Problem 30 - Essential Calculus

Question

Slope Fields, a differential equation, a point, and a slope field are given. (a) Sketch two approximate solutions of the differential equation on the slope field, one of which passes through the given point. (b) Use integration to find the particular solution of the differential equation and use a graphing utility to graph the solution. Compare the result with the sketches in part (a). To print an enlarged copy of the graph, go to the website www.mathgraphs.com. $$ \frac{d y}{d x}=x \sqrt{y} \cos x,(0,4) $$

Step-by-Step Solution

Answer

This problem focuses on finding the solution to a specific differential equation and plotting its graphical representation. It requires understanding of calculus concepts including slope fields and integration for the separation of variables.

Step 1: Understand and Analyze the Differential Equation

The given differential equation is \(\frac{dy}{dx} = x\sqrt{y}\cos{x}\). This relates the slope of the function \(y\) at any point \(x\) to \(x\sqrt{y}\cos{x}\).

Step 2: Sketch the Slope Field and Appropriate Solutions

Sketch a slope field for the differential equation. The slope field gives a visual representation of the possible slopes the function must have at every point on the plane. You'll allow curves to follow these slope lines to find the curves that pass through the point (0,4). Draw two curve solutions into this slope field plot.

Step 3: Integrate the Differential Equation

In order to find a particular solution for the differential equation, use separation of variables and integrate. Rearrange the equation to be in the form \(\frac{dy}{\sqrt{y}}=x\cos{x} dx\), then integrate both sides of the equation. Implement the initial conditions (0,4) to find the constant of integration.

Step 4: Graph the Resulting Solution and Compare

After having integrated the equation and found the constant of integration, plot the resulting function. Compare how this curve matches up with the sketched curves from the slope field in Step 2. Usually, there should be a good congruence between these methods.

Key Concepts

Slope Fields

Slope fields are a visual tool that helps us understand the behavior of differential equations. They provide a diagram that shows the slope of the solution curve at various points in the xy-plane, acting like a road map for the expected behavior of solutions to the differential equation.

In the case of our exercise with the equation \(\frac{dy}{dx} = x\sqrt{y}\cos{x}\), slope fields allow us to draw small line segments that represent the slope, or derivative, of the solution at those points.

• This means that at each point \((x, y)\), the slope of the tangent line to the solution curve is equal to \(x\sqrt{y}\cos{x}\).
• The slope field helps visualize how the solution curves should flow through any given point.
• One key point in the example is \((0,4)\), which guides us in sketching the initial part of the solution as it passes through this point.

By drawing the slope field, you can literally see the behavior of the differential equation solutions and make educated guesses about what complete solution curves might look like.

Separation of Variables

Separation of variables is a technique used to solve some types of differential equations by separating the variables on opposite sides of the equation. Once separated, each side of the equation can be integrated to find a solution.

For our differential equation \(\frac{dy}{dx} = x\sqrt{y}\cos{x}\), we use separation of variables to reorganize it into:

• \(\frac{dy}{\sqrt{y}} = x\cos{x}\,dx\)

This makes it easier to integrate each side independently.

Let’s breakdown these steps:

• First, isolate the terms involving \(y\) on one side and \(x\) on the other side. This requires algebraic manipulation.
• Next, integrate both sides separately:
• The integration of \(\frac{1}{\sqrt{y}}\,dy\) results in a function of \(y\).
• The integration of \(x\cos{x}\,dx\) results in a function of \(x\).
• Both integrals contribute to the overall solution, which you will then adjust with a constant to fit any initial conditions.

This method highlights how complex differential equations can often be broken down into simpler calculus problems.

Initial Conditions Integration

Initial conditions are often provided in differential equations to find a unique particular solution to the problem. They help us determine the constant of integration, creating a solution relevant to a specific real-world scenario.

In our example, the initial condition is given as \((0, 4)\). This means that when \(x = 0\), the value of \(y\) should be 4.

• After integrating, this initial condition helps us solve for the constant of integration \(C\).
• Once we substitute \((0,4)\) into the integrated function, we find \(C\) such that the function passes through the given point.
• This ensures that the solution accurately reflects this specific scenario dictated by the initial condition.

Successfully applying these initial conditions provides a particular solution curve that adheres to the slope field and aligns with initial real-world constraints.