Question

Use the Laplace transform to solve the given initial value problem. $$ y^{\prime \prime}+\omega^{2} y=\cos 2 t, \quad \omega^{2} \neq 4 ; \quad y(0)=1, \quad y^{\prime}(0)=0 $$

Short answer

Question: Determine the solution to the initial value problem $$y^{\prime \prime}+\omega^{2} y=\cos 2 t, \quad \omega^{2} \neq 4 ; \quad y(0)=1, \quad y^{\prime}(0)=0. $$ Answer: The solution to the given initial value problem is $$y(t) = \frac{\omega^2}{2(\omega^2 - 4)}\sin(\omega t) + \frac{1}{2}\sin(2t).$$

Step-by-step solution

Apply the Laplace transform to the given differential equation

To begin solving the initial value problem, we need to apply the Laplace transform to both sides of the given differential equation. The Laplace transform of the given equation is: $$ \mathcal{L}\{y^{\prime\prime}\} + \omega^2\mathcal{L}\{y\} = \mathcal{L}\{\cos 2t\} $$ Using Laplace transform properties: $$ s^{2}Y(s) - sy(0) - y'(0) + \omega^{2}Y(s) = \frac{s}{s^2 + 4} $$ Now substitute the initial conditions \(y(0) = 1\) and \(y'(0) = 0\), and solve for Y(s).

Substitute the initial conditions and solve for Y(s)

Plug in the initial conditions \(y(0) = 1\) and \(y'(0) = 0\) into the equation: $$ s^{2}Y(s) - s + \omega^{2}Y(s) = \frac{s}{s^2 + 4} $$ Next, combine the terms with Y(s): $$ Y(s)(s^{2} + \omega^{2}) = s + \frac{s}{s^2 + 4} $$ Now, solve for Y(s): $$ Y(s) = \frac{s + \frac{s}{s^2 + 4}}{s^{2} + \omega^{2}} $$ $$ Y(s) = \frac{s(s^2 + 4) + s}{(s^2 + \omega^2)(s^2 + 4)} $$

Apply inverse Laplace transform to solve for y(t)

Finally, we apply the inverse Laplace transform to Y(s) to obtain the solution y(t): $$ y(t) = \mathcal{L}^{-1}\left\{ \frac{s(s^2 + 4) + s}{(s^2 + \omega^2)(s^2 + 4)} \right\} $$ Apply partial fraction decomposition to Y(s) and then apply inverse Laplace transform to each term: $$ y(t)=\frac{1}{\omega^2-4}[(\omega^2-4)\mathcal{L}^{-1}\left\{\frac{s}{s^{2}+4} \right\}-\omega^{2} \mathcal{L}^{-1}\left\{\frac{s}{s^{2}+\omega^{2}} \right\}] $$ Finally, evaluate the inverse Laplace transforms of the terms: $$ y(t) = \frac{1}{\omega^2-4}[(\omega^2-4)\frac{1}{2}\sin(2t) - \omega^{2}\frac{1}{\omega}\sin(\omega t)] $$ Simplify the expression: $$ y(t) = \frac{\omega^2}{2(\omega^2 - 4)}\sin(\omega t) + \frac{1}{2}\sin(2t) $$ So the solution to the given initial value problem is: $$ y(t) = \frac{\omega^2}{2(\omega^2 - 4)}\sin(\omega t) + \frac{1}{2}\sin(2t) $$

Key concepts

Initial Value Problem

In mathematics, an initial value problem (IVP) involves solving a differential equation with specific conditions at the start point. These conditions specify the value of the function, and possibly its derivatives, at a given point, often time \( t = 0 \). In this exercise, the initial value conditions are \( y(0) = 1 \) and \( y'(0) = 0 \). These conditions ensure the uniqueness of the solution and guide us in using the Laplace Transform technique effectively.

The goal is to find a function \( y(t) \) that not only satisfies the differential equation \( y'' + \omega^2 y = \cos 2t \) but also meets the given initial requirements. It is like helping a ship start at a specific location and direction on its journey across the ocean of solutions.

Inverse Laplace Transform

The inverse Laplace Transform is a powerful mathematical tool used to revert a function from its Laplace Transform back to the time domain. This process is essential in solving differential equations, especially after transforming the function into a function of \( s \), denoted as \( Y(s) \).

After solving for \( Y(s) \), the next step is to find \( y(t) \) by applying the inverse Laplace Transform. We use tables and properties of Laplace Transforms to handle each part separately, ensuring we obtain the function of time correctly.

In our exercise, the expression \( Y(s) \) was transformed back to \( y(t) \), leading to the final solution: \( y(t) = \frac{\omega^2}{2(\omega^2 - 4)}\sin(\omega t) + \frac{1}{2}\sin(2t) \). Each term in \( Y(s) \) is carefully matched to its time-domain counterpart to construct an accurate solution.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions. This process is particularly beneficial when handling inverse Laplace Transforms as it allows for straightforward transformations of simpler fractions.

In the given solution, the function \( Y(s) = \frac{s(s^2 + 4) + s}{(s^2 + \omega^2)(s^2 + 4)} \) is split into simpler components. Each of these components corresponds to a term that we can easily inverse transform using standard Laplace Transform tables.

By decomposing \( Y(s) \) into simpler fractions, it becomes more manageable and allows us to apply the inverse Laplace Transform term-by-term, making the process simpler and the solution more transparent.

Differential Equations

Differential equations describe relationships involving rates of change and are fundamental in modeling numerous scientific phenomena. In this exercise, we deal with a second-order linear differential equation with constant coefficients: \( y^{\prime\prime} + \omega^2 y = \cos 2t \).

The challenge lies in finding a function \( y(t) \) that satisfies this equation and the initial conditions. Applying the Laplace Transform simplifies the differential equation into an algebraic one, reducing the complexity and handling the conditions easily.

Differential equations are powerful tools for describing various systems such as mechanical vibrations, electrical circuits, and even population models, making them indispensable in both pure and applied mathematics.