Question

Solve the following initial-value problems by using integrating factors. $$ (2+x) y^{\prime}=y+2+x, y(0)=0 $$

Short answer

The solution is \( y = (2+x)\ln\left(\frac{|2+x|}{2}\right) \).

Step-by-step solution

Write the Equation in Standard form

First, we need to express the given differential equation in the standard linear form \( y' + P(x)y = Q(x) \). Divide both sides of the equation \((2+x) y' = y + 2 + x\) by \(2+x\):\[ y' = \frac{y}{2+x} + 1\]Rearrange it to standard form:\[ y' - \frac{1}{2+x}y = 1 \]

Identify P(x) and Determine the Integrating Factor

In the equation \( y' - \frac{1}{2+x}y = 1 \), identify \(P(x) = -\frac{1}{2+x} \).The integrating factor \( \mu(x) \) is given by the formula:\[ \mu(x) = e^{\int P(x)\, dx} = e^{-\int \frac{1}{2+x}\, dx}\]

Solve for the Integrating Factor

Calculate the integral:\[ \int \frac{-1}{2+x} \, dx = -\ln|2+x| \]Hence, the integrating factor becomes:\[ \mu(x) = e^{-\ln|2+x|} = \frac{1}{2+x} \]

Multiply the Equation by the Integrating Factor

Multiply the entire differential equation \(y' - \frac{1}{2+x}y = 1 \) by the integrating factor \( \frac{1}{2+x} \):\[ \frac{1}{2+x}y' - \frac{1}{(2+x)^2}y = \frac{1}{2+x} \]

Express the Left Side as a Derivative

Identify that the left side of the equation is the derivative of a product:\[ \frac{d}{dx}\left(\frac{y}{2+x}\right) = \frac{1}{2+x}y' - \frac{1}{(2+x)^2}y \]So, the equation becomes:\[ \frac{d}{dx}\left(\frac{y}{2+x}\right) = \frac{1}{2+x} \]

Integrate Both Sides

Integrate both sides of the equation with respect to \(x\):\[ \int \frac{d}{dx}\left(\frac{y}{2+x}\right) \, dx = \int \frac{1}{2+x} \, dx \]This yields:\[ \frac{y}{2+x} = \ln|2+x| + C \]

Solve for \( y \)

Solve the equation \( \frac{y}{2+x} = \ln|2+x| + C \) for \( y \):\[ y = (2+x)(\ln|2+x| + C) \]

Apply the Initial Condition

Use the initial condition \( y(0) = 0 \) to find \( C \):Substitute \( x = 0 \) and \( y = 0 \) into the equation:\[ 0 = (2 + 0)(\ln|2+0| + C) \]\[ 0 = 2(\ln(2) + C) \]\[ \ln(2) + C = 0 \]Thus, \( C = -\ln(2) \).

Substitute the Constant Back to General Solution

Substitute \( C = -\ln(2) \) back into the expression for \( y \):\[ y = (2+x)(\ln|2+x| - \ln(2)) \]Simplify:\[ y = (2+x)\ln\left(\frac{|2+x|}{2}\right) \]

Conclude the Solution

The solution to the initial value problem is:\[ y = (2+x)\ln\left(\frac{|2+x|}{2}\right) \]

Key concepts

Initial-Value Problems

In calculus and differential equations, an initial-value problem is a type of problem where we determine a function's value based on given values at a specific point. It's like solving a puzzle where the initial condition guides us to the correct solution.

Initial-value problems involve one or more differential equations along with specified values, called the initial conditions. These conditions are crucial as they ensure that the solution is unique to the problem at hand. Without them, a differential equation can have numerous possible solutions, leading to ambiguity.

• An initial-value problem typically looks like: \( y'(x) = f(x, y), \ y(x_0) = y_0 \)
• The initial condition \( y(x_0) = y_0 \) provides a starting point for finding the specific solution.

Consider an example: We have \( y' = y + 2 + x \) with \( y(0) = 0 \). Here, the initial condition \( y(0) = 0 \) helps us find the constant, \( C \), after integrating the differential equation.

Linear Differential Equations

Linear differential equations are a group of equations where the unknown function and its derivatives appear linearly. These equations do not multiply or exponentiate the unknown function. Understanding such equations is key, as they form the basis for more complex differential equations.

The general form of a first-order linear differential equation is given by:

• \( y' + P(x)y = Q(x) \)

Here, \( P(x) \) and \( Q(x) \) are given functions. In these equations, the main goal is to find the function \( y(x) \) that satisfies the equation. Linear differential equations are essential in various scientific fields, such as physics and engineering, as they describe processes like heat transfer and electrical circuits.

For the equation \( (2+x) y^{\prime} = y + 2 + x \), it needs to be transformed into the form \( y' + P(x)y = Q(x) \) before solving, which ensures we can apply structured methods like integrating factors to find the solution.

Integrating Factor Method

The integrating factor method is a powerful technique used to solve linear differential equations. This method involves multiplying the entire differential equation by a specially chosen function, the integrating factor, to make the equation easier to solve.

Here's how it works:

• Convert the differential equation to the form \( y' + P(x)y = Q(x) \).
• Calculate the integrating factor, \( \mu(x) = e^{\int P(x) \, dx} \).
• Multiply the whole equation by \( \mu(x) \) to transform it into an easily integrable form.

In the provided exercise, for example, once the integrating factor \( \mu(x) \) is calculated as \( \frac{1}{2+x} \), it simplifies the differential equation, allowing easy integration to solve for \( y \).

Using the integrating factor method consolidates the equation into a derivative of a product form \( \frac{d}{dx} \left( \frac{y}{2+x} \right) \). This transformation simplifies the process, allowing quick integration to find the solution and later applying the initial condition to pinpoint the constant \( C \).