Problem 4 - Advanced Engineering Mathematics

Question

Use the Laplace transform to solve the given differential equation subject to the indicated initial conditions. $$ y^{\prime \prime}+16 y=\delta(t-2 \pi), \quad y(0)=0, y^{\prime}(0)=0 $$

Step-by-Step Solution

Answer

The solution is \( y(t) = u(t - 2\pi)\sin(4(t - 2\pi)) \).

Step 1: Apply Laplace Transform to the Equation

Begin by taking the Laplace transform of both sides of the differential equation. The equation is \( y'' + 16y = \delta(t - 2\pi) \). The Laplace transform of the left side is \( \mathcal{L}\{y''\} + 16\mathcal{L}\{y\} \) and the right side is \( \mathcal{L}\{\delta(t - 2\pi)\} \). Use the initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \) to simplify \( \mathcal{L}\{y''\} \).

Step 2: Simplify Using Initial Conditions

The Laplace transform of the second derivative with initial conditions is \( s^2Y(s) - sy(0) - y'(0) \). With \( y(0) = 0 \) and \( y'(0) = 0 \), this becomes \( s^2 Y(s) \). Thus, the transform of the equation becomes \( s^2 Y(s) + 16 Y(s) = e^{-2\pi s} \).

Step 3: Solve for \( Y(s) \)

Factor out \( Y(s) \) from the left-hand side: \( (s^2 + 16) Y(s) = e^{-2\pi s} \). Solve for \( Y(s) \) by dividing both sides by \( (s^2 + 16) \): \( Y(s) = \frac{e^{-2\pi s}}{s^2 + 16} \).

Step 4: Apply the Inverse Laplace Transform

To find \( y(t) \), apply the inverse Laplace transform to \( Y(s) = \frac{e^{-2\pi s}}{s^2 + 16} \). Use the shifting property of the Laplace transform, \( \mathcal{L}^{-1}\left\{ \frac{e^{-as}}{s^2 + \omega^2} \right\} = u(t-a)\sin(\omega(t-a)) \), yielding \( y(t) = u(t - 2\pi)\sin(4(t - 2\pi)) \).

Step 5: Write the Final Solution

The solution to the differential equation is \( y(t) = u(t - 2\pi)\sin(4(t - 2\pi)) \), where \( u(t - 2\pi) \) is the Heaviside step function, ensuring \( y(t) = 0 \) for \( t < 2\pi \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are used in various fields such as physics, engineering, and biology to model complex systems. In the context of this exercise, we have a second-order linear differential equation, which means it involves a second derivative and is linear in terms of the unknown function and its derivatives.
This specific equation is given by \[ y'' + 16y = \delta(t - 2\pi) \].
The equation includes a delta function, which often represents a sudden force acting at a specific point in time. Solving these equations often requires determining a function, \( y(t) \), that satisfies the equation for all values of \( t \) and adheres to any given initial conditions.

Initial Conditions

Initial conditions are crucial in solving differential equations because they allow us to find a unique solution. When we apply the Laplace transform to a differential equation, these conditions help us solve for the constants that arise.
In this problem, the initial conditions are \( y(0) = 0 \) and \( y'(0) = 0 \). These conditions specify the state of the system at time \( t = 0 \).

• \( y(0) = 0 \) indicates the initial position is zero.
• \( y'(0) = 0 \) indicates the initial velocity is zero.

Using these conditions, we simplify the Laplace transform of the second derivative, which leads us to the solution of the problem. These conditions also prevent arbitrary shifts in the solution ensuring it is uniquely defined.

Heaviside Step Function

The Heaviside step function, denoted as \( u(t - a) \), is a simple mathematical function used to switch a signal on at a particular time. In our solution, it appears as part of the inverse Laplace transform. The function jumps from 0 to 1 at \( t = a \), effectively "turning on" the solution from this point forward.
In our problem, it is used to control the function \( y(t) = u(t - 2\pi)\sin(4(t - 2\pi)) \). This representation ensures that \( y(t) \) remains at zero for times before \( 2\pi \), which aligns with the initial impact occurring at \( t = 2\pi \) due to the delta function \( \delta(t - 2\pi) \).
The Heaviside function is integral in solutions involving sudden changes, acting as a switch to incorporate discontinuities within the solution seamlessly.