Problem 29 - Advanced Engineering Mathematics

Question

For \(y^{\prime}+\frac{1}{x+1} y=\frac{\ln x}{x+1}\) an integrating factor is \(e^{\int[1 /(x+1)] d x}=x+1\) so that \(\frac{d}{d x}[(x+1) y]=\ln x\) and \(y=\frac{x}{x+1} \ln x-\frac{x}{x+1}+\frac{c}{x+1}\) for \(0< x<\infty .\) If \(y(1)=10\) then \(c=21\) and \(y=\frac{x}{x+1} \ln x-\frac{x}{x+1}+\frac{21}{x+1}\).

Step-by-Step Solution

Answer

The particular solution is \(y = \frac{x}{x+1} \ln x - \frac{x}{x+1} + \frac{21}{x+1}\).

Step 1: Identify the differential equation

The given differential equation is \(y' + \frac{1}{x+1} y = \frac{\ln x}{x+1}\). This is a first-order linear ordinary differential equation.

Step 2: Determine the integrating factor

The integrating factor is found using the formula \(e^{\int P(x) \, dx}\), where \(P(x) = \frac{1}{x+1}\). Thus, the integrating factor becomes \(e^{\int \frac{1}{x+1} \, dx} = e^{\ln|x+1|} = x+1\).

Step 3: Rewrite the equation using the integrating factor

By multiplying the entire differential equation by the integrating factor \(x+1\), we have: \((x+1) y' + y = \ln x\). This can be rewritten as \(\frac{d}{dx}[(x+1)y] = \ln x\).

Step 4: Integrate both sides

Integrate both sides of the equation \(\frac{d}{dx}[(x+1)y] = \ln x\) with respect to \(x\). The left side integrates to \((x+1)y\) and the integral of \(\ln x\) with respect to \(x\) is \(x \ln x - x + C\). Therefore, we have \((x+1)y = x \ln x - x + C\).

Step 5: Solve for \(y\)

To find \(y\), divide the equation \((x+1)y = x \ln x - x + C\) by \(x+1\), giving \(y = \frac{x}{x+1} \ln x - \frac{x}{x+1} + \frac{C}{x+1}\). This is the general solution of the differential equation.

Step 6: Apply initial condition to find \(C\)

Given the initial condition \(y(1) = 10\), substitute \(x = 1\) and \(y = 10\) into the equation \(y = \frac{x}{x+1} \ln x - \frac{x}{x+1} + \frac{C}{x+1}\). This simplifies to \(10 = 0 - \frac{1}{2} + \frac{C}{2}\). Solve for \(C\): \(10 = -\frac{1}{2} + \frac{C}{2}\) leads to \(10.5 = \frac{C}{2}\). So, \(C = 21\).

Step 7: Write the particular solution

Substitute \(C = 21\) back into the expression for \(y\): \(y = \frac{x}{x+1} \ln x - \frac{x}{x+1} + \frac{21}{x+1}\). This is the particular solution that satisfies the given initial condition.

Key Concepts

Integrating Factor

An integrating factor is a powerful tool to solve linear ordinary differential equations (ODEs). In simple terms, it helps to transform a difficult ODE into one that can be easily integrated. The process involves multiplying every term of the differential equation by this special function, making the left side of the equation become an exact derivative.

For the ODE given: \[ y' + \frac{1}{x+1} y = \frac{\ln x}{x+1} \]the integrating factor is derived using the formula: \[ e^{\int P(x) \, dx} \]where \( P(x) = \frac{1}{x+1} \). Integrating \( P(x) \) gives:

• \( \int \frac{1}{x+1} \, dx = \ln|x+1| \)
• Therefore, the integrating factor is \( e^{\ln|x+1|} = x+1 \)

By applying \( x+1 \) as the integrating factor, the differential equation can be rewritten into a form where the left-hand side is \[ \frac{d}{dx}[(x+1)y] = \ln x \].This allows you to integrate both sides, simplifying the process of finding the solution.

Initial Value Problem

An initial value problem involves finding a particular solution to a differential equation that satisfies a specific initial condition. This means that along with the differential equation, you are provided with a starting point or value of the function, usually at a particular point in time.

In our exercise, the initial value condition is given as \( y(1) = 10 \). This condition is crucial because it allows you to compute the constant of integration, \( C \), in your solution.

Upon deriving the general solution from the ODE, which is: \[ y = \frac{x}{x+1} \ln x - \frac{x}{x+1} + \frac{C}{x+1} \]we substitute \( x = 1 \) and \( y = 10 \) to find \( C \). This step ensures that the solution fits the specific scenario defined at the starting point. Solving gives \( C = 21 \), resulting in the particular solution \[ y = \frac{x}{x+1} \ln x - \frac{x}{x+1} + \frac{21}{x+1} \].This solution now correctly reflects the situation where the initial condition \( y(1) = 10 \) holds true.

Ordinary Differential Equations (ODEs)

Ordinary Differential Equations, or ODEs, are equations that involve functions of one independent variable and their derivatives. They are called "ordinary" to distinguish them from partial differential equations, which deal with multiple independent variables.

The ODE in our problem is a **first-order linear ODE** because it involves only first derivatives of the function \( y \), and it can be expressed in the form:\[ y' + P(x) y = Q(x) \]where both \( P(x) \) and \( Q(x) \) are functions of \( x \). The task is to determine the function \( y(x) \) that satisfies this equation.

One common technique employed to solve first-order linear ODEs is the integrating factor method. This method converts the equation into a form where direct integration is possible, allowing for simpler solving of the equation. This enables you to systematically derive the function \( y \) that will satisfy not only the differential equation itself but also any initial value conditions if provided. Understanding ODEs is crucial in various fields, as they model many dynamically changing systems in physics, engineering, biology, and economics.