Chapter 15: Problem 2

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

Short Answer

Expert verified

The solution is $ y = 3e^t + 2te^t $.

Step by step solution

Apply the Laplace Transform to both sides

Take the Laplace Transform of both sides of the differential equation y^{\text{'} }-y=2e^{t}yields$ \mathcal{L}\{y^{\text{'}}-y\} = \mathcal{L}\{2e^{t}\} $

Use the properties of the Laplace Transform

Apply the properties to transform the differential equation into algebraic form. The Laplace Transform of a derivative is given by: $ \mathcal{L}\{ y^{\text{'}} \} = s Y(s) - y_0 $where $Y(s)$ is the Laplace Transform of $y(t)$. Additionally, the Laplace Transform $\mathcal{L}\{ y \} = Y(s) $.

Substitute the initial condition and the transforms

Insert the given initial condition $ y_0 = 3 $:\[s Y(s) - 3 - Y(s) = \mathcal{L}\{ 2e^{t} \}\]

Transform the non-homogeneous term

Calculate the Laplace Transform of the non-homogeneous term: $ \mathcal{L}\{ 2e^{t} \} = 2 \times \frac{1}{s-1} = \frac{2}{s-1} $

Combine the terms and solve for Y(s)

Combine all terms and simplify:\[ sY(s) - Y(s) - 3 = \frac{2}{s-1} \]\[ (s-1)Y(s) = 3 + \frac{2}{s-1} \]Solve for $Y(s)$:\[ Y(s) = \frac{3 + \frac{2}{s-1}}{s-1} \]Simplify further:\[ Y(s) = \frac{3(s-1) + 2}{(s-1)^2} \]\[ Y(s) = \frac{3s - 3 + 2}{(s-1)^2} \]\[ Y(s) = \frac{3s - 1}{(s-1)^2} \]

Apply the Inverse Laplace Transform

Use the Inverse Laplace Transform to find $y(t)$:\[ y(t) = \mathcal{L}^{-1}\left\{ \frac{3s - 1}{(s-1)^2} \right\} \]Splitting into partial fractions:\[ \frac{3s - 1}{(s-1)^2} = 3 \cdot \frac{s-1}{(s-1)^2} + \frac{2}{(s-1)^2} \]\[ y(t) = 3e^t + 2te^t \]

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Key Concepts

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

Differential Equations

Differential equations involve functions and their derivatives. They describe how a function changes and are foundational in fields like physics and engineering. In this exercise, the differential equation is given by ewline $ y' - y = 2e^t $. This equation is linear because it can be written as a linear combination of the function and its first derivative. The goal is to find a function $ y(t) $ that satisfies this relationship. Solving differential equations often involves transforming them into a simpler form, like using the Laplace transform, to make them easier to work with.

Initial Conditions

Initial conditions are constraints that specify the value of the function and/or its derivatives at a particular point. They are crucial for finding a unique solution to a differential equation. In this example, the initial condition is given by ewline $ y(0) = 3 $. This means that at time $ t = 0 $, the value of the function $ y $ is 3. Initial conditions help to create a specific solution that fits the problem, rather than a general family of solutions.

Inverse Laplace Transform

The Inverse Laplace Transform is used to convert a function from its Laplace transform back to the original time domain. After solving for $ Y(s) $ in the Laplace domain, we find that ewline$ Y(s) = \frac{3s - 1}{(s-1)^2} $. To find $ y(t) $, we split this expression into partial fractions: ewline $ \frac{3s - 1}{(s-1)^2} = 3 \cdot \frac{s-1}{(s-1)^2} + \frac{2}{(s-1)^2} $. Applying the inverse transform to each term separately, we get: ewline$ y(t) = 3e^t + 2te^t $. This provides the solution in the original time domain.

Partial Fractions

Partial fractions are a technique used to simplify the inverse Laplace transform process. By breaking down a complex fraction into simpler parts, we can easily apply the inverse transform to each part. For ewline $ \frac{3s - 1}{(s-1)^2} $, we decompose it into ewline $ 3 \cdot \frac{s-1}{(s-1)^2} + \frac{2}{(s-1)^2} $. Each of these partial fractions can then be inverted separately, using known Laplace Transform pairs, to find the solution ewline $ y(t) = 3e^t + 2te^t $. This step is essential for making the inverse transform straightforward and manageable.