Problem 23 - Advanced Engineering Mathematics

Question

In Problems, use the Laplace transform to solve the given initial-value problem. $$ y^{\prime \prime}+2 y^{\prime}+y=0, \quad y(0)=1, \quad y^{\prime}(0)=1 $$

Step-by-Step Solution

Answer

The solution to the initial-value problem is \( y(t) = e^{-t}(1 + t) \).

Step 1: Take the Laplace Transform of the Differential Equation

First, take the Laplace Transform of each term in the differential equation:\[ y'' + 2y' + y = 0 \]Using the properties of Laplace Transforms, we have:\[ Ligl\{ y'' \bigr\} = s^2 Y(s) - sy(0) - y'(0) \]\[ L\{ 2y' \} = 2(sY(s) - y(0)) = 2sY(s) - 2 \]\[ L\{ y \} = Y(s) \] Substitute these into the Laplace Transform of the original equation:

Step 2: Substitute Initial Conditions and Simplify

Substitute the initial conditions \( y(0) = 1 \) and \( y'(0) = 1 \) into the transformed equation:\[ s^2 Y(s) - s(1) - 1 + 2(sY(s) - 1) + Y(s) = 0 \]\[ s^2 Y(s) - s - 1 + 2sY(s) - 2 + Y(s) = 0 \]Simplify the equation to:\[ (s^2 + 2s + 1)Y(s) - s - 3 = 0 \]

Step 3: Solve for Y(s)

To solve for \( Y(s) \), isolate it:\[ (s^2 + 2s + 1)Y(s) = s + 3 \]\[ Y(s) = \frac{s + 3}{s^2 + 2s + 1} \]Note that \( s^2 + 2s + 1 \) can be factored as \( (s+1)^2 \):\[ Y(s) = \frac{s + 3}{(s + 1)^2} \]

Step 4: Perform Partial Fraction Decomposition

To perform the inverse Laplace Transform, break \( \frac{s + 3}{(s + 1)^2} \) into partial fractions if necessary. However, since it's already a single fraction, we don't need to decompose further. Therefore the inverse Laplace can be directly found using known transforms.

Step 5: Perform the Inverse Laplace Transform

Identify the inverse Laplace transformation formulas that match:\[ L^{-1}\left\{\frac{s}{(s+a)^2}\right\} = e^{-at}(1 + at) \]Since we have \( a = 1 \), translate:\[ y(t) = e^{-t}(1 + t) \]

Step 6: Write the Final Solution

The final solution after performing the inverse Laplace transformation gives the solution to the initial-value problem:\[ y(t) = e^{-t}(1 + t) \]

Key Concepts

Differential Equation

A differential equation is an equation that involves an unknown function and its derivatives. In the exercise provided, the differential equation is second-order, given by \[ y^{\prime \prime} + 2y^{\prime} + y = 0. \]Differential equations can model a wide range of physical phenomena, such as motion, growth, decay, and more.

• **Order**: The highest derivative in the equation. Here, it's second-order because of the term \(y^{\prime \prime}\).
• **Linear**: If the function and its derivatives appear linearly without any products or powers. Our equation is linear.
• **Homogeneous**: If the equation equals zero, like in this exercise.

The key goal is to find the unknown function, in this case, \(y(t)\), that satisfies the equation. Using tools like Laplace transforms can simplify this process, particularly for linear differential equations.

Inverse Laplace Transform

The inverse Laplace transform is a tool used to convert a function from the Laplace domain back to the time domain. In solving differential equations, it's a crucial step to find the original function from its transformed version.In the exercise, we started with the Laplace transform \(Y(s) = \frac{s + 3}{(s + 1)^2}\). To find \(y(t)\), we identify and apply the inverse Laplace transform that matches this expression:\[ y(t) = e^{-t}(1 + t). \]The key to this process is recognizing standard inverse transform formulas:

• \( L^{-1}\left\{\frac{s}{(s+a)^2}\right\} = e^{-at}(1 + at). \)

Here, \(a = 1\), and by applying this formula to the transformed function, we successfully retrieve our time domain function \(y(t)\). Being familiar with common Laplace and inverse Laplace transforms drastically eases this procedure.

Initial-Value Problem

Initial-value problems are differential equations coupled with specific initial conditions, which define the values of the function and some of its derivatives at a starting point, usually time \(t = 0\). The purpose is to find a particular solution that satisfies these conditions.In the given problem, we had:

• \(y(0) = 1\)
• \(y^{\prime}(0) = 1\)

These conditions mean that at time \(t = 0\), our function \(y\) equals 1, and its first derivative \(y{\prime}\) is also 1.Applying these initial conditions when transforming the differential equation into the Laplace domain allows us to solve specifically for \(Y(s)\), and subsequently, determine \(y(t)\) accurately. Initial values ensure that the solution is not just general, but specific to the context or conditions given.

Partial Fraction Decomposition

Partial fraction decomposition is often a tool used to simplify complex fractions, especially when taking the inverse Laplace transform. Though not used in our particular exercise, it's important to understand its role.When the Laplace transform of a function results in a fraction where the denominator cannot be easily matched with known inverse transforms, partial fraction decomposition breaks it into simpler fractions:

• Easier to manage with standard inverse transforms.
• Transforms complex expressions into a sum of simpler, known ones.

In this exercise, the expression \(\frac{s + 3}{(s + 1)^2}\) was straightforward enough, having a known inverse form already. But had it been more complex, decomposing it would be key to proceed with solving the inverse transform. Understanding when and how to use this technique is crucial for efficiently working with Laplace transforms.