Question

Solve each differential equation. $$ y^{\prime}+\frac{2 y}{x+1}=(x+1)^{3} $$

Short answer

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

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is \( y' + \frac{2y}{x+1} = (x+1)^3 \). This is a first-order linear differential equation because it can be written in the form \( y' + P(x)y = Q(x) \), where \( P(x) = \frac{2}{x+1} \) and \( Q(x) = (x+1)^3 \).

Find the Integrating Factor

To solve the equation, we need to find the integrating factor \( \mu(x) \). The integrating factor is given by \( \mu(x) = e^{\int P(x) \, dx} \). Substitute \( P(x) = \frac{2}{x+1} \), the integral becomes \( \int \frac{2}{x+1} \, dx = 2\ln|x+1| = \ln|(x+1)^2| \). Thus, \( \mu(x) = e^{\ln|(x+1)^2|} = (x+1)^2 \).

Multiply the Equation by the Integrating Factor

Multiply every term of the original differential equation by the integrating factor \( (x+1)^2 \). This results in:\( (x+1)^2 y' + (x+1)^2 \cdot \frac{2y}{x+1} = (x+1)^5 \).Simplify to:\( (x+1)^2 y' + 2(x+1)y = (x+1)^5 \).

Rewrite as an Exact Differential

Notice that the left-hand side of the equation \( (x+1)^2 y' + 2(x+1)y \) is the derivative of \( (x+1)^2 y \).Thus, the equation becomes: \( \frac{d}{dx}[(x+1)^2 y] = (x+1)^5 \).

Integrate Both Sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}[(x+1)^2 y] \, dx = \int (x+1)^5 \, dx \].The left side simplifies to \((x+1)^2 y\) and the right side results in \( \frac{(x+1)^6}{6} + C \), with \( C \) as the constant of integration.

Solve for \( y \)

Now solve for \( y \):\[ (x+1)^2 y = \frac{(x+1)^6}{6} + C \].Divide both sides by \((x+1)^2\):\[ y = \frac{(x+1)^6}{6(x+1)^2} + \frac{C}{(x+1)^2} \].Simplify the expression:\[ y = \frac{(x+1)^4}{6} + \frac{C}{(x+1)^2} \].

Key concepts

first-order linear differential equations

First-order linear differential equations are equations that involve the first derivative of a function and can be expressed in a specific linear form. This form is typically represented as:\[ y' + P(x)y = Q(x) \]Here, \(y'\) is the derivative of \(y\) with respect to \(x\), \(P(x)\) and \(Q(x)\) are functions of \(x\) only. This type of equation is called "linear" because it does not involve powers or products of \(y\) or its derivative.Understanding and identifying the structure of first-order linear differential equations is crucial when solving them. If you can rewrite a given differential equation in this form, you are halfway to solving it! This structure ensures that certain methods, like using an integrating factor, can be applied effectively. When faced with a non-homogeneous first-order linear differential equation, the goal is to find either a general or particular solution for \(y\).

integrating factor

To solve first-order linear differential equations, we often employ the integrating factor technique. The integrating factor \(\mu(x)\) is a function that, when multiplied by the entire differential equation, transforms it into an easily integrable form.### How to Find the Integrating Factor- Identify \(P(x)\) from the standard form \(y' + P(x)y = Q(x)\).- Compute the integrating factor: \( \mu(x) = e^{\int P(x) \, dx} \).This step involves finding the integral of \(P(x)\) and then exponentiating the result. Once the integrating factor is determined, multiply it through the original equation. This process takes advantage of the product rule for differentiation, consolidating the left-hand side as a derivative, which then easily lends itself to integration.By applying the integrating factor, the differential equation can be rewritten such that it becomes straightforward to solve. This simplification allows you to find a general solution in terms of an integral that can be evaluated to find the specific solution needed. Remember, the magic of the integrating factor lies in its ability to transform a complex equation into something much simpler, making the process of integration seamless.

exact differential equations

Exact differential equations have a very distinct form where the entire equation can be derived from a single function's total differential. For an equation to be exact, there should exist a function \(F(x,y)\) such that:\[ \frac{\partial F}{\partial x} = M(x,y) \quad \text{and} \quad \frac{\partial F}{\partial y} = N(x,y) \]### Characteristics of Exact Differential Equations- The equation can be written as: \( M(x,y) \, dx + N(x,y) \, dy = 0 \).- For it to be exact, \(M_y = N_x\), where \(M_y\) is the partial derivative of \(M\) with respect to \(y\) and \(N_x\) is the partial derivative of \(N\) with respect to \(x\).In the specific context of the solved exercise, after multiplying the differential equation by the integrating factor, it becomes effectively exact. This means it can be reformulated to utilize the nice property where one side of the equation is the derivative of a function, making it directly integrable.Exact differential equations simplify taking derivatives to solve. They help in contextualizing the problem into finding a function whose derivative matches the form, providing a direct route to the solution. This approach works beautifully as it leverages natural calculus processes to solve differential statements elegantly.