Problem 31 - Calculus Concepts

Question

Use the figures to sketch the particular solution for each given initial condition. \(\frac{d y}{d x}=\cos x\) a. \(x=0, y=0\) b. \(x=2, y=5\) c. \(x=-1, y=2\)

Step-by-Step Solution

Answer

Solve the integral, use initial conditions to find \(C\), and sketch \(y = \sin x + C\).

Step 1: Understand the Differential Equation

The differential equation given is \(\frac{dy}{dx} = \cos x\). This means the rate of change of \(y\) with respect to \(x\) is the cosine of \(x\).

Step 2: Integrate the Differential Equation

To find \(y\), integrate both sides of the equation with respect to \(x\). The integral of \(\cos x\) is \(\sin x + C\), where \(C\) is the constant of integration. Therefore, \(y = \sin x + C\).

Step 3a: Solve for Particular Solution (Condition a)

Apply the initial condition \((x=0, y=0)\). Substitute these values into the equation: \[ 0 = \sin(0) + C \] Since \(\sin(0) = 0\), we have:\[ 0 = 0 + C \]Thus, \(C = 0\). The particular solution is \(y = \sin x\) for condition a.

Step 3b: Solve for Particular Solution (Condition b)

Apply the initial condition \((x=2, y=5)\). Substitute these values into the equation: \[ 5 = \sin(2) + C \] Thus, \[ C = 5 - \sin(2) \].The particular solution is \(y = \sin x + (5 - \sin 2)\) for condition b.

Step 3c: Solve for Particular Solution (Condition c)

Apply the initial condition \((x=-1, y=2)\). Substitute these values into the equation: \[ 2 = \sin(-1) + C \] Thus, \[ C = 2 - \sin(-1) \].The particular solution is \(y = \sin x + (2 - \sin(-1))\) for condition c.

Step 4: Sketch the Solutions (Overview)

Each particular solution can be sketched based on its general form, \(y = \sin x + C\), where the value of \(C\) shifts the graph of \(\sin x\) vertically. Condition a has no vertical shift, condition b is shifted up by \(5 - \sin 2\), and condition c is shifted up by \(2 - \sin(-1)\).

Key Concepts

Initial Conditions

Initial conditions are vital when dealing with differential equations, as they help us determine specific solutions, known as particular solutions, to these equations. When we have a differential equation, it typically involves an unknown function and its derivatives. In order to find a unique and precise function that satisfies the differential equation, initial conditions are used. These are values given at certain points that provide specific insight into the solution. For example, in our exercise, we have three different initial conditions:

• For condition (a), when \(x = 0\) and \(y = 0\), the function passes through the point \((0,0)\).
• In condition (b), where \(x = 2\) and \(y = 5\), the function is tailored to pass through \((2,5)\).
• And for condition (c), with \(x = -1\) and \(y = 2\), the function intersects the point \((-1,2)\).

These conditions assist us in pinpointing the constant of integration, \(C\), to yield a particular solution unique to each scenario.

Integration

Integration is a fundamental operation in calculus that is used to solve differential equations. This process involves finding a function given its derivative. In the context of our exercise, integration allows us to determine the function \(y\) from the given derivative \(\frac{dy}{dx} = \cos x\). To perform the integration, we take the integral of both sides with respect to \(x\). The integral of \(\cos x\) is \(\sin x\), leading to the expression \(y = \sin x + C\), where \(C\) is the constant of integration. Integration is crucial for transforming a differential equation into a solvable equation, as it gives us the general form of the solution. The constant \(C\) forms the bridge between the general solution and the particular solutions we find using the initial conditions.

Particular Solutions

Particular solutions are derived from the general solution of a differential equation by applying initial conditions. The general solution contains an arbitrary constant \(C\), which must be determined for each specific initial condition to find the particular solution. Here's how we find the particular solutions using our initial conditions:

• For condition (a): With \(x = 0\) and \(y = 0\), we substitute into \(y = \sin x + C\) to solve for \(C\), leading to no vertical shift with \(C = 0\), giving \(y = \sin x\).
• For condition (b): With \(x = 2\) and \(y = 5\), substituting these into the general solution gives \(C = 5 - \sin(2)\), resulting in \(y = \sin x + (5 - \sin 2)\).
• For condition (c): With \(x = -1\) and \(y = 2\), substituting gives \(C = 2 - \sin(-1)\), resulting in \(y = \sin x + (2 - \sin(-1))\).

Each particular solution describes a distinct curve tailored to pass through the points specified by the initial conditions.

Graph Sketching

Graph sketching is an important skill because it visually represents the behavior of solutions to differential equations. Through our exercise, we depicted different particular solutions, each stemming from a general form \(y = \sin x + C\). Graph sketching turns abstract solutions into visual aids, enhancing understanding.To sketch these:

• Condition (a): Since \(C = 0\), the graph is the standard sine wave, \(y = \sin x\), with no vertical shift.
• Condition (b): The graph shifts vertically upwards by \(5 - \sin(2)\), creating a new sine wave, transformed accordingly.
• Condition (c): The sine wave again shifts vertically by \(2 - \sin(-1)\), portraying a distinct sine curve based on the given point.

When sketching these graphs, note the changes in amplitude and position due to the different initial conditions, showcasing how solutions logically spread throughout the plane.