = -2\).

Step 3: Determine the General Solution

Using the solutions from the characteristic equation, the general solution of the differential equation is given by \(y(x) = C_1 x^r_1 + C_2 x^r_2\), where \(r_1 = 0\) and \(r_2 = -2\). Thus, \(y(x) = C_1 x^0 + C_2 x^{-2} = C_1 + \frac{C_2}{x^2}\).

Step 4: Apply Initial Conditions

Apply the initial conditions \(y(1) = 0\) and \(y'(1) = 4\) to the general solution. Start by applying \(y(1) = 0\):\[C_1 + \frac{C_2}{1^2} = 0\]\which simplifies to \(C_1 + C_2 = 0\). This yields the relation \(C_1 = -C_2\).

Step 5: Find the Derivative and Solve for Constants

Differentiate \(y(x) = C_1 + \frac{C_2}{x^2}\) to get:\[y'(x) = -2C_2 x^{-3}\]. Apply the initial condition \(y'(1) = 4\): \[-2C_2 \cdot 1^{-3} = 4\]. Solving gives \(C_2 = -2\). Since \(C_1 = -C_2\), we have \(C_1 = 2\).

Step 6: Write the Particular Solution

Substitute \(C_1 = 2\) and \(C_2 = -2\) into the general solution \(y(x) = C_1 + \frac{C_2}{x^2}\). The particular solution is:\[y(x) = 2 - \frac{2}{x^2}\].

Key Concepts

Second-Order Linear Differential Equation

A second-order linear differential equation contains the second derivative, which essentially means it involves terms up to the second derivative of a function. These types of equations are widely used in various scientific fields because they describe systems with accelerating components, such as mechanical or electrical systems.
A second-order differential equation will generally look like this:

• Has the form: \(a(x) y'' + b(x) y' + c(x) y = g(x)\)
• Key component: the presence of the second derivative \(y''\)
• Can be linear or nonlinear depending on how the function and its derivatives appear

In the given problem, we see the equation \(x^2 y'' + 3x y' = 0\), which is a clear example of a second-order linear equation, where \(c(x) = 0\) and \(g(x) = 0\). This particular form indicates it is a special kind of differential equation known as the Cauchy-Euler equation.

Characteristic Equation

The characteristic equation is a critical tool in solving homogeneous linear differential equations, especially those with constant coefficients. It is derived from assuming a solution of a specific functional form and is fundamental in determining the behavior of the solutions.
For the Cauchy-Euler type equation, the characteristic equation is crucial. It is given by:

• Formed as \(ar(r-1) + br + c = 0\)
• Derived from substituting \(y = x^r\) into the differential equation
• Allows us to find the roots \(r_1\), \(r_2\), ..., which give the nature of solutions

In our problem, substituting the coefficients \(a = 1\), \(b = 3\), and \(c = 0\) results in the characteristic equation \(r(r-1) + 3r = 0\). Simplifying this leads us to the factorized form \(r(r + 2) = 0\), providing the roots \(r_1 = 0\) and \(r_2 = -2\). These roots dictate the form of the general solution for the differential equation.

Initial-Value Problem

An initial-value problem provides specific conditions at a given point, which allows us to find a particular solution to a differential equation. This is essential because, often, differential equations have families of solutions. Initial conditions help isolate the exact solution out of these possibilities.
Here's what you need to know about initial-value problems:

• Given as values of the function and/or its derivatives at a particular point, such as \(y(a) = y_0\) and \(y'(a) = y'_0\)
• They "fix" the constants in the general solution by using known values
• Essential for modeling real-world scenarios, ensuring the solution matches specifics

In our task, the initial conditions are \(y(1) = 0\) and \(y'(1) = 4\). Using these, we derive expressions for the constants in the general solution form \(y(x) = C_1 + \frac{C_2}{x^2}\), leading to the specific solution using the constants \(C_1\) and \(C_2\) found through the conditions.

Homogeneous Differential Equation

A homogeneous differential equation is one where the terms all involve the function or its derivatives. This often means that if the differential equation is set to zero, it includes no additional forcing term or disturbance influences.
Here are some characteristics of homogeneous differential equations:

• The right-hand side of the equation equals zero
• Formally written as \(a(x) y'' + b(x) y' + c(x) y = 0\)
• The solutions tend to follow particular patterns or behaviors

In this exercise, the given equation is homogeneous because it includes only terms of the function \(y\) and its derivatives and equates to zero, i.e., \(x^2 y'' + 3x y' = 0\). Solving such equations typically involves analyzing the characteristic equation, giving us a way to form general solutions which are then adjusted to fit specific initial conditions or boundary requirements.