Question

Show that each function is a solution to the given differential equation. $$y^{\prime} \cot x+3+y=0, y=C \cos x-3$$

Short answer

After differentiating and substituting the function \(y = C \times \text{cos}(x) - 3\) into the differential equation, simplifying it yields 0, which confirms that the function is indeed a solution.

Step-by-step solution

Write down the original function

Identify the function given in the problem, which is expressed as: \(y = C \times \text{cos}(x) - 3\).

Differentiate the original function

Calculate the derivative of the given function \(y = C \times \text{cos}(x) - 3\). The derivative of \(\text{cos}(x)\) is \(-\text{sin}(x)\), thus \(y' = -C \times \text{sin}(x)\).

Substitute into the differential equation

Substitute \(y'\) and \(y\) into the given differential equation \(y' \times \text{cot}(x) + 3 + y = 0\). This gives us \((-C \times \text{sin}(x)) \times \text{cot}(x) + 3 + (C \times \text{cos}(x) - 3)\).

Simplify the equation

Simplify the substituted equation by using the trigonometric identity \(\text{sin}(x) \times \text{cot}(x) = \text{cos}(x)\). We get \(-C \times \text{cos}(x) + 3 + C \times \text{cos}(x) - 3\).

Check if the simplified equation satisfies the differential equation

The terms \(-C \times \text{cos}(x)\) and \(C \times \text{cos}(x)\) cancel each other out, and the constants 3 and -3 also cancel each other. Thus, the simplified expression becomes 0. This shows the function \(y = C \times \text{cos}(x) - 3\) satisfies the given differential equation \(y' \times \text{cot}(x) + 3 + y = 0\).

Key concepts

Solving Differential Equations

Understanding how to solve differential equations is crucial for students diving into advanced mathematics and scientific applications. These equations describe relationships between functions and their derivatives, expressing how a quantity changes. In the context of our exercise, we're asked to confirm a function is a solution to a given differential equation.

The method we use here is direct substitution. This approach involves taking the presumed solution, differentiating it, and then substituting the function and its derivative back into the original differential equation. If, after simplifying, we get a true statement (usually zero equals zero), then the function is indeed a solution to the differential equation. Mastery of solving differential equations can aid in understanding physics, engineering, and economics where these tools model real-world phenomena.

Trigonometric Functions

Trigonometric functions, including sine and cosine, are at the heart of the problem we're exploring. These functions are periodic and represent the relationship between the angles and side lengths of a right triangle extended to the unit circle. Cosine illustrates the x-coordinate and sine the y-coordinate of a point on the unit circle.

In differential calculus, knowing the derivatives of trigonometric functions is key. As demonstrated in our exercise, when we differentiate \( \cos(x) \) we obtain \( -\sin(x) \). Trigonometric identities are also invaluable in simplifying expressions within equations. For instance, \( \sin(x) \cot(x) \) simplifies to \( \cos(x) \) because \( \(\cot(x) = \dfrac{\cos(x)}{\sin(x)} \)\). Thus, understanding the properties of these functions plays a pivotal role in both solving differential equations and in various applications like signal processing, electromagnetism, and even quantum mechanics.

Derivatives in Calculus

In calculus, derivatives represent the rate at which a function is changing at any point and are foundational to differential equations. When we talk about \( y' \), we're discussing the first derivative of the function \( y \) with respect to \( x \). Calculating the derivative involves applying rules such as the power rule, product rule, and chain rule, to name a few. However, for trigonometric functions specifically, there are standard derivatives to remember.

Knowing the derivatives of basic trigonometric functions by heart is essential: the derivative of \( \sin(x) \) is \( \cos(x) \) and the derivative of \( \cos(x) \) is \( -\sin(x) \), as shown in our exercise step. Furthermore, understanding how to apply these derivatives and manipulate them through trigonometric identities is key in solving more complex calculus problems, whether they be in pure mathematics or applied in fields like physics and engineering.