Problem 25 - Advanced Engineering Mathematics

Question

For \(y^{\prime}+\frac{1}{x} y=\frac{1}{x} e^{x}\) an integrating factor is \(e^{\int(1 / x) d x}=x\) so that \(\frac{d}{d x}[x y]=e^{x}\) and \(y=\frac{1}{x} e^{x}+\frac{c}{x}\) for \(0< x<\infty\). If \(y(1)=2\) then \(c=2-e\) and \(y=\frac{1}{x} e^{x}+\frac{2-e}{x}\).

Step-by-Step Solution

Answer

The solution is \(y = \frac{1}{x} e^{x} + \frac{2-e}{x}\).

Step 1: Identify the Differential Equation

The given differential equation is: \(y^{\prime} + \frac{1}{x} y = \frac{1}{x} e^{x}\). This is a first-order linear differential equation.

Step 2: Determine the Integrating Factor

For the equation \(y^{\prime} + P(x)y = Q(x)\), determine the integrating factor. Here, \(P(x) = \frac{1}{x}\) and thus the integrating factor \(\mu(x) = e^{\int \frac{1}{x} \, dx} = e^{\ln{x}} = x\).

Step 3: Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by \(x\), the integrating factor: \(x y^{\prime} + y = e^{x}\).

Step 4: Rewriting the Equation

Recognize that the left-hand side of the equation \(x y^{\prime} + y = e^{x}\) is the derivative of the product \(\frac{d}{dx}[xy]\). Therefore, rewrite the equation as: \(\frac{d}{dx}[xy] = e^{x}\).

Step 5: Integrate Both Sides

Integrate both sides of the equation: \(\int \frac{d}{dx}[xy] \, dx = \int e^{x} \, dx\). This gives: \(x y = e^{x} + c\), where \(c\) is the constant of integration.

Step 6: Solve for y

Solve for \(y\) by dividing both sides by \(x\): \(y = \frac{1}{x} e^{x} + \frac{c}{x}\).

Step 7: Apply the Initial Condition

Use the initial condition \(y(1) = 2\) to find \(c\). Substitute \(x = 1\) and \(y = 2\) into the equation: \(2 = e + c\). Thus, solving for \(c\), we get \(c = 2 - e\).

Step 8: Write the Final Solution

Substitute \(c = 2 - e\) back into the expression for \(y\), yielding: \(y = \frac{1}{x} e^{x} + \frac{2 - e}{x}\).

Key Concepts

Integrating Factor

In the world of first-order linear differential equations, the integrating factor is a powerful tool that simplifies solving them. An integrating factor is a function, often denoted as \( \mu(x) \), that we multiply through the differential equation to make it easier to solve. Given a linear differential equation of the form \( y^{\prime} + P(x)y = Q(x) \), the integrating factor is defined as:

• \( \mu(x) = e^{\int P(x) \, dx} \)

This approach is particularly useful because multiplying the entire differential equation by this factor turns the left side into the derivative of a product: \( \frac{d}{dx}[\mu(x)y] \). In our exercise, \( P(x) = \frac{1}{x} \), which leads to:

• \( \mu(x) = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x \)

Hence, we transform our original equation to a more manageable form by focusing on this integrating factor.

Initial Conditions

Initial conditions are specific values provided to help find the particular solution of a differential equation. They are very important because they allow us to identify the constant of integration and understand the behavior of the solution specific to the problem's context.
When we solve differential equations, especially in practical applications, merely having a general solution is insufficient. Initial conditions, like \( y(1) = 2 \), give us a starting point.
By substituting \( x = 1 \) into our general solution \( y = \frac{1}{x} e^{x} + \frac{c}{x} \), we get:

• \( 2 = e + c \)
• Solving this, \( c = 2 - e \)

This helps to tailor the solution \( y \) to fit the specific initial condition provided.

Constant of Integration

The constant of integration, denoted as \( c \), arises when we integrate a function. In the context of differential equations, when integrating to find solutions, the constant states that there are infinitely many potential solutions. Each different constant gives a different line or curve.
After finding \( \, x y = e^{x} + c\), and solving for \( y \), we introduce the constant of integration \( c \):

• \( y = \frac{1}{x} e^{x} + \frac{c}{x} \)

To determine \( c \), initial conditions are used, as mentioned earlier. Here, once we know \( c = 2 - e \), we have a specific solution among the many possibilities offered by the general solution.

First-Order Differential Equations

First-order differential equations involve derivatives of the first degree (i.e., only the first derivative of the function appears). These are the simplest kind of differential equations to solve and have numerous applications in modeling real-world phenomena like heat transfer and population growth.
The general form of a linear first-order differential equation is:

• \( y^{\prime} + P(x)y = Q(x) \)

These equations can often be solved using the integrating factor method. For our exercise, we rewrote it in this standard form and employed an integrating factor to streamline the solution.
The steps involve identifying \( P(x) \), calculating the integrating factor \( \mu(x) \), making the left side a derivative, and then solving through integration. Despite their simplicity, they form the basis for understanding more complex differential equations.