Chapter 15: Problem 11

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$ y^{\prime \prime}-4 y=4 e^{2 t}, \quad y_{0}=0, \quad y_{0}^{\prime}=1 $$

Short Answer

Expert verified

y(t) = \text{sinh}(2t)

Step by step solution

Take the Laplace Transform of Both Sides

Apply the Laplace transform to the differential equation: \[ \text{L}\big\brace{y''(t) - 4y(t)} = \text{L}\big\brace{4e^{2t}} \].

Use Laplace Transform Properties

Use the properties of the Laplace transform for derivatives and the given initial conditions: \[ \text{L}\big\brace{y''(t)} - 4\text{L}\big\brace{y(t)} = \text{L}\big\brace{4e^{2t}} \]. Given $ y(0) = 0 $ and $ y'(0) = 1 $, we get: \[ s^2Y(s) - sy(0) - y'(0) - 4Y(s) = \frac{4}{s-2} \].

Substitute Initial Conditions

Substitute $ y(0) = 0 $ and $ y'(0) = 1 $: \[ s^2Y(s) - 1 - 4Y(s) = \frac{4}{s-2} \].

Solve for $ Y(s) $

Combine like terms and solve for $ Y(s) $: \[ (s^2 - 4)Y(s) = \frac{4}{s-2} + 1 \]. Simplify to: \[ Y(s) = \frac{4}{(s-2)(s^2-4)} + \frac{1}{s^2-4} \].

Partial Fraction Decomposition

Perform partial fraction decomposition on $ Y(s) $: \[ Y(s) = \frac{A}{s-2} + \frac{B}{s+2} + \frac{C}{s-2} \]. Combine terms: \[ Y(s) = \frac{4(s+2) + 1}{(s-2)(s+2)} = \frac{4s + 8 + 1}{(s-2)(s+2)} \].

Simplify and Inverse Laplace Transform

Simplify the combined terms and find the inverse Laplace transform: \[ Y(s) = \frac{4s + 9}{(s-2)(s+2)} \] becomes: \[ Y(s) = \frac{A}{s-2} + \frac{B}{s+2} \]. Find $A$ and $B$: Solve to get: $ A = 1 $ and $ B = 1 $, thus: \[ Y(s) = \frac{1}{s-2} + \frac{1}{s+2} \].

Apply Inverse Laplace Transform to Each Term

Apply the inverse Laplace transform to each term: \[ y(t) = \text{L}^{-1}\big\brace{\frac{1}{s-2}}\big + \text{L}^{-1}\big\brace{\frac{1}{s+2}}\big \]. This yields: \[ y(t) = e^{2t} - e^{-2t} \].

Write Final Solution

Combine the terms to write the final solution: \[ y(t) = \frac{e^{2t} - e^{-2t}}{2} \], which simplifies to: \[ y(t) = \text{sinh}(2t) \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Laplace transform properties

Laplace transform is an integral transform used to switch a function from the time domain, denoted as $t$, to the frequency domain, denoted as $s$. This is particularly useful in solving differential equations. For example, the Laplace transform of a function $y(t)$ is represented by $Y(s)$.
Some key properties of Laplace transforms include:

Linearity: \text{L}\big\brace{a f(t) + b g(t)} = a \text{L}\big\brace{f(t)} + b \text{L}\big\brace{g(t)}
Differentiation: \text{L}\big\brace{y'}(t) = sY(s) - y(0)
Laplace of exponential functions: \text{L}\big\brace{e^{at}} = \frac{1}{s-a}

This property makes handling differential equations easier, as it turns derivatives into algebraic equations.

partial fraction decomposition

Partial fraction decomposition is a method used to express a complex rational function as a sum of simpler fractions. This technique is especially handy when we need to take the inverse Laplace transform. In a nutshell, it involves breaking down a fraction like $\frac{4}{(s-2)(s^2-4)}$ into simpler parts that can be easily inverted.
Here are the steps involved:

Factor the denominator: $(s^2 - 4)$ can be written as $(s-2)(s+2)$.
Set up the decomposition: \frac{4}{(s-2)(s+2)} = \frac{A}{s-2} + \frac{B}{s+2}
Solve for $A$ and $B$ by clearing the denominators and equating coefficients.

This simplifies the expression and makes it easier to find the inverse Laplace transform for each term individually.

initial conditions in differential equations

Initial conditions are the values of the function and its derivatives at the starting point (usually $t=0$). These conditions are essential in finding a unique solution to a differential equation. Given $y'(0)=1$ and $y(0)=0$, the initial conditions influence the transformed equations greatly. For example, from the property of Laplace transforms: \text{L}\big\brace{y'}(t) = sY(s) - y(0), we substitute $y(0)=0$ leading to $sY(s) - y'(0)$ for $y''(t)$, resulting in $s^2Y(s) - sy(0) - y'(0)$. This application of initial conditions is crucial for accurate and specific solutions.

inverse Laplace transform

The inverse Laplace transform converts the function from the frequency domain back to the time domain. In this exercise, once we have $Y(s)$ expressed in simpler fractions, we can find $y(t)$ by taking the inverse transform of each part. For instance, solving $\text{L}^{-1}\big\brace{\frac{1}{s-2}}\big$ gives $e^{2t}$ and $\text{L}^{-1}\big\brace{\frac{1}{s+2}}\big$ gives $e^{-2t}$. These inversions are key steps in arriving at the final solution. Hence, the final function $y(t) = \text{sinh}(2t)$ is obtained by combining these inverse transforms correctly.