Problem 48 - Advanced Engineering Mathematics

Question

(a) An integrating factor for \(y^{\prime}-2 x y=-1\) is \(e^{-x^{2}} .\) Thus \\[ \begin{aligned} \frac{d}{d x}\left[e^{-x^{2}} y\right] &=-e^{-x^{2}} \\ e^{-x^{2}} y &=-\int_{0}^{x} e^{-t^{2}} d t=-\frac{\sqrt{\pi}}{2} \operatorname{erf}(x)+c \end{aligned} \\] From \(y(0)=\sqrt{\pi} / 2,\) and noting that erf \((0)=0,\) we get \(c=\sqrt{\pi} / 2 .\) Thus \\[ y=e^{x^{2}}\left(-\frac{\sqrt{\pi}}{2} \operatorname{erf}(x)+\frac{\sqrt{\pi}}{2}\right)=\frac{\sqrt{\pi}}{2} e^{x^{2}}(1-\operatorname{erf}(x))=\frac{\sqrt{\pi}}{2} e^{x^{2}} \operatorname{erfc}(x) \\] (b) Using a CAS we find \(y(2) \approx 0.226339\)

Step-by-Step Solution

Answer

The solution is \( y = \frac{\sqrt{\pi}}{2} e^{x^2} \operatorname{erfc}(x) \), with \( y(2) \approx 0.226339 \).

Step 1: Identify the Differential Equation

The given differential equation is \( y' - 2xy = -1 \). Our goal is to solve this first-order linear differential equation using the integrating factor method.

Step 2: Integrating Factor Calculation

The given integrating factor is \( e^{-x^2} \). An integrating factor is generally expressed as \( e^{\int P(x) \, dx} \) for an equation of the form \( y' + P(x)y = Q(x) \). Here, \( P(x) = -2x \), so the integrating factor is indeed \( e^{-x^2} \).

Step 3: Multiply the Equation by the Integrating Factor

We multiply the entire differential equation by the integrating factor, \( e^{-x^2} \), to obtain \( e^{-x^2} y' - 2xe^{-x^2}y = -e^{-x^2} \). This allows the left-hand side to be re-written as the derivative of a product.

Step 4: Write as Derivative of a Product

Express the left-hand side as the derivative of a product: \( \frac{d}{dx}(e^{-x^2}y) = -e^{-x^2} \). This step simplifies the integration process.

Step 5: Integrate Both Sides

Integrate both sides with respect to \( x \): \( e^{-x^2}y = -\int_{0}^{x} e^{-t^2} \ dt + C \). The integral on the right-hand side cannot be expressed in elementary functions, hence it involves the error function \( \operatorname{erf}(x) \).

Step 6: Simplify Using Error Function

The integral \( -\int_{0}^{x} e^{-t^2} \ dt \) results in \( -\frac{\sqrt{\pi}}{2} \operatorname{erf}(x) \), based on the integration limits and the definition of the error function.

Step 7: Apply Initial Condition to Find Constant

Given \( y(0) = \frac{\sqrt{\pi}}{2} \) and \( \operatorname{erf}(0) = 0 \), substitute in the equation to solve for \( C \): \( \frac{\sqrt{\pi}}{2} = C \). Thus, \( C = \frac{\sqrt{\pi}}{2} \).

Step 8: Find General Solution

Substitute \( C \) back into the equation: \( y = e^{x^2}\left(-\frac{\sqrt{\pi}}{2} \operatorname{erf}(x) + \frac{\sqrt{\pi}}{2}\right) \). Simplify to get \( y = \frac{\sqrt{\pi}}{2} e^{x^2} (1 - \operatorname{erf}(x)) = \frac{\sqrt{\pi}}{2} e^{x^2} \operatorname{erfc}(x) \), where \( \operatorname{erfc}(x) = 1 - \operatorname{erf}(x) \).

Step 9: Evaluate Particular Solution Using CAS

Using a Computer Algebra System (CAS), evaluate the expression at \( x = 2 \) for numerical approximation. The solution \( y(2) \approx 0.226339 \) is confirmed.

Key Concepts

Integrating Factor

An integral part of solving first-order linear differential equations is the use of an integrating factor. This is a function that helps transform a differential equation so that it becomes easier to solve. In our exercise, we have the differential equation \( y' - 2xy = -1 \). To solve it, we need an integrating factor that can convert the left-hand side into the derivative of a specific product. The general formula for the integrating factor \( \ \) is \( e^{\int P(x) \, dx} \), where \( P(x) \) is the coefficient of \( y \) in the standard form of the equation \( y' + P(x)y = Q(x) \). In our problem, \( P(x) = -2x \), which makes the integrating factor \( e^{-x^2} \). This simple step paves the way for the equation to be solved as the product of a derivative.

Error Function (erf)

The error function, denoted as \( \operatorname{erf}(x) \), plays a crucial role in this particular exercise. While performing the integration of \( e^{-t^2} \), we encounter a function that cannot be expressed in terms of elementary functions. Here, the error function comes into play. It is defined as:

• \( \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt \)

The error function is important in probability, statistics, and differential equations. In this problem, it arises naturally when integrating \( e^{-x^2} \), resulting in a term \( -\frac{\sqrt{\pi}}{2} \operatorname{erf}(x) \). Understanding and using \( \operatorname{erf}(x) \) is integral for finding the function \( y \) that satisfies the initial conditions.

Initial Value Problem

When solving differential equations, initial value problems provide specific conditions that must be satisfied by the solution. It involves finding a particular solution to a differential equation that passes through a given initial point. In our example, the initial condition is \( y(0) = \frac{\sqrt{\pi}}{2} \).This condition helps determine the constant \( C \) after integration. Substituting \( x = 0 \) into \( y = e^{x^2} \left(-\frac{\sqrt{\pi}}{2} \operatorname{erf}(x) + C\right) \), we find that since \( \operatorname{erf}(0) = 0 \), the equation simplifies, leading us to \( C = \frac{\sqrt{\pi}}{2} \). This establishes the complete solution for the differential equation.

First-Order Linear Differential Equation

A first-order linear differential equation is characterized by the highest derivative being of the first order. They are typically expressed in the form \( y' + P(x)y = Q(x) \). In this exercise, we have the equation \( y' - 2xy = -1 \).These equations can often be solved using techniques like the integrating factor to simplify and integrate. The significance of these equations is in modeling real-world phenomena such as cooling processes, electrical circuits, and various physical systems. In the provided problem, applying the integrating factor turns the left-hand side into \( \frac{d}{dx}(e^{-x^2}y) \), making it straightforward to integrate both sides and solve. Such methods emphasize the practicality and power of first-order linear differential equations in computational mathematics.