Question

Finding a General Solution In Exercises \(41-52,\) use integration to find a general solution of the differential equation. $$ \frac{d y}{d x}=\sin 2 x $$

Short answer

The general solution of the given differential equation is \( y = -\frac{1}{2} \cos 2x + C \).

Step-by-step solution

Write down the differential equation

For the given differential equation, we have \(\frac{d y}{d x} = \sin 2x\).

Integrate both sides of the equation

Given the differential equation \(\frac{d y}{d x}=\sin 2x\), perform integration on both sides. When integrate the left side, we simply get \(y\). To integrate the right side, remember that the integral of \( \sin ax\) is \(-\frac{1}{a} \cos ax + C\). Following this rule, we get: \[ y= -\frac{1}{2} \cos 2x + C \], where \( C \) represents the integration constant.

Key concepts

Integration Techniques

Integration is a fundamental concept in calculus and is used to solve differential equations by finding antiderivatives. For the equation \( \frac{d y}{d x} = \sin 2x \), integration allows us to find a function, or a family of functions, that relates \( y \) to \( x \).
Integration on both sides of a differential equation must be performed accurately to ensure the solution is correct. On the left side, integrating \( \frac{d y}{d x} \) with respect to \( x \) simply gives us \( y \). This is because differentiation and integration are inverse operations.
On the right side, we need to integrate \( \sin 2x \). Here, we apply the substitution rule, recognizing the integral of \( \sin ax \) as \(-\frac{1}{a} \cos ax + C \). For \( \sin 2x \), \( a = 2 \), and thus the integral becomes \( -\frac{1}{2} \cos 2x + C \).

• Always remember to add a constant \( C \) when completing indefinite integrals.
• This constant represents an infinite number of antiderivatives, reflecting the different possible vertical positions of the cosine curve along the y-axis.

Mastering these integration techniques is crucial for solving first-order differential equations.

General Solution

The general solution of a differential equation is a family of solutions that includes an arbitrary constant, usually represented by \( C \). It encompasses all possible specific solutions of the equation.
In our exercise, after integrating \( \frac{d y}{d x} = \sin 2x \), we reach the general solution \( y = -\frac{1}{2} \cos 2x + C \).

This solution indicates:

• The function \( -\frac{1}{2} \cos 2x \) is the core part of the solution representing one particular antiderivative of \( \sin 2x \).
• The constant \( C \) accounts for any vertical shift, showing that the solution can adjust for initial conditions or points specific to different problems.

In practice, to find a particular solution, additional conditions (like \( y(x_0) = y_0 \), known as initial conditions) are applied to solve for \( C \). Thus, the general solution provides the framework from which specific solutions are tailored.

Constant of Integration

The constant of integration, \( C \), is an integral part of finding general solutions in calculus. It appears when performing indefinite integrals, reflecting any constant difference between antiderivatives of a function.
Why is it important? Because when you differentiate a function, any constant term becomes zero. Therefore, during integration, every possible constant value is represented by \( C \).

In our exercise with \( \frac{d y}{d x} = \sin 2x \), when we integrate to achieve \( y = -\frac{1}{2} \cos 2x + C \), \( C \) represents all values that can shift the cosine wave vertically. Without \( C \):

• We would only have one specific antiderivative for \( \sin 2x \), potentially missing other solutions satisfying the differential equation under various conditions.
• In application, \( C \) allows solutions to encompass initial conditions and specific requirements.

Understanding the nature of \( C \) is vital for working with differential equations, as it ensures all possible solutions are accounted for and applicable to various real-world scenarios.