Problem 27 - Calculus - AP Edition

Question

(a) Show that \(y=x \sin x\) is a solution to \(y^{\prime \prime}+y=2 \cos x\) (b) Show that \(y=x \sin x\) is a solution of the equation \(y^{(4)}+y^{\prime \prime}=-2 \cos x\)

Step-by-Step Solution

Answer

Both equations are satisfied with the given function.

Step 1: Differentiate the Function

Find the first and second derivatives of the function \( y = x \sin x \). Let \( y' = \frac{d}{dx}(x \sin x) = x \cos x + \sin x \) using the product rule. The product rule formula is \( (uv)' = uv' + vu' \). Differentiating again gives \( y'' = \frac{d}{dx}(x \cos x + \sin x) = x (-\sin x) + \cos x + \cos x = -x \sin x + 2\cos x \).

Step 2: Substitute into First Equation Check

Substitute \( y = x \sin x \) and \( y'' = -x \sin x + 2\cos x \) into the equation \( y'' + y = 2 \cos x \). Then \( (-x \sin x + 2\cos x) + x \sin x = 2 \cos x \), which simplifies to \( 2 \cos x = 2 \cos x \). The equation holds true.

Step 3: Find Higher Derivatives

Find the third derivative, \( y''' = \frac{d}{dx}(-x \sin x + 2 \cos x) = -x \cos x - \sin x - 2 \sin x = -x \cos x - 3\sin x \), and then the fourth derivative, \( y^{(4)} = \frac{d}{dx}(-x \cos x - 3 \sin x) = -x (-\sin x) - \cos x - 3 \cos x = x \sin x - 4\cos x \).

Step 4: Substitute into Second Equation Check

Substitute \( y^{(4)} = x \sin x - 4 \cos x \) and \( y'' = -x \sin x + 2 \cos x \) into the equation \( y^{(4)} + y^{\prime\prime} = -2 \cos x \). Then \( (x \sin x - 4 \cos x) + (-x \sin x + 2 \cos x) = -2 \cos x \), which simplifies to \( -2 \cos x = -2 \cos x \). The equation holds true.

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus used for differentiating products of two functions. When you have a function that is the product of two simpler functions, say \( u(x) \) and \( v(x) \), the derivative of their product is not simply the derivative of the first function times the second and vice versa. Instead, the product rule provides a formula:

• \( (uv)' = u'v + uv' \)

This means you multiply the derivative of the first function \( u' \) by the second function \( v \), and add it to the product of the first function \( u \) and the derivative of the second function \( v' \).
In the given exercise, the function \( y = x \sin x \) is differentiated using this rule. Here:

• \( u(x) = x \) with derivative \( u'(x) = 1 \)
• \( v(x) = \sin x \) with derivative \( v'(x) = \cos x \)
• So, \( y' = x \cos x + \sin x \)

Understanding when and how to apply the product rule is essential for dealing with more advanced calculus problems, especially those involving differential equations.

Higher Derivatives

Higher derivatives extend the concept of differentiation beyond just the first derivative. The first derivative \( y' \) measures the rate of change of the function, while the second derivative \( y'' \) can indicate how this rate of change itself is changing, often associated with the concept of acceleration if \( y \) is a position function.
The exercise progresses to finding higher derivatives because the function \( y = x \sin x \) must satisfy differential equations involving second and fourth derivatives.

• First, the second derivative: \( y'' = -x \sin x + 2 \cos x \)
• Then, the third derivative: \( y''' = -x \cos x - 3 \sin x \)
• Finally, the fourth derivative: \( y^{(4)} = x \sin x - 4 \cos x \)

These derivatives are used to formulate and solve differential equations by substitution into the equation itself. Recognizing patterns in derivatives and knowing when to stop differentiating is crucial in solving high-order differential equations.

Solution Verification

Solution verification involves checking a proposed solution against the differential equation. In this context, for part (a) of the exercise, one needs to show that \( y = x \sin x \) satisfies the equation \( y'' + y = 2 \cos x \). This is done by substituting \( y \) and its second derivative \( y'' \) into the equation:

• Substitute \( y = x \sin x \) and \( y'' = -x \sin x + 2 \cos x \).
• Simplify to show: \( 2 \cos x = 2 \cos x \).

When both sides of the equation equate after substitution, it confirms that the proposed function indeed satisfies the differential equation.
In part (b), similarly verify \( y = x \sin x \) within \( y^{(4)} + y'' = -2 \cos x \) by:

• Substituting \( y^{(4)} = x \sin x - 4 \cos x \) and \( y'' = -x \sin x + 2 \cos x \).
• Simplify to confirm: \( -2 \cos x = -2 \cos x \).

Verification ensures that the function is not just an arbitrary solution, but a correct one that meets all conditions of the differential equation.