Question

Use the window \([-2,2] \times[-2,2]\) to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. A detailed direction field is not needed. $$y^{\prime}(x)=\sin x, y(-2)=2$$

Short answer

Question: Sketch a direction field for the first-order differential equation \(y^{\prime}(x) = \sin x\) on the interval \([-2, 2]\) and find the solution for \(y(x)\) that satisfies the initial condition \(y(-2) = 2\). Briefly describe the solution curve. Answer: The solution curve for the given differential equation with the initial condition \(y(-2) = 2\) is represented by the equation \(y(x) = -\cos x + 2 + \cos(2)\). The curve passes through the point \((-2, 2)\) and follows the direction field determined by the slopes of lines at different points \((x, y)\) calculated using \(y^{\prime}(x) = \sin x\).

Step-by-step solution

Plot the direction field for \(y^{\prime}(x) = \sin x\)

On the interval \([-2, 2]\), we'll calculate the slope of lines at different points \((x, y)\) using the equation \(y^{\prime}(x) = \sin x\). Since \(y^{\prime}(x)\) only depends on \(x\) and not \(y\), the direction field will have all the horizontal lines having the same slope at the points on the same vertical line (constant slopes).

Solve the differential equation

To solve the differential equation, we'll find the antiderivative of the function \(y^{\prime}(x)\). $$y^{\prime}(x)= \sin x $$ Integrating both sides, we get: $$y(x) = \int \sin x\, dx = -\cos x + C$$ Where \(C\) is the constant of integration.

Determine the particular solution using the initial condition

Now we have the general solution of the differential equation, \(y(x) = -\cos x + C\). We'll use the initial condition \(y(-2) = 2\) to find the value of \(C\): $$2 = -\cos(-2) + C$$ Solving for \(C\), we get: $$C = 2 + \cos(2)$$ Therefore, the particular solution is: $$y(x) = -\cos x + 2 + \cos(2)$$

Sketch the solution curve

Now we have the particular solution which represents the solution that passes through the initial condition \(y(-2) = 2\). We'll plug this equation into the direction field sketched in Step 1, and draw the curve that follows the slopes in the direction field and satisfies the initial condition: The curve will pass through the point \((-2, 2)\) and follow the direction field as it follows the equation $$y(x) = -\cos x + 2 + \cos(2)$$ By following these steps, you can visualize how the solution curve changes based on the direction field and initial condition.

Key concepts

Differential Equation

Understanding a differential equation is crucial when dealing with dynamic systems, where change is constant. A differential equation is a mathematical equation that relates some function with its derivatives. In simple terms, it describes how a particular quantity changes with respect to another, usually time. The equation \( y'(x) = \sin x \) from our exercise is a first-order differential equation, indicating that it involves the first derivative of the unknown function y(x) with respect to x.

The beauty of differential equations lies in their ability to model various phenomenon such as population growth, heat conduction, and motion of particles. Solving a differential equation means finding the function \( y(x) \) that satisfies the equation for all values within a certain domain.

Initial Condition

To uniquely determine the solution to a differential equation, we often need an initial condition. An initial condition is a value that the solution must satisfy at a certain point, typically when \( x = x_0 \). In the exercise, the initial condition is given as \( y(-2) = 2 \).

This constraint serves to anchor our solution, providing us with a starting point. Without it, there could be an infinite number of functions that satisfy the differential equation, each differing by a constant. But in the presence of an initial condition, there is generally a unique solution, known as the particular solution, which corresponds to this specific condition.

Antiderivative

The antiderivative, also known as the indefinite integral, is the reverse process of differentiation. If we have a function f(x), its antiderivative F(x) is a function whose derivative is f(x), denoted as \( F'(x) = f(x) \). The general solution to a simple differential equation such as the one in our exercise involves finding an antiderivative.

In the case of \( y'(x) = \sin x \), the antiderivative is \( y(x) = -\cos x + C \), where C represents the constant of integration. Each value of C provides a different function, and this family of functions is called the general solution to the differential equation.

Particular Solution

Upon integrating the general solution, we encounter the constant of integration, C, which represents an infinite set of possible functions. The particular solution is the one function within this set that passes through a given point, complying with the initial condition.

In our exercise, we found the particular solution by inserting the initial condition \( y(-2) = 2 \) into the general solution \( y(x) = -\cos x + C \). By solving for C, we narrowed down the infinite possibilities to the one specific solution that fits the given scenario. This is the solution that not only works everywhere in the equation but also hits the 'target' of the initial condition, making it the unique solution to our differential equation problem.