Problem 4 - Calculus: Early Transcendentals

Question

(a) For what values of \( k \) does the function \( y = \cos kt \) satisfy the differential equation \( 4y^{"} = - 25y? \) (b) For those values of \( k. \) verify that every member of the family functions \( y = A \sin kt + B \cos kt \) is also a solution.

Step-by-Step Solution

Answer

The values of \(k\) are \(\pm\frac{5}{2}\). Both functions satisfy the differential equation.

Step 1: Find the Second Derivative

The given function is \(y = \cos kt\). First, we need to find the second derivative of this function. First derivative: \(y' = -k \sin kt\).Second derivative: \(y'' = -k^2 \cos kt\).

Step 2: Substitute into Differential Equation

The differential equation is given by \(4y'' = -25y\). We substitute the second derivative:\[4(-k^2 \cos kt) = -25 \cos kt.\]

Step 3: Simplify the Equation

Simplify the equation:\[-4k^2 \cos kt = -25 \cos kt.\] Since \(\cos kt\) is not zero, we can divide both sides by \(\cos kt\):\[-4k^2 = -25.\]

Step 4: Solve for k

Now solve the equation \(-4k^2 = -25\):Divide by \(-4\), we get:\[k^2 = \frac{25}{4}.\]Therefore, \(k = \pm\frac{5}{2}.\)

Step 5: Verify General Solution

For the given values of \(k = \pm\frac{5}{2}\), verify that every member of the family of functions \(y = A \sin kt + B \cos kt\) satisfies the differential equation. Take the second derivative of \(y\): \[y = A \sin kt + B \cos kt\]\[y' = Ak \cos kt - Bk \sin kt\]\[y'' = -Ak^2 \sin kt - Bk^2 \cos kt\]Substitute into \(4y'' = -25y\):\[4(-Ak^2 \sin kt - Bk^2 \cos kt) = -25(A \sin kt + B \cos kt).\]

Step 6: Equate and Verify for Each Component

Compare the components:For \(\sin kt\):\[-4Ak^2 = -25A\]\[k^2 = \frac{25}{4}\] \For \(\cos kt\):\[-4Bk^2 = -25B\]\[k^2 = \frac{25}{4}\] \Both satisfy the equation with \(k = \pm\frac{5}{2}\). Thus, the family of functions satisfies the differential equation as every constant \(A\) and \(B\) results in a true statement.

Key Concepts

Second Derivative

The second derivative focuses on finding the rate of rate changes within a function. When you first hear about the second derivative, think of it as the 'acceleration' of a graph. It tells us not just how a function is changing, but how its rate of change itself is evolving. Take the simple function: \( y = \cos kt \).

To find the second derivative, follow these steps:

• First, find the first derivative: \( y' = -k \sin kt \). This tells us how \(y\) changes relative to \(t\).
• Then, take the derivative of the first derivative to get the second derivative: \( y'' = -k^2 \cos kt \).

The second derivative \( y'' = -k^2 \cos kt \) indicates that for the function \( y = \cos kt \), its rate of change, or slope, is modified by the factor \(-k^2\).

This is instrumental in analyzing how solutions behave within a given differential equation, showing the deeper nuances of function behaviors.

Trigonometric Functions

Trigonometric functions are mathematical functions that relate to angles and triangles. In our exercise, these functions play a central role. The primary trigonometric functions we deal with are the sine and cosine functions.

With the function \( y = \cos kt \), we are working with the cosine function.

• The cosine function \( \cos(\theta) \) represents the horizontal coordinate of a point on a unit circle as it rotates around its center.
• The sine function \( \sin(\theta) \) corresponds to the vertical coordinate.

In the context of the differential equation, the cosine function’s properties, specifically its periodic nature, become relevant. This means that as you solve the equation, this periodic nature must be taken into account to achieve a valid solution. Such properties ensure consistency and periodicity, vital for proving that a function satisfies an equation over a continuous range of values.

General Solution

Finding a general solution means determining a formula that includes all possible solutions to a differential equation. This is where constants like \( A \) and \( B \) come into play, which symbolically represent any required values based on initial conditions.

For our exercise, the differential equation is tested against the general solution \( y = A \sin kt + B \cos kt \). This generalized form can adapt to a wide range of initial conditions and, for our calculated values of \( k = \pm\frac{5}{2} \), meets the structure of the differential equation.

The inclusion of both sine and cosine functions in the general solution allows for:

• Flexibility in solving differential equations.
• Accommodation for various curves or wave patterns.
• The ability to construct complex wave-like functions, showcasing a harmony between sine and cosine components.

This idea demonstrates how differential solutions can momentously transition from mathematics to practical applications in physics, engineering, and beyond.

Homogeneous Equation

A homogeneous differential equation is characterized by its right-hand side being zero when the terms are summed up, often resulting solely from the presence of derivatives and equal to zero. Here’s the intriguing part: homogeneous equations suggest that the solution components can be scaled proportionately to find families of solutions.

In our example of the equation \( 4y'' = -25y \), when restructured, leads to a standard form reflecting homogeneous characteristics without any non-zero constant term.

When verifying solutions like \( y = A \sin kt + B \cos kt \) for homogeneity, notice how:

• The form inherently reflects scalar multiplication properties, such as altering \( A \) and \( B \) still giving permissible solutions because the overall structure remains intact.
• It reveals that both sine and cosine solutions share a tightly interwoven relationship defined wholly by the governing principles of the equation.

Understanding this is beneficial as it illustrates broader classes that fulfill the equation criteria, signifying numerous possible functions that share this homogeneous quality.