Problem 12 - 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. $$ y^{\prime \prime}+y=\sin t, \quad y(0)=1, y^{\prime}(0)=-1 $$

Step-by-Step Solution

Answer

The solution is \( y(t) = \frac{1}{2}\sin t + \cos t \).

Step 1: Take the Laplace Transform

Apply the Laplace transform to every term in the differential equation: \( y'' + y = \sin t \). The Laplace transform of derivatives involves initial conditions. We get: \( ext{L}[y''] + ext{L}[y] = ext{L}[ ext{sin } t] \). Using properties of Laplace transforms, this becomes: \( s^2Y(s) - sy(0) - y'(0) + Y(s) = \frac{1}{s^2+1} \).

Step 2: Substitute Initial Conditions

Plug in the initial conditions \( y(0) = 1 \) and \( y'(0) = -1 \) into the transformed equation: \( s^2Y(s) - s(1) - (-1) + Y(s) = \frac{1}{s^2+1} \). Simplify to: \( (s^2 + 1)Y(s) - s + 1 = \frac{1}{s^2+1} \).

Step 3: Solve for Y(s)

Solve for \( Y(s) \) by moving the constant terms to the right side and combining like terms: \( (s^2 + 1)Y(s) = \frac{1}{s^2+1} + s - 1 \). Divide both sides by \( s^2 + 1 \): \( Y(s) = \frac{1}{(s^2+1)^2} + \frac{s-1}{s^2+1} \).

Step 4: Inverse Laplace Transform

Break down \( Y(s) \) into parts: \( Y(s) = \frac{1}{(s^2+1)^2} + \frac{s}{s^2+1} - \frac{1}{s^2+1} \). Recognize these as known Laplace transforms. The inverse transforms are: \( ext{sin } t/2 + ext{cos } t - ext{sin } t \). Combine terms to get the solution: \( y(t) = \frac{1}{2}\text{sin } t + \text{cos } t \).

Key Concepts

Initial Value Problem

An initial value problem in mathematics involves finding a function that satisfies a given differential equation and fulfills specific initial conditions. This means that at some initial time, usually denoted as \( t = 0 \), the function is given specific values. These values are used to determine a unique solution that matches the behavior specified by the conditions. In the problem we are working with, we have the differential equation \( y'' + y = \sin t \) and initial conditions \( y(0) = 1 \) and \( y'(0) = -1 \). The initial conditions mean at time \( t = 0 \), the solution \( y(t) \) starts with the values \( y(0) = 1 \) for the function itself, and \( y'(0) = -1 \) for the derivative or the rate of change of the function. These conditions help us find the particular solution to the differential equation that perfectly fits our initial scenario.

Differential Equation

A differential equation is a mathematical equation that relates a function to its derivatives, expressing how the function changes over time. These equations are fundamental in modeling real-world systems where processes evolve over time, such as physics, engineering, and economics.In this exercise, the differential equation is \( y'' + y = \sin t \). Here, \( y'' \) represents the second derivative of the function \( y \), indicating how the rate of change of \( y' \), the derivative of \( y \), evolves. The term \( y \) itself is the function we want to find. The right-hand side, \( \sin t \), is a non-homogeneous term representing an external influence or input into the system. Solving this equation involves integrating the information about \( y \) and its derivatives to find a function that satisfies the equation over time.

Inverse Laplace Transform

The inverse Laplace transform is a mathematical process used to retrieve a function from its Laplace transform. Essentially, it reverses the Laplace transformation. The Laplace transform is a powerful tool for solving differential equations as it simplifies these equations into algebraic ones.In this solution, after computing \( Y(s) \), the expression of the Laplace-transformed function, the inverse Laplace transform process is applied to bring us back to the time domain. The expression obtained was \( Y(s) = \frac{1}{(s^2+1)^2} + \frac{s}{s^2+1} - \frac{1}{s^2+1} \). Each component corresponds to known inverse transforms:- \( \frac{1}{(s^2+1)^2} \) corresponds to \( \frac{1}{2}\sin t \),- \( \frac{s}{s^2+1} \) corresponds to \( \cos t \), and- \( \frac{1}{s^2+1} \) is \( \sin t \).By using a table of Laplace transforms, we can identify these components, ensuring we find the original function \( y(t) \) in the time domain: \( y(t) = \frac{1}{2}\sin t + \cos t \).

Solving ODEs Step by Step

Solving ordinary differential equations (ODEs) step by step involves a systematic approach that simplifies finding solutions. For many ODEs, especially with initial value problems, using the Laplace transform adds clarity and ease. Here's a breakdown of the steps: - **Take the Laplace Transform**: Begin by applying the Laplace transform to each term of the differential equation. This turns derivatives into algebraic expressions, incorporating any initial conditions into the calculation.- **Substitute Initial Conditions**: Any initial conditions provided are plugged into the transformed version of the equation, which simplifies the algebra.- **Solve for the Transform**: Rearrange the equation to solve for the Laplace-transformed function, \( Y(s) \). This might involve moving terms around to isolate \( Y(s) \).- **Find the Inverse Transform**: Finally, use inverse Laplace transforms to move from the frequency (s) domain back to the time domain. This last step gives us the solution for \( y(t) \) that meets both the differential equation and the initial conditions.Following these steps systematically results in not only solving the ODE but also understanding how the solution behaves according to the initial conditions provided.