Chapter 15: Problem 12

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

Short Answer

Expert verified

The solution is $ y(t) = e^{-t} - 2 t e^{-t} + \cosh(t) + 2 \sinh(t) $.

Step by step solution

- Take Laplace Transform of the Differential Equation

Apply the Laplace transform to both sides of the differential equation:\[ y'' - y = e^{-t} - 2te^{-t} \]Using the Laplace transform properties, we get:\[ L\{y''\} - L\{y\} = L\{e^{-t}\} - L\{-2te^{-t}\} \]

- Apply Initial Conditions

Using the initial conditions $ y(0) = 1 $ and $ y'(0) = 2 $, express the Laplace transforms:\[ L\{y''\} = s^2 Y(s) - s y(0) - y'(0) = s^2 Y(s) - s - 2 \]\[ L\{y\} = Y(s) \]Substitute the transformed expressions and initial conditions:\[ s^2 Y(s) - s - 2 - Y(s) = \frac{1}{s+1} - \frac{2}{(s+1)^2} \]

- Rearrange the Transformed Equation

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

- Simplify the Solution

Express each term separately and simplify:\[ Y(s) = \frac{1}{s+1} - \frac{2}{(s+1)^2} + \frac{s}{s^2 - 1} + \frac{2}{s^2 - 1} \]Recognizing the partial fractions and inverse Laplace transforms:\[ \frac{1}{s+1} = e^{-t}, \quad - \frac{2}{(s+1)^2} = -2 t e^{-t}, \quad \frac{s}{s^2 - 1} = \cosh(t), \quad \frac{2}{s^2 - 1} = 2 \sinh(t) \]

- Apply Inverse Laplace Transform

Combine terms appropriately to get the inverse Laplace transform of the function:\[ y(t) = e^{-t} - 2 t e^{-t} + \cosh(t) + 2 \sinh(t) \]

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Key Concepts

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

Differential Equations

A differential equation involves functions and their derivatives. In our exercise, we work with a second-order differential equation, meaning it has a second derivative: \[ y'' - y = e^{-t} - 2te^{-t} \]
Such equations are essential in modeling real-world phenomena. Here, we solve the equation using the Laplace transform technique. Understanding differential equations helps in fields like physics, engineering, and economics.

Initial Conditions

Initial conditions specify the state of a system at the start (t=0). For this problem, the initial conditions are:

$y(0) = 1$
$y'(0) = 2$

These conditions are crucial because they allow us to find unique solutions to differential equations. Applying initial conditions in Laplace transforms includes substituting known values to solve for the unknown function effectively.

Laplace Transform Properties

Laplace transform properties simplify solving differential equations by converting them into algebraic equations. Key properties used are:

$L\{y''\} = s^2Y(s) - sy(0) - y'(0)$
$L\{y\}= Y(s)$

Transforming the given differential equation helps in managing complex parts seamlessly. After applying Laplace transforms, we often combine like terms to solve for $Y(s)$, the transformed function we seek.

Inverse Laplace Transform

The inverse Laplace transform converts back from the algebraic domain to the time domain. After we derive the algebraic solution for $Y(s)$, we use known inverse transforms:

$\frac{1}{s+1} = e^{-t}$
$-\frac{2}{(s+1)^2} = -2te^{-t}$
$\frac{s}{s^2-1} = \cosh(t)$
$\frac{2}{s^2-1} = 2\sinh(t)$

This step is essential to interpret the solution back in the context of the original problem. The final solution is: \[ y(t) = e^{-t} - 2te^{-t} + \cosh(t) + 2\sinh(t) \]