Question

Show that the solution of the initial-value problem $$ \begin{array}{r} y^{\prime}-2 x y=1 \\ y(0)=1 \end{array} $$ is $$ \phi(x)=e^{x^{2}}\left(1+\frac{\sqrt{\pi}}{2} \operatorname{erf}(x)\right) $$ where the emor function is defined by $$ \operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} d t $$ and is tabulated.

Short answer

Based on the step-by-step solution provided, determine if the given function \(\phi(x) = e^{x^2}\left(1 + \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)\right)\) is a solution to the initial-value problem \(y'(x) - 2xy(x) = 1\) with initial condition \(y(0) = 1\). Answer: Yes, the given function \(\phi(x)\) is a valid solution for the initial-value problem since it satisfies both the differential equation and the initial condition.

Step-by-step solution

Confirm that the given function satisfies the differential equation

The given differential equation is \(y'(x) - 2xy(x) = 1\). Let's denote the given function as \(\phi(x) = e^{x^2}\left(1 + \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)\right)\). We need to find its derivative and plug it back into the equation to see if it satisfies the equation. Differentiate \(\phi(x)\) with respect to \(x\): $$ \begin{aligned} \phi'(x) &= \frac{d}{dx} \left( e^{x^2}\left(1 + \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)\right) \right) \\ &= 2xe^{x^2}\left(1 + \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)\right) + e^{x^2}\frac{\sqrt{\pi}}{\sqrt{\pi}} e^{-x^2} \\ &= 2xe^{x^2}\left(1 + \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)\right) + e^{x^2 - x^2}\\ \end{aligned} $$ Plugging \(\phi(x)\) and \(\phi'(x)\) back into the equation: $$ \begin{aligned} \phi'(x) - 2x\phi(x) &= \left(2xe^{x^2}\left(1 + \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)\right) + 1\right) - 2x\left(e^{x^2}\left(1 + \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)\right)\right) \\ &= 1 \\ \end{aligned} $$ Since the above equality is true, the function \(\phi(x)\) satisfies the given differential equation.

Check if the given function satisfies the initial condition

The initial condition given is \(y(0) = 1\). Let's substitute \(x=0\) in \(\phi(x)\) to check if the initial condition is satisfied. $$ \begin{aligned} \phi(0) &= e^{0^2}\left(1 + \frac{\sqrt{\pi}}{2}\operatorname{erf}(0)\right) \\ &= \left(1 + \frac{\sqrt{\pi}}{2} \cdot 0\right) \\ &= 1 \\ \end{aligned} $$ Since \(\phi(0)\) evaluates to 1, the function \(\phi(x)\) satisfies the initial condition.

The given function is a valid solution of the IVP

Since the given function \(\phi(x)\) satisfies both the differential equation and the initial condition, we can conclude that it is indeed a valid solution of the given initial-value problem.

Key concepts

Initial-Value Problem

An initial-value problem in differential equations involves finding a function that satisfies a given first-order differential equation along with an initial condition. The differential equation provides the rate of change of a function, while the initial condition gives the specific value of the function at a certain point. This combination allows us to find a particular solution rather than a general one, which is crucial in applying the mathematical model to real-world scenarios.

In the given exercise, the differential equation is \[y' - 2xy = 1\]. The initial condition is expressed as \(y(0) = 1\).

The initial condition ensures that the solution curve passes through the point \((0,1)\). Solving such problems typically involves integrating the differential equation and applying the initial condition to determine any constants of integration.

Differential Equation Solution

To solve a differential equation, we need to find a function that satisfies the given equation under provided conditions. In this exercise, the function\[\phi(x) = e^{x^2}\left(1 + \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)\right)\]is suggested as the solution to the initial-value problem. A typical procedure to verify this involves differentiating the proposed solution and substituting it into the original differential equation.

By computing the derivative \(\phi'(x)\) and substituting \(\phi(x)\) and \(\phi'(x)\) back into the differential equation, we can check if both sides of the equation match. In this case, they do, confirming that \(\phi(x)\) is the solution that satisfies the differential equation and the initial condition.

Erf Function

The error function, denoted as \(\operatorname{erf}(x)\), is a special function used in probability, statistics, and partial differential equations. It is defined as:
\[\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} dt\]

This non-elementary function cannot be expressed using elementary functions, but it is widely tabulated and used. The function grows to approximate one for large positive \(x\) and negative one for large negative \(x\).

In the given solution, \(\operatorname{erf}(x)\) is part of the proposed solution function \(\phi(x)\), contributing to solving the differential equation by integrating across the Gaussian curve.

Integration Techniques

Solving differential equations often requires integrating functions, whether it's evaluating indefinite integrals, solving through substitution, or recognizing integrable forms like those involving the error function. In our context, integrating forms like \(e^{-t^2}\) resulting in the \(\operatorname{erf}(x)\) requires acknowledgment of its special status, as it doesn’t have standard antiderivatives.

Techniques used might involve:

• Recognizing standard integral forms.
• Substitution to simplify the integrals.
• Using tabulated integral values, especially with special functions like \(\operatorname{erf}(x)\).

Understanding the role of these functions and techniques is crucial for validating solutions to initial-value problems.