Question

Solve each differential equation. $$ x y^{\prime}+(1+x) y=e^{-x} ; y=0 \text { when } x=1 \text { . } $$

Short answer

The solution is \( y = \frac{1}{x} - \frac{e^{-x}}{x} \).

Step-by-step solution

Recognize the type of differential equation

The given differential equation \( xy' + (1+x)y = e^{-x} \) is a first-order linear differential equation. We can solve it using an integrating factor.

Standard form of linear differential equation

Rearrange the equation to standard linear form: \( y' + \frac{(1+x)}{x} y = \frac{e^{-x}}{x} \). Now, identify \( P(x) = \frac{1+x}{x} \) and \( Q(x) = \frac{e^{-x}}{x} \).

Calculate the integrating factor

The integrating factor \( \mu(x) \) is found by calculating \( e^{\int P(x) \, dx} \). Compute the integral \( \int \frac{(1+x)}{x} \, dx = \int (1 + \frac{1}{x}) \, dx = x + \ln |x| \). Thus, the integrating factor is \( \mu(x) = e^{x + \ln |x|} = x e^x \).

Multiply through by the integrating factor

Multiply the entire differential equation by the integrating factor \( x e^x \): \( e^x x y' + e^x (1+x) y = e^x \). This simplifies the left-hand side to the derivative of \( y \mu(x) \): \( \frac{d}{dx}[y x e^x] = e^x \).

Integrate both sides

Now, integrate both sides with respect to \( x \): \( \int \frac{d}{dx}[y x e^x] \, dx = \int e^x \, dx \). The left side becomes \( y x e^x \), and the right side becomes \( e^x \). So, \( y x e^x = e^x + C \).

Solve for y

Solve for \( y \): \( y = \frac{e^x + C}{x e^x} = \frac{1}{x} + \frac{C e^{-x}}{x} \).

Apply initial conditions

Apply the initial condition \( y = 0 \) when \( x = 1 \): \( 0 = \frac{1}{1} + C e^{-1} \). Thus, \( C = -e \).

Substitute C back into the general solution

Substitute \( C = -e \) back into the equation for \( y \): \( y = \frac{1}{x} - \frac{e^{-x}}{x} \).

Key concepts

Integrating factor

An integrating factor is a crucial tool in solving linear differential equations. It transforms a non-exact differential equation into an exact one, making it easier to integrate. To find the integrating factor, we typically look at the standard form of a first-order linear differential equation:

• \( y' + P(x) y = Q(x) \).

The integrating factor \( \mu(x) \) is given by \( e^{\int P(x) \, dx} \). In the exercise, we calculated this as \( e^{x + \ln |x|} = x e^x \).
This factor allows us to simplify our given equation, making the left-hand side a derivative of a product, which is much simpler to integrate. Once we have this integrating factor, we multiply through the entire differential equation by it. This leads to an equation where we can straightforwardly integrate both sides to solve for the function \( y \).

Initial conditions

In differential equations, initial conditions are used to determine a specific solution from a family of solutions. Every solution obtained from a differential equation has a constant of integration (denoted as \( C \)).
By applying the initial conditions, we can solve for this constant. For instance, in our exercise, the initial condition given was \( y=0 \) when \( x=1 \). We substituted these values into our general solution to find \( C = -e \).

• Initial conditions help to "pin" the solution to a specific real-world scenario or initial state.

By using them, we ensure the solution is not just a general or theoretical form but is applicable to the specific problem we have at hand.

Linear differential equation

A linear differential equation is an equation involving an unknown function and its derivatives. In its simplest form, a first-order linear differential equation can be written as:

• \( a(x) y' + b(x) y = c(x) \).

For our problem, the equation \( xy' + (1+x)y = e^{-x} \) is already in this form.
To make it easier to solve, we convert it to standard form: \( y' + \frac{(1+x)}{x} y = \frac{e^{-x}}{x} \). From here, we can identify the functions \( P(x) \) and \( Q(x) \) that help determine the integrating factor.
Linear differential equations are incredibly common in various fields such as physics, engineering, and economics because they can model many natural processes.

Separation of variables

Separation of variables is a mathematical method used to solve differential equations by separating functions of different variables onto opposite sides of the equation. Although not directly used in our example, understanding this technique is important when dealing with different types of differential equations.

• The basic goal is to rearrange the equation so that each variable and its differential are on different sides, typically in the form \( f(y) \, dy = g(x) \, dx \).

After separation, both sides are integrated independently. This method is particularly useful for simpler equations or those that can be neatly separated.
Separation of variables highlights the versatility of strategies available when solving differential equations and can sometimes be easier than employing integrating factors when applicable.
Being aware of multiple methods allows for flexibility and efficiency in approaching differential equations.