Question

Use the method of isoclines to sketch the approximate integral curves of each of the differential equations. \(y^{\prime}=\frac{\sin x}{y}\).

Short answer

Rewriting the given differential equation as \(y \cdot y' = \sin x\), we find the isoclines by substituting \(y'\) with constant slope \(k\): \(y = \frac{\sin x}{k}\). We then draw the isoclines for different values of \(k\) (excluding \(k=0\)), and use them to sketch the approximate integral curves. This provides a visual representation of the family of solutions to the differential equation \(y' = \frac{\sin x}{y}\).

Step-by-step solution

Rewrite the Differential Equation

Given the differential equation, \(y' = \frac{\sin x}{y}\), we can rewrite it as: \[y \cdot y' = \sin x\] #Step 2: Find the Isoclines#

Find the Isoclines

An isocline is the set of all points on the coordinate plane where the slope of the tangent to the integral curve is constant. To find the isoclines for this problem, we will substitute \(y'\) in the equation with a constant slope \(k\): \[y \cdot k = \sin x\] Now, we can solve for \(y\) to get the isoclines equation: \[y = \frac{\sin x}{k}\] #Step 3: Draw the Isoclines on the Coordinate Plane#

Draw the Isoclines

Choose some different values for the constant slope \(k\), such as \(k = -2, -1, 0, 1, 2\), and compute the corresponding equations for the isoclines. Then, draw the isoclines on the coordinate plane: 1. For \(k = -2\), we have \[y = \frac{\sin x}{-2}\]. 2. For \(k = -1\), we have \[y = \frac{\sin x}{-1}\]. 3. For \(k = 0\), as sin(x) is between -1 and 1 (Inclusive). It will never become 0, thus there will not be any isocline for k=0. 4. For \(k = 1\), we have \[y = \frac{\sin x}{1}\]. 5. For \(k = 2\), we have \[y = \frac{\sin x}{2}\]. #Step 4: Sketch the Approximate Integral Curves#

Sketch the Integral Curves

Using the isoclines as guides, sketch the approximate integral curves on the coordinate plane. Remember that the integral curves must be tangent to the isoclines at every point, so they will follow the slope given by the value of \(k\). This will give you a visual representation of the family of solutions to the given differential equation. By following these steps, we have successfully used the method of isoclines to sketch the approximate integral curves of the given differential equation: \(y' = \frac{\sin x}{y}\).

Key concepts

Integral Curves

The concept of an 'integral curve' is one of the key notions when dealing with differential equations. Integral curves are simply the graphs of solutions to differential equations. In the exercise above, we are looking at the differential equation \(y' = \frac{\sin x}{y}\), where the goal is to sketch the integral curves using the method of isoclines.

An integral curve represents a real-world process or system's behavior modeled by the differential equation; for example, it could depict the growth of a population over time or the cooling of an object. To sketch them, we use information about the slopes of the tangents to these curves at various points, which is given by the differential equation itself. The key is to remember that at every point on an integral curve, the slope of the tangent to the curve must match the value given by the differential equation at that point.

Ordinary Differential Equations

Next, let's explore 'ordinary differential equations' (ODEs), which are equations involving derivatives of a function and are vital in expressing the rate of change of phenomena. The given exercise considers an ODE \(y' = \frac{\sin x}{y}\). ODEs are called 'ordinary' because they involve derivatives with respect to only one independent variable, as opposed to 'partial' differential equations, which involve derivatives with respect to multiple variables.

In many scientific fields, these equations help describe natural phenomena such as motion, growth, decay, and other time-dependent processes. The solution to an ODE is a function or a set of functions that satisfy the equation, and often this solution can't be expressed in a simple formula, hence the need for methods like isoclines to visualize the behaviors of solutions.

Slope Fields

Another visualization tool we use in our exercise is the 'slope field,' also known as a direction field. It's a graphical representation of the slopes of tangents of integral curves of a differential equation at a span of points in the plane. For the ODE \(y' = \frac{\sin x}{y}\), a slope field would consist of small line segments (or arrows) that represent the slope \(y'\) at various \(x\) and \(y\) coordinates.

The slope field helps us to see the overall pattern and direction trends of integral curves without actually solving the equation. It's like sketching out a path in a field of grass by indicating the direction you need to go at every point. The method of isoclines can be considered a part of creating a slope field, where we focus on lines that represent constant slopes.

Differential Equation Solution

Finally, let's address the 'differential equation solution.' A solution to a differential equation is a function that satisfies the equation when substituted into it. For our example, the solution would be a function \(y(x)\) such that when \(y'\) is computed as \(\frac{dy}{dx}\), the original equation \(y' = \frac{\sin x}{y}\) holds true for all \(x\) in the domain of \(y\).

However, it's often complicated, sometimes impossible, to find an exact analytic form for this solution, which is why methods like sketching integral curves using isoclines are so valuable. They offer us a way to understand the behavior of the solutions without requiring an exact formula, providing insight into the nature of the system described by the ODE. By understanding isoclines, slope fields, and the relationship between solutions and ODEs, students can gain a deeper appreciation for the power of differential equations in modeling and problem-solving in many fields of science and engineering.