Problem 37 - Advanced Engineering Mathematics

Question

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 \\ y &=\frac{\sqrt{\pi}}{2} e^{x^{2}} \operatorname{erf}(x)+c e^{x^{2}} \end{aligned} \\] From \(y(1)=(\sqrt{\pi} / 2) e \operatorname{erf}(1)+c e=1\) we get \(c=e^{-1}-\frac{\sqrt{\pi}}{2} \operatorname{erf}(1) .\) The solution of the initial-value problem is \\[ y=\frac{\sqrt{\pi}}{2} e^{x^{2}} \operatorname{erf}(x)+\left(e^{-1}-\frac{\sqrt{\pi}}{2} \operatorname{erf}(1)\right) e^{x^{2}} \\] \\[ =e^{x^{2}-1}+\frac{\sqrt{\pi}}{2} e^{x^{2}}(\operatorname{erf}(x)-\operatorname{erf}(1)) \\]

Step-by-Step Solution

Answer

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

Step 1: Write the Differential Equation

The given differential equation is \(y' - 2xy = 1\). We are given that the integrating factor for this equation is \(e^{-x^2}\).

Step 2: Verify the Integrating Factor

Multiply both sides of the differential equation by the integrating factor \(e^{-x^2}\). This gives \(e^{-x^2}y' - 2xe^{-x^2}y = e^{-x^2}\).

Step 3: Reformulate as an Exact Differential

Notice that the left side can be rewritten as the derivative of \(e^{-x^2}y\) with respect to \(x\): \(\frac{d}{dx}(e^{-x^2}y) = e^{-x^2}\).

Step 4: Integrate Both Sides

Integrate both sides with respect to \(x\). The left becomes \(e^{-x^2}y\), and the right side becomes \(\int e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}\operatorname{erf}(x) + C\).

Step 5: Solve for \(y\)

Multiply through by \(e^{x^2}\) to solve for \(y\): \(y = \frac{\sqrt{\pi}}{2} e^{x^2} \operatorname{erf}(x) + Ce^{x^2}\).

Step 6: Apply Initial Condition

Use the initial condition \(y(1) = 1\). Plug \(x = 1\) into the solution: \(1 = \frac{\sqrt{\pi}}{2} e \operatorname{erf}(1) + Ce\).

Step 7: Solve for Constant \(C\)

Rearrange the equation from Step 6 to solve for \(C\): \(C = e^{-1} - \frac{\sqrt{\pi}}{2} \operatorname{erf}(1)\).

Step 8: Write the Final Solution

Substitute \(C\) back into the expression for \(y\) found in Step 5 to get the final solution:\[ y = \frac{\sqrt{\pi}}{2} e^{x^2} \operatorname{erf}(x) + (e^{-1} - \frac{\sqrt{\pi}}{2} \operatorname{erf}(1)) e^{x^2} \]

Key Concepts

Integrating Factor

An integrating factor is a function that, when multiplied by both sides of a given differential equation, transforms it into an exact differential equation (a form that can be easily integrated). In the given exercise, we see that the equation is not in a readily integrable form: \( y' - 2xy = 1 \).
When you introduce the integrating factor, \( e^{-x^2} \), the equation is multiplied throughout by this factor, making the left-hand side a perfect derivative: \( e^{-x^2} y' - 2xe^{-x^2} y = e^{-x^2} \).

The whole point of using an integrating factor is to facilitate integrating the equation by recognizing it as a derivative of a product. This transforms complex equations into more manageable ones. Essentially, if you multiply your differential equation by an exponential function like \( e^{\int P(x) \, dx} \), where \( P(x) \) is a component of the original equation, it often simplifies the equation significantly.

Initial Value Problem

An initial value problem involves finding a solution to a differential equation that also satisfies a specific condition or conditions at a particular point. These conditions are typically given as the value of the unknown function, say \( y \), at some specific point \( x \).
For instance, in this exercise, the condition is \( y(1) = 1 \). This means that when \( x = 1 \), the function \( y \) must equal 1.

Finding the correct solution of the differential equation means not only solving for \( y \) but also determining the constant \( C \) in the general solution so that the initial condition is satisfied. These initial conditions are crucial because they ensure the uniqueness of the solution among the possible solutions that a differential equation might have.

Error Function

The error function, denoted as \( \operatorname{erf}(x) \), is a special function used in probability, statistics, and partial differential equations related to the Gaussian distribution. It's especially important in the field of differential equations that involve exponential functions of squared terms.
In the exercise, we integrate \( e^{-x^2} \), which results in a function expressed involving the error function: \( \int e^{-t^2} \, dt = \frac{\sqrt{\pi}}{2} \operatorname{erf}(x) + C \).

The error function ranges between -1 and 1, and is used as a standard mathematical function for describing distributions that stray from the normal curve, providing solutions to complex integrals that arise in differential equations. Understanding its properties allows us to manipulate and solve equations that involve Gaussian-type expressions.

Exact Differential Equation

An exact differential equation is one that can be expressed in the form \( \frac{d}{dx}(f(x, y)) = g(x) \). This form enables easy integration. The process of reformulating the differential equation using an integrating factor is ideally meant to assist in making the equation exact.
In this exercise, by applying the integrating factor \( e^{-x^2} \), the left-hand side transforms into \( \frac{d}{dx}(e^{-x^2} y) \), making it exact because it represents a complete derivative that can be directly integrated.

Converting a differential equation into an exact one simplifies the problem greatly, as it allows us to proceed directly to integrating both sides. Mastering this technique is a fundamental skill in solving linear differential equations and is widely used because it reduces the problem to the application of basic calculus concepts.