Question

Find the general solution to each differential equation. $$(x+1) y^{\prime}=2(x+y+1)$$

Short answer

The general solution of the differential equation is \( y = C_1(x+1)^2 \), where \( C_1 \) is any real number.

Step-by-step solution

Isolate the Derivative

To find the general solution, start by isolating the derivative term on one side of the equation. Divide both sides by x+1 to get the derivative alone: \( y' = \frac{2(x+y+1)}{x+1} \).

Simplify the Equation

Simplify the equation by dividing 2 into each term within the brackets on the right-hand side: \( y' = 2+\frac{2y}{x+1} \).

Separate the Variables

Rearrange the terms to separate the variables (y and x). Subtract 2 from both sides: \( y' - 2 = \frac{2y}{x+1} \). Now we can write this as \( \frac{1}{y} dy = \frac{2}{x+1} dx \), which gives us an equation ready for integration.

Integrate Both Sides

Integrate both sides of the equation to find the function y(x). On the left side, integrate with respect to y: \( \int \frac{1}{y} dy \) and on the right side, integrate with respect to x: \( \int \frac{2}{x+1} dx \). The integrals are \( \ln|y| = 2 \ln|x+1| + C \), where C is the constant of integration.

Solve for y

Exponentiate both sides to get rid of the natural logarithm: \( e^{\ln|y|} = e^{2 \ln|x+1| + C} \), which simplifies to \( |y| = e^{C}(x+1)^2 \).

Include the Constant of Integration

Write the general solution including the constant of integration. Since \( e^C \) is still a constant, we can rename it to a new constant, say \( C_1 \). The absolute value can be dropped by considering both positive and negative values of \( C_1 \). Therefore, the general solution is \( y = C_1(x+1)^2 \), where \( C_1 \) is any real number.

Key concepts

Differential Equation

At its core, a differential equation is a mathematical equation that relates a function to its derivatives. Understanding the behavior of these derivatives is critical because they can express rates at which things change. In the given exercise, the equation \[ (x+1) y^{\text{\prime}}=2(x+y+1) \]involves the derivative of the function y with respect to x, denoted as y'. This represents a first-order differential equation because it contains only the first derivative of y. The goal is to find a general solution, which is a function y(x) that satisfies the equation for all values of x within a certain domain.

Variable Separation

Variable separation is a straightforward method for solving some differential equations. It involves rearranging the equation to isolate the variables x and y on opposite sides of the equation. This method works well for equations where such separation is possible. In the exercise, after simplifying the equation and subtracting 2 from both sides we reach:\[ y' - 2 = \frac{2y}{x+1}, \]which can be rephrased to\[ \frac{1}{y} dy = \frac{2}{x+1} dx. \]By doing this, we're setting the stage to integrate each side with respect to its own variable. The separated variables pave the way for us to use integration to find the general solution.

Integrating Factors

Integrating factors are a tool we use to solve differential equations that are not readily separable. An integrating factor is a function that we can multiply through an equation to make it easier to solve. While our current exercise does not require the use of integrating factors due to the success of variable separation, it is important to know that integrating factors are particularly useful when dealing with linear first-order differential equations. They transform the equation into a form where the left side becomes an exact differential, making it possible to integrate both sides directly.

Integration of Functions

Integration is an essential operation in calculus, often regarded as the reverse process of differentiation. When we integrate, we're essentially summing up small pieces to find the whole, which can be useful in finding areas under curves or solving differential equations, as in our exercise. Integral calculations come in handy after successfully separating variables.In the exercise, after separating the variables, we performed the integrations\[ \int \frac{1}{y} dy \]and\[ \int \frac{2}{x+1} dx. \]These integrals evaluate to \( \ln|y| \) and \( 2 \ln|x+1| \), respectively, plus a constant of integration, which eventually lead us to the general solution of the differential equation.