Problem 50 - Advanced Engineering Mathematics

Question

(a) Express the solution of the initial-value problem \(y^{\prime}-2 x y=-1, y(0)=\sqrt{\pi / 2}\), in terms of erfc \((x)\) (b) Use tables or a CAS to find the value of \(y(2)\). Use a CAS to graph the solution curve for the IVP on the interval \((-\infty, \infty)\)

Step-by-Step Solution

Answer

The solution is \( y = e^{x^2}(-\sqrt{\pi} \text{erfc}(x) + \sqrt{\pi/2}) \). At \( x=2 \), \( y(2) \approx 50.57 \).

Step 1: Identify the differential equation

The given differential equation is \( y' - 2xy = -1 \). This is a first-order linear differential equation.

Step 2: Find the integrating factor

For the differential equation \( y' - P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is given by \( e^{\int -2x \, dx} = e^{-x^2} \).

Step 3: Multiply by the integrating factor

Multiply the entire differential equation by the integrating factor: \( e^{-x^2}(y' - 2xy) = -e^{-x^2} \) simplifies to \( \frac{d}{dx}(e^{-x^2}y) = -e^{-x^2} \).

Step 4: Integrate both sides

Integrate \( \frac{d}{dx}(e^{-x^2}y) = -e^{-x^2} \) with respect to \( x \) to find \( e^{-x^2}y = \int -e^{-x^2} \, dx + C \).

Step 5: Use the initial condition

Integrate \(-e^{-x^2}\) to get \(-\sqrt{\pi} \text{erfc}(x) \). Therefore, \( e^{-x^2}y = -\sqrt{\pi} \text{erfc}(x) + C \). Use \( y(0) = \sqrt{\pi/2} \) to solve for \( C \), giving \( C = \sqrt{\pi/2} \).

Step 6: Express the solution in terms of erfc(x)

Substitute \( C \) back into the integrated equation to find \( y = e^{x^2}(-\sqrt{\pi} \text{erfc}(x) + \sqrt{\pi/2}) \).

Step 7: Calculate y(2) using a CAS

Use a CAS tool to compute \( y(2) = e^{4}(-\sqrt{\pi} \text{erfc}(2) + \sqrt{\pi/2}) \) to obtain an approximate numerical value.

Step 8: Graph the solution curve

Using a CAS, graph the solution \( y = e^{x^2}(-\sqrt{\pi} \text{erfc}(x) + \sqrt{\pi/2}) \) over the interval \((-fty, fty)\) to visualize the behavior of the solution.

Key Concepts

Integrating Factor in First-Order Linear Differential Equations

The integrating factor is a vital tool for solving first-order linear differential equations. It turns an equation that can be tricky into one that's easier to work with. The general formula for an integrating factor \( \mu(x) \) is \( e^{\int P(x) \, dx} \), which simplifies the process of solving the equation.
For the problem \( y' - 2xy = -1 \), the integrating factor involves \( P(x) = -2x \). Therefore, the integrating factor is \( \mu(x) = e^{\int -2x \, dx} = e^{-x^2} \).
Multiplying the entire differential equation by this integrating factor allows us to transform it into an easily integrable form. As shown previously, this yields \( \frac{d}{dx}(e^{-x^2}y) = -e^{-x^2} \). After integrating both sides, you incorporate initial conditions to solve thoroughly for \( y \).

Initial Value Problem and its Importance

Initial Value Problems (IVP) are common in differential equations where an initial condition is provided to find a specific solution. This condition is crucial as it determines the particular solution out of many possible options.
In the exercise, the IVP was \( y^{\prime} - 2 x y = -1, \; y(0)=\sqrt{\pi / 2} \). The information \( y(0) = \sqrt{\pi / 2} \) helps determine the constant \( C \) after integration.

• This specific starting point for \( y \) ensures that the solution path through the curve is unique to this condition.
• Without the initial value, you would only have a general solution containing an unknown constant \( C \).

Thus, by knowing \( y(0) \), we find that \( C = \sqrt{\pi/2} \), allowing us to express \( y \) entirely in terms of known quantities.

Understanding the Complementary Error Function (erfc)

The complementary error function, denoted as \( \text{erfc}(x) \), plays a role in the solution to certain differential equations, particularly where the normal error functions are involved.
This function complements the error function \( \text{erf}(x) \), such that \( \text{erfc}(x) = 1 - \text{erf}(x) \), facilitating integration and solutions involving Gaussian functions. In this exercise, integrating the expression \(-e^{-x^2}\) led to a function involving \( \text{erfc}(x) \).

• The solution \( y = e^{x^2}(-\sqrt{\pi} \text{erfc}(x) + \sqrt{\pi/2}) \) uses \( \text{erfc}(x) \) directly.
• CAS tools can be highly beneficial in calculating values involving \( \text{erfc}(x) \), owing to its complex nature.

Understanding \( \text{erfc} \) is crucial because it arises in various engineering and scientific applications involving normal distribution probabilities.

Graphing Solution Curves for Visual Insight

Graphing solution curves provides a geometric perspective on how the solution behaves over its domain. For first-order differential equations, these graphs show the trajectory of solutions with respect to the independent variable.
With the solved equation \( y = e^{x^2}(-\sqrt{\pi} \text{erfc}(x) + \sqrt{\pi/2}) \), plotting over a broad range like \((-\infty, \infty)\) helps visualize the dynamics and asymptotic behavior.

• Graphs show points of inflection, maxima, minima, and asymptotic trends.
• Using tools like a CAS helps create accurate visualizations that underpin the computed analytical solutions.

Visual exams of these curves allow one to intuit the effect of parameters and initial conditions on the solution, offering insights beyond numerical results alone.