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 the direction field for the given equation \(y^{\prime}(x) = \sin{x}\) on the window \([-2,2] \times [-2,2]\), and sketch the solution curve corresponding to the initial condition \(y(-2) = 2\). Answer: The direction field can be roughly sketched by following the sign of sine values and plotting arrows accordingly within the window. The solution curve for the initial condition \(y(-2)=2\) is \(y(x) = -\cos{x} + (2+\cos{(-2)})\), and can be sketched on the same window as well.

Step-by-step solution

Sketch the direction field

To sketch the direction field, first, we plot arrows to represent the derivative value at different points in the window provided. Here, the arrows will indicate how the solution curve changes with respect to x at different values of x and y. 1. For \(y^{\prime}(x) = \sin x\), we find that we only need to focus on the values of x as the equation does not require the values of y for the calculation. 2. Sketch the arrows along the x-axis on the window according to the values of \(\sin x\). Remember that \(\sin x\) oscillates between -1 and 1. 3. On the interval \([-2,2]\), we just need to indicate the direction of the arrows on the window according to the sign of the \(\sin x\) values.

Find the solution curve for the initial condition

To find the solution curve corresponding to the initial condition \(y(-2) = 2\), we need to integrate the given equation and plot the curve. 1. Integrate the equation \(y^{\prime}(x) = \sin x\) with respect to x to obtain the general solution for y: \(y(x) = -\cos{x} + C\) 2. Use the initial condition \(y(-2) = 2\) to find the value of the constant C: \(2 = -\cos{(-2)} + C\). This gives us the value of C: \(C = 2 + \cos{(-2)}\) 3. The solution curve for the given initial condition becomes: \(y(x) = -\cos{x} + (2 + \cos{(-2)})\) 4. Sketch the curve on the window \([-2,2] \times [-2, 2]\). After completing these steps, you should have a rough sketch of the direction field and the solution curve for the given initial condition.

Key concepts

Differential Equations

Differential equations are equations that relate a function to its derivatives. They provide a way to describe phenomena where there's a relationship between varying quantities. For example, the rate at which a population grows might be proportional to the current population size, making a differential equation a suitable model.
In this particular exercise, we have a differential equation given by \(y'(x) = \sin x\). This equation denotes the slope of the function \(y(x)\) at any point \(x\). To solve a differential equation, one often needs to integrate the equation to find the function \(y(x)\) itself, possibly considering any given initial conditions.

Solution Curves

Solution curves, often referred to as integral curves, are graphical representations of the solutions to differential equations. These curves indicate how the solution behaves over a range of values. In the context of our exercise, the solution curve is determined by integrating the differential equation \(y'(x) = \sin x\) to get \(y(x) = -\cos x + C\).
The direction field helps visualize multiple possible solution curves depending on different initial conditions. The arrows in the direction field show the slope of the solution curve at various points, providing insight into the behavior of the curve before it's fully sketched. By considering the initial condition \(y(-2) = 2\), we trace the particular solution curve fitting through this condition.

Initial Conditions

Initial conditions are critical in determining the specific solution to a differential equation. They provide specific information to pinpoint which curve out of the family of solutions is the correct one for a given problem. In exercises like the one provided, an initial condition such as \(y(-2) = 2\) tells us the exact position the solution curve must pass through.
This initial condition allows us to solve for the constant \(C\) in the integrated solution \(y(x) = -\cos x + C\). By plugging \(-2\) into the equation for \(x\) and \(2\) for \(y(x)\), we solve for \(C\). This grants us a precise solution: \(y(x) = -\cos x + (2 + \cos(-2))\), ensuring the curve passes through \((-2, 2)\) on the plot.

Integration in Calculus

Integration is a fundamental concept in calculus used to find antiderivatives or area under curves. When dealing with differential equations, integration is typically used to find a function from its derivative.
In our exercise, integrating \(y'(x) = \sin x\) gives us \(y(x) = \int \sin x \, dx = -\cos x + C\), where \(C\) is the constant of integration. This antiderivative represents a family of functions which are possible solutions to the original differential equation. The initial condition is used to determine the precise value for \(C\), isolating the specific solution curve needed for the exercise. Integration, therefore, transitions our focus from the rate of change (or slope) to constructing the full function that models the situation described by the differential equation.