Question

Verify that the following functions are solutions to the given differential equation.\(y=\pi e^{-\cos x}\) solves \(y^{\prime}=y \sin x\)

Short answer

Yes, \( y = \pi e^{-\cos x} \) is a solution to \( y' = y \sin x \).

Step-by-step solution

Differentiate the Given Function

To verify if the function is a solution to the differential equation, first find its derivative. The function is \( y = \pi e^{-\cos x} \). Using the chain rule, the derivative is:\[y' = \pi e^{-\cos x} \cdot \frac{d}{dx}(-\cos x) \] \[y' = \pi e^{-\cos x} \cdot \sin x \] Thus, the derivative \( y' = \pi e^{-\cos x} \sin x \).

Substitute into the Differential Equation

Now substitute \( y = \pi e^{-\cos x} \) and \( y' = \pi e^{-\cos x} \sin x \) into the differential equation \( y' = y \sin x \). Start with the left side:\[y' = \pi e^{-\cos x} \sin x\]Now, the right side:\[y \sin x = (\pi e^{-\cos x}) \sin x = \pi e^{-\cos x} \sin x\]Both sides equal \( \pi e^{-\cos x} \sin x \).

Confirm Solution Equality

Since both sides of the differential equation are equal:\[y' = y \sin x \]This confirms that the given function \( y = \pi e^{-\cos x} \) is indeed a solution to the differential equation.

Key concepts

Chain Rule Simplified

To differentiate the function \( y = \pi e^{-\cos x} \), we utilize the **chain rule**. The chain rule is a fundamental calculus technique used to differentiate composite functions. It allows us to take the derivative of a function by differentiating its outer and inner components separately. Here, our outer function is \( e^{-\cos x} \) and the inner function is \(-\cos x\).

• First, differentiate the outer function \( e^{u} \) with respect to \( u \) which gives \( e^{u} \).
• Then, differentiate the inner function \(-\cos x\) with respect to \( x \), resulting in \( \sin x \).

By applying the chain rule: \[ y' = \pi e^{-\cos x} \cdot \sin x \].
This shows how the chain rule helps solve complex differentiation problems with ease.

Understanding Derivatives

The **derivative** is a powerful tool in calculus that represents the rate of change of a function with respect to a variable. When differentiating the function \( y = \pi e^{-\cos x} \), we are essentially finding how \( y \) changes as \( x \) changes.

• The derivative \( y' \) computes the slope of the tangent line to the curve at any given point.
• It provides critical information about the function's behavior, such as increasing or decreasing.

In this context, differentiating accurately using the chain rule yields \( y' = \pi e^{-\cos x} \sin x \), which shows us how \( y \) changes in response to variations in \( x \). This derivative is vital for verifying solutions of differential equations.

Solution Verification Techniques

**Solution verification** in differential equations involves confirming that a particular function satisfies the equation. For this exercise, the goal is to verify if \( y = \pi e^{-\cos x} \) is indeed a solution of the given differential equation \( y' = y \sin x \).

• First, obtain the derivative \( y' \) of the proposed solution.
• Substitute \( y \) and \( y' \) back into the original differential equation.
• Check if both sides of the equation match.

If both sides equate, then the function is a valid solution. Here, upon substitution, both sides become \( \pi e^{-\cos x} \sin x \), confirming the solution's validity.

The Art of Substitution

**Substitution** is a key method used in calculus and differential equations to prove that a proposed solution is correct. For verification purposes, substitution involves replacing the variables in the differential equation with their corresponding expressions from the function and its derivative.

• Here we substitute \( y = \pi e^{-\cos x} \) and \( y' = \pi e^{-\cos x} \sin x \) into the equation \( y' = y \sin x \).
• After substitution, both the left and right sides must reconcile to validate the solution.

In this problem, the substitution confirms that the expressions match, ensuring that the function is indeed a solution.

Trigonometric Functions Insight

**Trigonometric functions** such as sine and cosine play a significant role in calculus, especially in problems involving differential equations. In this exercise, the trigonometric function \( \cos x \) appears in the exponent, and \( \sin x \) is involved in the derivative calculation. Understanding these functions is crucial.

• \( \cos x \) is the adjacent side over the hypotenuse in a right-angled triangle and affects the phase in trigonometric identities.
• \( \sin x \) depicts the opposite side over the hypotenuse and is frequently seen when differentiating trigonometric expressions.

In deriving solutions, these functions provide necessary transformations and behaviors that are pivotal to solving and verifying differential equations.