Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, solve each differential equation. $$ y^{\prime}+\frac{2 y}{x+1}=(x+1)^{3} $$

Step-by-Step Solution

Answer

The solution is \( y = \frac{(x+1)^4}{6} + \frac{C}{(x+1)^2} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( y^{\prime} + \frac{2y}{x+1} = (x+1)^{3} \). This is a first-order linear differential equation of the form \( y^{\prime} + P(x)y = Q(x) \) where \( P(x) = \frac{2}{x+1} \) and \( Q(x) = (x+1)^{3} \).

Step 2: Calculate Integrating Factor

The integrating factor \( \mu(x) \) for a differential equation \( y^{\prime} + P(x)y = Q(x)\) is given by \( \mu(x) = e^{\int P(x) \, dx} \). Here, \( \int P(x) \, dx = \int \frac{2}{x+1} \, dx = 2 \ln|x+1| \). Therefore, the integrating factor is \( \mu(x) = e^{2 \ln|x+1|} = (x+1)^2 \).

Step 3: Multiply Through by Integrating Factor

Multiply the entire differential equation by the integrating factor \((x+1)^2 \):\[ (x+1)^2 y^{\prime} + (x+1)^2 \frac{2y}{x+1} = (x+1)^2 \cdot (x+1)^3 \].This simplifies to:\[ (x+1)^2 y^{\prime} + 2(x+1)y = (x+1)^5 \].

Step 4: Rewrite the Left Side as a Derivative

The left side of the equation \( (x+1)^2 y^{\prime} + 2(x+1)y \) can be rewritten as a derivative:\[ \frac{d}{dx}((x+1)^2 y) = (x+1)^5 \].This is possible because the derivative of \((x+1)^2 y\) with respect to \(x\) involves the product rule, which matches the left side of the equation.

Step 5: Integrate Both Sides

Integrate both sides with respect to \(x\):\[ \int \frac{d}{dx}((x+1)^2 y) \, dx = \int (x+1)^5 \, dx \].This gives:\[ (x+1)^2 y = \frac{(x+1)^6}{6} + C \], where \(C\) is the constant of integration.

Step 6: Solve for y

Solve the equation \( (x+1)^2 y = \frac{(x+1)^6}{6} + C \) for \(y\):\[ y = \frac{(x+1)^6}{6(x+1)^2} + \frac{C}{(x+1)^2} \].This simplifies to:\[ y = \frac{(x+1)^4}{6} + \frac{C}{(x+1)^2} \].

Key Concepts

First-order Linear Differential Equation

A first-order linear differential equation is a type of differential equation that can be expressed in the general form \( y' + P(x)y = Q(x) \). Here, \( y' \) denotes the derivative of \( y \) with respect to a variable, often \( x \). This form is linear because it involves no powers or products of \( y \) or its derivatives beyond the first order.

In the problem at hand, the equation \( y^{\prime} + \frac{2y}{x+1} = (x+1)^{3} \) is recognized as a first-order linear differential equation by identifying \( P(x) = \frac{2}{x+1} \) and \( Q(x) = (x+1)^{3} \). These equations are crucial in modeling real-world phenomena where changes depend linearly on the current state.

The process of solving these equations often involves transforming or manipulating them using algebraic techniques and calculus. The identified coefficients \( P(x) \) and \( Q(x) \) guide us to apply specific methods for finding the function \( y \). Estimating a solution involves integrating functions with respect to the independent variable, a skill essential for many fields, including physics and engineering.

Integrating Factor Method

The Integrating Factor Method is a technique used to solve first-order linear differential equations. Its main goal is to render the problem's left side into a single derivative, thus simplifying integration. The method involves finding an integrating factor, a function, which when multiplied by the whole equation, achieves this simplification.

For the equation \( y^{\prime} + \frac{2y}{x+1} = (x+1)^{3} \), the integrating factor \( \mu(x) \) is calculated as \( \mu(x) = e^{\int P(x) \, dx} \). Here, \( P(x) = \frac{2}{x+1} \), and integrating it gives us \( 2 \ln|x+1| \). Hence, the integrating factor is \( (x+1)^2 \).

Employing the integrating factor transforms the differential equation into a format where the left side is the derivative of a product: \( \frac{d}{dx}((x+1)^2 y) = (x+1)^5 \). The beauty of this step is how it leverages the product rule, making the equation simpler to integrate and solve. This approach is critical for efficiently handling linear equations and underpins many mathematical models used in dynamic systems.

Product Rule in Calculus

The Product Rule is a fundamental concept in calculus, used to differentiate products of two functions. If you have two functions of \( x \), say \( u(x) \) and \( v(x) \), the product rule states that the derivative of their product \( u(x)v(x) \) is given by \( u'(x)v(x) + u(x)v'(x) \).

In the differential equation solution process, the product rule appears when rewriting the equation's left side after being multiplied by the integrating factor. For instance, after obtaining \( (x+1)^2 y^{\prime} + 2(x+1)y \), the use of the product rule allows this expression to be perceived as \( \frac{d}{dx}((x+1)^2 y) \).

This step is pivotal because it streamlines the process of integration. Once we identify an expression as a derivative, we can directly integrate it with respect to \( x \) to find \( y(x) \). The product rule thus not only aids in derivatives but also simplifies solving differential equations, highlighting its integral role in mathematical computations.