Problem 1 - Advanced Engineering Mathematics

Question

\(\frac{d y}{d x}=\sin 5 x\)

Step-by-Step Solution

Answer

The solution is \( y = -\frac{1}{5} \cos(5x) + C \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{d y}{d x} = \sin(5x) \). It's a first-order differential equation that can be solved using integration.

Step 2: Integrate the Right-Hand Side

To solve \( \frac{d y}{d x} = \sin(5x) \), we need to integrate \( \sin(5x) \) with respect to \( x \). Set up the integral: \[ y = \int \sin(5x) \, dx \]

Step 3: Use Substitution for Integration

Let \( u = 5x \), then \( du = 5 \, dx \) or \( dx = \frac{du}{5} \). This changes the integral to: \[ y = \int \sin(u) \frac{du}{5} = \frac{1}{5} \int \sin(u) \, du \]

Step 4: Solve the Integral

The integral of \( \sin(u) \, du \) is \(-\cos(u)\). Therefore,\[ \frac{1}{5} \int \sin(u) \, du = -\frac{1}{5} \cos(u) + C \] Substituting back for \( u = 5x \), we get:\[-\frac{1}{5} \cos(5x) + C \]

Step 5: Write the Solution

Thus, the solution to the differential equation \( \frac{d y}{d x} = \sin(5x) \) is:\[ y = -\frac{1}{5} \cos(5x) + C \]

Key Concepts

Understanding Integration

Integration is a fundamental concept in calculus used to find functions that describe accumulated quantities, like area under a curve. When dealing with differential equations, integration helps us find the original function given its rate of change. In the context of solving the given first-order differential equation \( \frac{d y}{d x} = \sin(5x) \), our goal is to reverse the differentiation process by integrating the sine function.

The integral of a function represents the collection of all possible solutions and includes a constant \( C \), which accounts for the indefinite nature of anti-differentiation. In our exercise, integrating \( \sin(5x) \) gives us the solution:

• \( y = \int \sin(5x) \, dx = -\frac{1}{5} \cos(5x) + C \).

This result shows us the function whose derivative is \( \sin(5x) \). Understanding integration is crucial as it equips us with the tools to solve complex differential equations by essentially "undoing" differentiation.

The Substitution Method: Simplifying Integration

The substitution method is a powerful technique in integration, especially useful for functions that are composed or where a direct approach seems complex. In our differential equation, we faced the integral \( \int \sin(5x) \, dx \), which required simplification through substitution.

Here's a step-by-step breakdown of how substitution works in this context:

• First, we set \( u = 5x \), transforming the variable inside the sine function and simplifying our task.
• Next, calculate \( du = 5 \, dx \), which leads to \( dx = \frac{du}{5} \).
• The integral becomes \( \int \sin(u) \frac{du}{5} = \frac{1}{5} \int \sin(u) \, du \), a much simpler form.

By substituting, we made the integration process more manageable and unfolded a path towards the correct solution. Once integration is complete, we substitute back \( u = 5x \) to revert to the original variable. This approach is essential in tackling integrals that initially appear challenging, greatly increasing efficiency and accuracy in finding solutions.

Exploring Trigonometric Functions

Trigonometric functions like sine and cosine are periodic and appear often in various mathematical contexts, including solving differential equations.

In this exercise, understanding their properties is key for integration and substitution. For the equation \( \frac{d y}{d x} = \sin(5x) \), we deal directly with the sine function, whose integral results in a cosine function, specifically:

• \( \int \sin(u) \, du = -\cos(u) + C \).

This key relationship allows us to solve the differential equation by connecting the rate of change with the function generating that rate.

Understanding these fundamental trigonometric functions involves knowing their derivatives and antiderivatives, as well as their behavior in rotational and periodic movements. By mastering these functions, we can efficiently approach not only simple integrals but also more complex differential equations, enhancing our problem-solving toolkit significantly.