Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, solve each differential equation. $$ x y^{\prime}+(1+x) y=e^{-x} ; y=0 \text { when } x=1 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Differential Equation

The differential equation given is \( x y^{\prime} + (1+x) y = e^{-x} \). This is a first-order linear differential equation.

Step 2: Write the Equation in Standard Form

Rearrange the given equation to the standard form for linear differential equations: \( y^{\prime} + P(x) y = Q(x) \), where \( P(x) = \frac{1+x}{x} \) and \( Q(x) = \frac{e^{-x}}{x} \).

Step 3: Find the Integrating Factor

Calculate the integrating factor \( \mu(x) \) using the formula: \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{1+x}{x} \, dx} \]. This simplifies to \( \mu(x) = e^{x} x \).

Step 4: Multiply the Equation by the Integrating Factor

Multiply every term in the differential equation by \( \mu(x) = e^{x} x \): \[ e^{x} x y^{\prime} + e^{x} (1+x) y = 1 \].

Step 5: Integrate and Solve

Notice that the left side is the derivative of \( e^{x} x y \). Therefore, integrate both sides:\[ \int d(e^{x} x y) = \int 1 \, dx \].This gives \( e^{x} x y = x + C \).

Step 6: Apply the Initial Condition

Use the condition \( y(1) = 0 \):\[ e^{1} \cdot 1 \cdot 0 = 1 + C \rightarrow C = -1 \].

Step 7: Solve for y

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

Key Concepts

First-order Linear Differential Equation

The equation given in the exercise is a first-order linear differential equation. This type of equation involves two main components: the function we are trying to solve for (often written as \( y \)), and its first derivative \( y' \).
In a standard form, the equation appears as \( y' + P(x) y = Q(x) \). Here, \( P(x) \) and \( Q(x) \) are functions of the same variable \( x \).
These equations are called 'linear' because all terms involving \( y \) and its derivatives are of the first degree. Recognizing the form of a first-order linear differential equation is the first step in solving it because it dictates the method we'll use. Solving such equations often involves transforming them into this standard format, which we'll discuss further in the next sections.

Integrating Factor

The integrating factor is a powerful tool used to solve first-order linear differential equations. It is essentially a function that, when multiplied by the entire differential equation, simplifies the left-hand side to the derivative of a product of functions.
For our equation, the integrating factor \( \mu(x) \) is calculated using the formula \( \mu(x) = e^{\int P(x) \, dx} \). For the problem we are working on, \( P(x) = \frac{1+x}{x} \), leading to an integrating factor of \( \mu(x) = e^{x} x \).
By multiplying both sides of the equation with this integrating factor, the equation becomes easier to integrate, as the left-hand side can be expressed as a single derivative: \( \frac{d}{dx}(e^{x} x y) \). This simplification is crucial for finding the general solution of the differential equation.

Initial Condition

The initial condition is an additional piece of information often provided with differential equations. It specifies the value of the solution at a particular point, ensuring a unique solution.
In this problem, the initial condition given is \( y(1) = 0 \). This implies that when \( x = 1 \), the value of \( y \) must be zero.
Initial conditions are vital because they determine the constant of integration that emerges when solving differential equations. After integrating the differential equation using the integrating factor, we find a general solution with an arbitrary constant \( C \). By substituting the initial condition into the general solution, we can solve for \( C \), resulting in a specific solution that satisfies both the differential equation and the initial condition.

Solution of Differential Equations

Solving a differential equation involves determining a function that satisfies the given relationship between its values and derivatives. For first-order linear differential equations like the one in this problem, the process involves several steps.
After finding the integrating factor and multiplying it through the equation, we transform the left-hand side into a derivative form. By integrating both sides, we obtain an equation involving a constant of integration \( C \).
Using the initial condition given, we can identify the value of \( C \), which allows us to express the particular solution. For this problem, after applying \( y(1) = 0 \), we find \( C = -1 \), leading to the specific solution \( y = \frac{x - 1}{e^{x} x} \). This solution meets all the requirements of the original differential equation and the initial condition, providing a complete answer to the problem.