Problem 57 Express the solution of the init... [FREE SOLUTION]

Chapter 2: Problem 57

Express the solution of the initial-value problem $$ \frac{d y}{d x}-2 x y=1, \quad y(1)=1 $$ in terms of $\operatorname{erf}(x)$.

Short Answer

Expert verified

$ y = \left(\frac{\sqrt{\pi}}{2}\operatorname{erf}(x) + e^{-1} - \frac{\sqrt{\pi}}{2}e^{-1}\operatorname{erf}(1)\right)e^{x^2} $.

Step by step solution

Identify the Type of Differential Equation

The given differential equation $ \frac{d y}{d x} - 2xy = 1 $ is a first-order linear differential equation. It can be expressed in the standard form $ \frac{d y}{d x} + P(x)y = Q(x) $ with $ P(x) = -2x $ and $ Q(x) = 1 $.

Calculate the Integrating Factor

The integrating factor $ \mu(x) $ is given by $ e^{\int P(x) \, dx} = e^{-\int 2x \, dx} = e^{-x^2} $. This integrating factor will be used to solve the differential equation.

Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor: $ e^{-x^2} \frac{dy}{dx} - 2xe^{-x^2}y = e^{-x^2} $. This simplifies to $ \frac{d}{dx}(ye^{-x^2}) = e^{-x^2} $.

Integrate Both Sides

Integrate both sides of $ \frac{d}{dx}(ye^{-x^2}) = e^{-x^2} $ with respect to $ x $. The left side becomes $ ye^{-x^2} $ and the right side involves the error function: $ \int e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}\operatorname{erf}(x) + C $.

Solve for y(x)

Combine results into the equation $ ye^{-x^2} = \frac{\sqrt{\pi}}{2}\operatorname{erf}(x) + C $. Solving for $ y $ gives: $ y = \left(\frac{\sqrt{\pi}}{2}\operatorname{erf}(x) + C\right)e^{x^2} $.

Apply the Initial Condition

Use the initial condition $ y(1) = 1 $ to determine $ C $: $ 1 = \left(\frac{\sqrt{\pi}}{2}\operatorname{erf}(1) + C\right)e^{1^2} $. Simplifying: $ C = e^{-1} - \frac{\sqrt{\pi}}{2}e^{-1}\operatorname{erf}(1) $.

Final Solution

Substitute $ C $ back into the expression for $ y $: $ y = \left(\frac{\sqrt{\pi}}{2}\operatorname{erf}(x) + e^{-1} - \frac{\sqrt{\pi}}{2}e^{-1}\operatorname{erf}(1)\right)e^{x^2} $. This is the solution of the differential equation expressed with $ \operatorname{erf}(x) $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial-Value Problem

An initial-value problem involves finding a function that satisfies a differential equation and fulfills specific conditions at a starting point.
In our exercise, the equation given is:

$ \frac{d y}{d x} - 2 x y = 1 $
Along with the condition $ y(1) = 1 $

The importance of the initial condition $ y(1) = 1 $ is to help determine the constant $ C $ when solving the differential equation. Without this condition, the solution would contain an unknown constant, making it incomplete. The initial condition narrows down the possible solutions to a unique one that fits both the equation and the initial-value condition.

Integrating Factor

The integrating factor is a handy tool when solving first-order linear differential equations. This factor transforms the differential equation into something we can easily solve.
Here's how it works:

The differential equation is first put in the standard form: $ \frac{d y}{d x} + P(x)y = Q(x) $
The integrating factor $ \mu(x) $ is calculated as $ e^{\int P(x) \, dx} $

In our example, $ P(x) = -2x $, so the integrating factor becomes:

$ \mu(x) = e^{-x^2} $

Multiplying the entire differential equation by this integrating factor allows us to rewrite it in a form that can be integrated directly, simplified greatly due to its ability to transform the equation into the derivative of a product.

Error Function

The error function, often written as $ \operatorname{erf}(x) $, is a special function that arises in probability, statistics, and differential equations, especially relating to normal distribution and diffusion processes.
It is defined as:

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

This is particularly useful in integrating expressions involving $ e^{-x^2} $, as it gives a neat, usable form that appears frequently in solutions. In this problem, $ \operatorname{erf}(x) $ uniquely expresses the integrated solution of $ e^{-x^2} $, which is an otherwise complex integral to handle directly.

First-Order Linear Differential Equation

A first-order linear differential equation is one that involves the first derivative of a function and has the general form $ \frac{d y}{d x} + P(x)y = Q(x) $.
This class of equations is important because they can describe a wide range of physical phenomena, from population dynamics to circuit analysis.
Key features include:

Involves first derivatives only (hence 'first-order')
Linear in terms of the unknown function and its derivative

In the exercise problem, the equation $ \frac{d y}{d x} - 2 x y = 1 $ fits this mold with the form:

$ P(x) = -2x $
$ Q(x) = 1 $

Solving these equations typically involves techniques such as finding an integrating factor, allowing us to convert the equation into an easily integrable form.

91影视

Express the solution of the initial-value problem $$ \frac{d y}{d x}-2 x y=1, \quad y(1)=1 $$ in terms of \(\operatorname{erf}(x)\).

Short Answer

Step by step solution

Identify the Type of Differential Equation

Calculate the Integrating Factor

Multiply Through by the Integrating Factor

Integrate Both Sides

Solve for y(x)

Apply the Initial Condition

Final Solution

Key Concepts

Initial-Value Problem

Integrating Factor

Error Function

First-Order Linear Differential Equation

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