Problem 29 - Advanced Engineering Mathematics

Question

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$ (x+1) \frac{d y}{d x}+y=\ln x, \quad y(1)=10 $$

Step-by-Step Solution

Answer

The solution is valid on the interval \((0, \infty)\).

Step 1: Recognize the Type of Differential Equation

The given differential equation is \((x+1) \frac{dy}{dx} + y = \ln x\). This is a first-order linear differential equation in the form \(P(x)\frac{dy}{dx} + Q(x)y = R(x)\) with \(P(x) = x + 1\), \(Q(x) = 1\), and \(R(x) = \ln x\).

Step 2: Rewrite in Standard Form

To rewrite the equation in the standard form \(\frac{dy}{dx} + p(x)y = g(x)\), divide every term by \(x+1\), giving \(\frac{dy}{dx} + \frac{1}{x+1}y = \frac{\ln x}{x+1}\).

Step 3: Find the Integrating Factor

Calculate the integrating factor \(\mu(x)\) using \(\mu(x) = e^{\int \frac{1}{x+1} \, dx}\). The integral of \(\frac{1}{x+1}\) is \(\ln|x+1|\), so \(\mu(x) = e^{\ln|x+1|} = |x+1|\).

Step 4: Solve Using the Integrating Factor

Multiply each term in the equation by the integrating factor \(|x+1|\): \(|x+1| \frac{dy}{dx} + \frac{|x+1|}{x+1} y = \frac{|x+1| \ln x}{x+1}\). Simplifying gives \((x+1) \frac{dy}{dx} = \ln x\). Since the derivative of \(y(x+1)\) with respect to \(x\) is \(\ln x\), integrate both sides with respect to \(x\) to solve for \(y\).

Step 5: Integrate Both Sides

Integrating the left side \(\int d(y(x+1))\) gives \(y(x+1)\). The right side becomes \(\int \ln x \, dx\). Using integration by parts: let \(u = \ln x\) and \(dv = dx\), then \(du = \frac{1}{x}dx\) and \(v = x\). Therefore, \(\int u \, dv = uv - \int v \, du = x\ln x - \int x\frac{1}{x} \, dx = x\ln x - x+C\). After simplifying, \(y(x+1) = x\ln x - x + C\).

Step 6: Solve for y

Divide through by \(x+1\) to get \(y = \frac{x\ln x - x + C}{x+1}\).

Step 7: Apply Initial Condition

Use the initial condition \(y(1) = 10\) to find \(C\). Substitute \(x = 1\) and \(y = 10\) into the equation: \(10 = \frac{1 \cdot \ln 1 - 1 + C}{1+1} = \frac{-1 + C}{2}\). Solving for \(C\), we have: \(2 \cdot 10 = -1 + C\) gives \(C = 21\).

Step 8: Write the General Solution

Substituting \(C = 21\) back, the solution is \(y = \frac{x \ln x - x + 21}{x+1}\).

Step 9: Determine the Interval of Validity

The solution is defined where both the differential equation and the initial condition are valid: Since \(\ln x\) requires \(x > 0\) and the denominator \(x+1\) cannot be zero, \(x eq -1\), the largest interval \(I\) where the solution holds, given \(x = 1\) is included, is \((0, \infty)\).

Key Concepts

Initial-Value Problems

An initial-value problem in the context of differential equations involves finding a function that satisfies a given differential equation, while also meeting initial conditions specified at a particular point. This problem provides extra information, such as a specific value of the function at a certain point, that sets it apart from other types of differential equations. For example, if you're given the differential equation with the condition \( y(1) = 10 \), it means that when \( x = 1 \), \( y \) must equal \( 10 \). This initial condition helps in determining the constant of integration, ensuring a unique solution to the differential equation.

Understanding initial-value problems is essential because they model many real-world systems. Wherever the state of a system at a specific time or location is known, an initial-value problem can describe how the system evolves over time or space. For students, mastering these problems includes learning to apply conditions accurately to find specific solutions.

Integrating Factor

The integrating factor is a pivotal tool in solving linear differential equations. Its purpose is to simplify differential equations by making them easily integrable. In our specific problem, the integrating factor \( \mu(x) \) is determined as \( e^{\int \frac{1}{x+1} \, dx} \).

• This integrating factor essentially transforms the left-hand side of the differential equation into the derivative of a product of functions.
• For example, multiplying the differential equation by the integrating factor, we obtain a simplification where \( \frac{d}{dx}(y(x+1)) \) appears.

This simplification makes it possible to integrate both sides easily, leading to a solution. Using an integrating factor is a powerful technique because it provides a methodical approach to solving a broad class of linear differential equations.

Interval of Validity

The interval of validity for a solution to a differential equation is crucial. It determines within which values the solution holds true or is defined. The interval takes into account any restrictions placed by the differential equation itself and any initial conditions. For our example, there are two main conditions to consider:

• The natural logarithm \( \ln x \) is only defined for \( x > 0 \).
• The term \( x+1 \) shouldn't equal zero, thus \( x eq -1 \).

Therefore, considering the initial condition of \( x = 1 \), the largest interval where the solution is valid is \( (0, \infty) \).

Students must be careful when determining the interval of validity, as ignoring constraints can lead to incorrect conclusions about the behavior of the solution. Understanding and applying these constraints ensures the solution accurately describes the system modeled by the differential equation.

First-Order Differential Equation

First-order differential equations are those where the highest derivative involved is the first derivative. These types of equations are the simplest form of differential equations and are often used to model rates of change.

In our problem, the equation \((x+1) \frac{dy}{dx} + y = \ln x\) is in this category. It can be rewritten in the standard form \( \frac{dy}{dx} + p(x)y = g(x) \). This form is essential for employing an integrating factor.

First-order differential equations are pervasive in both theoretical and applied contexts due to their simplicity and relevance to real-world systems, such as modeling exponential growth, decay processes, and cooling laws. Mastery of solving these equations forms the foundation for tackling more complex higher-order differential equations in mathematics and engineering.