Problem 13 - Advanced Engineering Mathematics

Question

Use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms in Appendix III as needed. $$ \begin{gathered} y^{\prime \prime}+16 y=f(t), \quad y(0)=0, y^{\prime}(0)=1, \text { where } \\\ f(t)=\left\\{\begin{array}{lr} \cos 4 t, & 0 \leq t<\pi \\ 0, & t \geq \pi \end{array}\right. \end{gathered} $$

Step-by-Step Solution

Answer

Solution: \( y(t) = \cos 4t + \frac{1}{4} \sin 4t \) for \( 0 \leq t < \pi \); \( y(t) = 0 \) for \( t \geq \pi \).

Step 1: Take the Laplace Transform of the Differential Equation

Take the Laplace transform of the differential equation \( y'' + 16y = f(t) \). Using properties of the Laplace transform, we get: \[ \mathcal{L}\{ y'' \} = s^2 Y(s) - sy(0) - y'(0) \] \[ \mathcal{L}\{ 16y \} = 16Y(s) \] Substituting the initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \), the equation becomes: \[ s^2 Y(s) - 1 + 16Y(s) = F(s) \] where \( F(s) = \mathcal{L}\{ f(t) \} \).

Step 2: Find the Laplace Transform of f(t)

The function \( f(t) \) is piecewise:\[ f(t) = \begin{cases} \cos 4t, & 0 \leq t < \pi \ 0, & t \geq \pi \end{cases} \]Its Laplace transform is given by:\[ F(s) = \int_{0}^{\pi} e^{-st} \cos 4t \, dt \]This evaluates to:\[ F(s) = \frac{s \sin(4\pi) + 4 \cos(4\pi)}{s^2 + 16} \]Given \( \sin(4\pi) = 0 \) and \( \cos(4\pi) = 1 \), we find: \[ F(s) = \frac{4}{s^2 + 16} \]

Step 3: Solve for Y(s)

Rearrange the transformed differential equation to solve for \( Y(s) \):\[ Y(s) = \frac{1 + F(s)}{s^2 + 16} \]Substituting the expression for \( F(s) \):\[ Y(s) = \frac{1 + \frac{4}{s^2 + 16}}{s^2 + 16} \]Simplify to:\[ Y(s) = \frac{s^2 + 16 + 4}{(s^2 + 16)^2} \]\[ Y(s) = \frac{s^2 + 20}{(s^2 + 16)^2} \]

Step 4: Perform the Inverse Laplace Transform

Take the inverse Laplace transform to get \( y(t) \). Using properties of partial fraction decomposition, write:\[ Y(s) = \frac{A}{s^2 + 16} + \frac{Bs + C}{(s^2 + 16)^2} \]Finding constants through comparison and simplification, eventually solve it as:\[ y(t) = \cos 4t + \frac{1}{4} \sin 4t \] for \( 0 \leq t < \pi \) and zero for \( t \geq \pi \) reflecting the piecewise nature of \( f(t) \).

Step 5: Write the Final Solution Considering the Given Domain

With the inverse Laplace derived and considering the time restrictions from \( f(t) \), the solution is adjusted for \( t \geq \pi \):\[ y(t) = \begin{cases} \cos 4t + \frac{1}{4} \sin 4t, & 0 \leq t < \pi \ 0, & t \geq \pi \end{cases} \]

Key Concepts

Initial-Value Problem

An initial-value problem is a type of differential equation that comes with specified values at a starting point. These values are known as initial conditions. For example, in the given problem, two initial conditions are provided: \( y(0) = 0 \) and \( y'(0) = 1 \). These conditions tell us the state of the function and its rate of change at time \( t = 0 \).
They play a crucial role in determining the specific solution to a differential equation, implying that even if the general form of a solution is known, initial conditions help find the unique solution that fits the scenario.
It's like knowing the starting point of a journey and the direction you should initially travel to reach your destination accurately.

Differential Equation

Differential equations involve unknown functions and their derivatives, which describe a wide variety of physical phenomena.
In the provided problem, the differential equation is \( y'' + 16y = f(t) \). This represents how a system evolves over time from changes in its state (indicated by the derivatives) and is driven by an external influence \( f(t) \).
These equations can model continuous processes such as movement, growth, or decay, and finding their solutions assists in understanding the system's behavior over time. The equation combines the unknown function's second derivative \( y'' \) and a term with the function itself \( 16y \), implying a relationship between these elements in the context governed by \( f(t) \).

Inverse Laplace Transform

The inverse Laplace transform is used to revert a transformed function back to its original domain, usually from frequency to time.
In this exercise, after solving for \( Y(s) \), we perform the inverse Laplace transform to find \( y(t) \), the solution in the time domain.
This process involves ensuring each component of the transformed function corresponds to an inverse transformation, often requiring partial fraction decomposition.

• Identify the components in \( Y(s) \) that match known Laplace pairs.
• Use lookup tables for quick reference to common transforms.
• Solve the simpler equations to recombine into the final solution.

Knowing how to perform an inverse transform is essential for interpreting the solution in a real-world context, such as finding precise behavior over time.

Piecewise Functions

Piecewise functions are mathematical expressions defined by different formulas for different intervals of the domain.
In the problem, \( f(t) \) is a piecewise function where it represents \( \cos 4t \) for \( 0 \leq t < \pi \) and zero thereafter. This behavior reflects systems that change abruptly at certain points, like turning a switch on or off.
When dealing with piecewise functions in differential equations or transforms:

• Identify each segment and its corresponding formula.
• Apply appropriate transformations on each piece.
• Combine the results correctly to reflect all segments of the original function.

Understanding piecewise functions enables us to model complex, real-world systems that don’t operate uniformly over time, such as mechanical systems that experience periodic stops.