Question

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+9 y=\cos 3 t, \quad y_{0}=2, y_{0}^{\prime}=0$$

Short answer

Using Laplace transforms, the solution is: \( y(t)=2 \cos(3t) + \frac{19}{9} \sin(3t) \)

Step-by-step solution

Apply the Laplace Transform to the Differential Equation

Taking the Laplace transform of both sides of the equation y'' + 9y = \cos(3t)\,we get \(\mathcal{L}\{y'' + 9y\} = \mathcal{L}\{\cos(3t)\}\). Utilizing the linearity of the Laplace transform, this becomes \(\mathcal{L}\{y''\} + 9\mathcal{L}\{y\} = \mathcal{L}\{\cos(3t)\}\).

Use Laplace Transform Properties

Recall that \(\mathcal{L}\{y''\} = s^2\mathcal{L}\{y\} - sy(0) - y'(0)\). Using the initial conditions \(y(0)=2\) and \(y'(0)=0\), we substitute to obtain \(s^2Y(s) - 2s + 0 + 9Y(s) = \mathcal{L}\{\cos(3t)\}\).

Laplace Transform of \(\cos(3t)\)

Recall that \(\mathcal{L}\{\cos(3t)\} = \frac{s}{s^2+9}\). Thus, the equation becomes\(s^2Y(s) - 2s + 9Y(s) = \frac{s}{s^2 + 9}\).

Solve for \(Y(s)\)

Combine like terms:\[(s^2 + 9) Y(s) - 2s = \frac{s}{s^2 + 9}\]. Factor to isolate \(Y(s)\):\[(s^2 + 9)Y(s) = \frac{s}{s^2 + 9} + 2s\]. Thus,\[Y(s) = \frac{s/(s^2 + 9) + 2s}{s^2 + 9}\].

Simplify \(Y(s)\)

Simplify the right-hand side:\[Y(s) = \frac{s + 2s(s^2 + 9)}{(s^2 + 9)^2} = \frac{s + 2s^3 + 18s}{(s^2 + 9)^2}\].Combine like terms in the numerator:\[Y(s) = \frac{2s^3 + 19s}{(s^2 + 9)^2}\].

Use the Inverse Laplace Transform

Split into partial fractions and find the inverse Laplace transform:\[Y(s) = \frac{2s^3}{(s^2 + 9)^2} + \frac{19s}{(s^2 + 9)^2}\].Using the inverse Laplace transform properties, \[\mathcal{L}^{-1}\left\{ \frac{2s^3}{(s^2 + 9)^2} \right \} = 2 \cos(3t), \mathcal{L}^{-1}\left\{ \frac{19s}{(s^2 + 9)^2} \right \} = \frac{19}{9} \sin(3t)\].Thus,\[y(t) = 2 \cos(3t) + \frac{19}{9} \sin(3t)\].

Key concepts

Differential Equations

Differential equations involve equations that include derivatives of a function. They are essential for modeling various real-world phenomena, from physics to biology. In the context of the provided exercise, we are dealing with a second-order linear differential equation given by:

\( y'' + 9y = \cos(3t) \).

The goal is to find the function \( y(t) \) that satisfies the differential equation and the initial conditions.

Initial Conditions

Initial conditions are given values of the function and its derivatives at a specific point, often at \( t = 0 \). They help uniquely determine the solution to a differential equation. For example, in our exercise:

• \( y(0) = 2 \)
• \( y'(0) = 0 \)

These conditions are used in the Laplace transform steps to simplify and solve for \( Y(s) \).

Inverse Laplace Transform

The inverse Laplace transform is used to convert a function in the Laplace domain back to the time domain. After solving the transformed differential equation for \( Y(s) \), we need to find the original function \( y(t) \). The Laplace transform of a function \( f(t) \) is typically denoted as \( \mathcal{L} \{ f(t) \} \), and its inverse is denoted as \( \mathcal{L}^{-1} \{ F(s) \} \).

In our solution, we use inverse Laplace transforms to convert terms like

• \( \mathcal{L}^{-1} \left\{ \frac{2s^3}{(s^2 + 9)^2} \right\ \} \right = 2 \cos (3t) \)
• \( \mathcal{L}^{-1} \left\{ \frac{19s}{(s^2 + 9)^2} \right\ \} = \frac{19}{9} \sin(3t) \)

resulting in \( y(t) = 2 \cos(3t) +\ \frac{19}{9} \sin(3t) \).

Partial Fractions

Partial fractions are used to break down complex rational expressions into simpler fractions, making them easier to invert using inverse Laplace transforms. In our problem, to solve for \( Y(s) \), we express it as:

• \( Y(s) = \frac{2s^3}{(s^2 + 9)^2} + \frac{19s}{(s^2 + 9)^2} \)

This decomposition allows us to handle each term individually when finding the inverse Laplace transform.

Trigonometric Functions

Trigonometric functions often appear in differential equations and their solutions, particularly in problems involving oscillatory behavior. Common functions include sine and cosine. In our solution:

• \( \cos (3t) \)
• \( \sin (3t) \)

These functions are pivotal in forming the final solution \( y(t) \). Trigonometric identities and their properties are used effectively when dealing with Laplace and inverse Laplace transforms.