Question

Solving an Initial-Value Problem Using the method of separation of variables, solve the initial-value problem $$ y^{\prime}=(2 x+3)\left(y^{2}-4\right), \quad y(0)=-3 $$

Short answer

The implicit solution is \( \frac{1}{2} \ln|\frac{y-2}{y+2}| = x^2 + 3x + \frac{1}{2} \ln 5 \).

Step-by-step solution

Separate Variables

To separate variables, rewrite the equation \( y^{\prime} = (2x + 3)(y^2 - 4) \) as \( \frac{dy}{dx} = (2x + 3)(y^2 - 4) \). Move all terms involving \( y \) to one side and terms involving \( x \) to the other side: \( \frac{dy}{y^2 - 4} = (2x + 3)dx \).

Integrate Both Sides

Integrate both sides of the equation. The left side with respect to \( y \) and the right side with respect to \( x \):\( \int \frac{dy}{y^2 - 4} = \int (2x + 3) dx \). The integrals are \( \frac{1}{2} \ln|\frac{y-2}{y+2}| = x^2 + 3x + C \) where \( C \) is the integration constant.

Apply Initial Condition

Substitute \( y(0) = -3 \) into the equation to solve for \( C \): \( \frac{1}{2} \ln|\frac{-3-2}{-3+2}| = 0^2 + 3(0) + C \). This simplifies to \( \frac{1}{2} \ln|5| = C \). Therefore, \( C = \frac{1}{2} \ln 5 \).

Write the General Solution

Incorporate \( C \) back into the solution: \( \frac{1}{2} \ln|\frac{y-2}{y+2}| = x^2 + 3x + \frac{1}{2} \ln 5 \). Solve for \( y \) if necessary, the implicit solution satisfies the differential equation given the initial condition.

Key concepts

Initial-Value Problem

An initial-value problem in mathematics involves finding a function that satisfies a differential equation and meets a specific condition at a given point, known as the initial condition. The initial condition provides the values of the function and possibly its derivatives at a particular point. In our example, we begin with the equation \( y' = (2x + 3)(y^2 - 4) \) and the initial condition \( y(0) = -3 \). These initial conditions help us determine the particular solution from the family of possible solutions we obtain after solving the differential equation.

In essence, the goal is to find a function \( y(x) \) that not only makes the differential equation true for all \( x \) but also passes through the point given by the initial condition.

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They serve as a powerful tool for describing various physical and mathematical phenomena where rates of change are involved. For our initial-value problem, the differential equation is \( y' = (2x + 3)(y^2 - 4) \). Here, \( y' \) denotes the derivative of \( y \) with respect to \( x \).

The aim in solving differential equations is often to find an explicit expression for \( y \) in terms of \( x \), although implicit solutions can sometimes suffice, as demonstrated in our solution process.

• Separation of variables is one technique used here, which involves rearranging the equation so that each variable and its derivative are on opposite sides of the equation.
• This method makes it easier to integrate and find the solution.

Integration

Integration is the process of finding the antiderivative or the reverse of differentiation. When we separate the variables in our differential equation, we end up integrating both sides to obtain the general solution. In the previous problem, this step appears as \( \int \frac{dy}{y^2 - 4} = \int (2x + 3) dx \).

To integrate, we transform the left side using partial fraction decomposition to simplify the integral, and the right side is straightforward polynomial integration.

• The solution gives us for the left side: \( \frac{1}{2} \ln|\frac{y-2}{y+2}| \).
• On the right: \( x^2 + 3x + C \), where \( C \) represents the constant of integration that arises from indefinite integration.

The role of integration here is to undo the differentiation present in the original differential equation.

Calculus

Calculus, the mathematical study of continuous change, encompasses both differentiation and integration, core concepts applied in solving our initial-value problem. The differential equation \( y' = (2x + 3)(y^2 - 4) \) lies in the realm of calculus as we deal with derivatives to describe the function \( y \).

By separating variables and integrating, we employ fundamental techniques of calculus to find the solution to the problem. - In this context, calculus enables a transition from a dynamic description (differential equation) to a static description (the integral) of the mathematical problem. - Calculus also provides the framework to analyze and interpret solutions, such as identifying specific solutions from families of curves through initial conditions.

Ultimately, calculus serves as the foundation for solving differential equations and related problems, highlighting its essential place in the mathematical toolkit.