Problem 3 - Advanced Engineering Mathematics

Question

In Problems \(1-4\), the given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem. $$ \begin{aligned} &y=c_{1} x+c_{2} x \ln x,(0, \infty) ; x^{2} y^{\prime \prime}-x y^{\prime}+y=0, y(1)=3 \\ &y^{\prime}(1)=-1 \end{aligned} $$

Step-by-Step Solution

Answer

The specific solution is \( y = 3x - 4x \ln x \).

Step 1: Verify the general solution

The given family of functions is \( y = c_1 x + c_2 x \ln x \). We need to verify that it satisfies the differential equation \( x^2 y'' - x y' + y = 0 \) on the interval \( (0, \infty) \).

Step 2: Compute the derivatives

First, compute the first derivative: \( y' = c_1 + c_2 (\ln x + 1) \). Next, compute the second derivative: \( y'' = \frac{c_2}{x} \).

Step 3: Substitute into the differential equation

Substitute \( y, y', y'' \) into the differential equation:\[ x^2 \left( \frac{c_2}{x} \right) - x (c_1 + c_2 (\ln x + 1)) + (c_1 x + c_2 x \ln x) = 0. \] Simplifying, we see the equation holds, verifying that \( y = c_1 x + c_2 x \ln x \) is indeed a solution.

Step 4: Apply initial conditions

Apply \( y(1) = 3 \) to get: \( c_1 (1) + c_2 (1) \ln 1 = 3 \). Simplifying: \( c_1 = 3 \) because \( \ln 1 = 0 \).

Step 5: Solve for the second initial condition

Now, apply \( y'(1) = -1 \) to the first derivative: \( c_1 + c_2 (0) + c_2 = -1 \) or \( c_1 + c_2 = -1 \). Substitute \( c_1 = 3 \) to find \( c_2 = -4 \).

Step 6: Final expression for the specific solution

Substitute \( c_1 = 3 \) and \( c_2 = -4 \) back into the general solution to get: \( y = 3x - 4x \ln x \). This is the member of the family that satisfies the initial value problem.

Key Concepts

Initial Value Problem

An initial value problem in differential equations involves finding a specific solution from a family of solutions that satisfies given initial conditions. These conditions typically specify the value of the function and/or its derivatives at a particular point.

• The goal is to find a unique solution that passes through a given point, unlike the general solution which represents an entire family of curves.
• Typically, you find the values of integration constants that fulfill the initial conditions provided.

For example, in the exercise, the goal was to find specific values of the constants in the general solution so that the function meets the criteria at a specific point in its domain. This means solving
the differential equation with extra criteria to pinpoint one specific solution.

General Solution

A general solution of a differential equation contains all possible solutions and includes arbitrary constants that can take various values. In the exercise we looked at, the function:\[y = c_{1} x + c_{2} x \ln x\]is the general solution. This means it contains all behaviors of the system described by the differential equation.

• The constants \( c_1 \) and \( c_2 \) represent these arbitrary constants reflecting different members of the family of solutions.
• To convert a general solution into a specific solution, initial or boundary conditions are provided.

This general solution, when verified, should satisfy the original differential equation for any values of the constants within specified intervals.

Derivatives

Derivatives are central to solving differential equations. They represent the rate of change of functions.

• The first derivative, \( y' \), is the rate of change of \( y \) with respect to \( x \).
• The second derivative, \( y'' \), represents the rate of change of the rate of change, essentially capturing the curvature or acceleration of \( y \).

In the context of the exercise, each derivative is calculated to help verify if the general solution meets the differential equation.To illustrate:- First derivative from the exercise: \[y' = c_1 + c_2 (\ln x + 1) \]- Second derivative: \[y'' = \frac{c_2}{x} \]These derivatives were then plugged into the differential equation to check the solution's validity across the given range.

Integration Constant

The integration constant arises naturally when solving differential equations, especially when integrating.When you integrate a function, there are potentially many different functions that can result, all differing by a constant. This is because differentiation of a constant results in zero, making them indistinguishable from one another through derivatives alone.

• Each integration stage introduces an unknown constant, referred to in our exercise as \( c_1 \) and \( c_2 \).
• These constants are ambiguous until specific conditions, like initial values, are applied to determine them precisely.

In the exercise, these constants were determined by applying the conditions \( y(1)=3 \) and \( y'(1)=-1 \). Solving these conditions gave us precise values for \( c_1 \) and \( c_2 \), making the solution specific rather than general.