Chapter 15: Problem 19

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$ y^{\prime \prime}+y^{\prime}-5 y=e^{2 t}, \quad y_{0}=1, \quad y_{0}^{\prime}=2 $$

Short Answer

Expert verified

Solve using Laplace transforms and then use partial fractions as needed to find $ y(t) $.

Step by step solution

Take the Laplace Transform of Both Sides

Apply the Laplace transform to each term of the differential equation. Recall the transforms $\text{L}\big\{y''\big\} = s^2Y(s) - sy(0) - y'(0)$, $\text{L}\big\{y' \big\} = sY(s) - y(0)$ and $\text{L}\big\{e^{at}\big\} = \frac{1}{s-a}$.

Substitute Initial Conditions

Insert the initial conditions $ y(0) = 1 $ and $ y'(0) = 2 $ into the Laplace-transformed equation to get $(s^2Y(s) - s \times 1 - 2) + (sY(s) - 1) - 5Y(s) = \frac{1}{s-2}$.

Simplify the Equation

Combine like terms: $ s^2Y(s) + sY(s) - 5Y(s) - (s + 2) = \frac{1}{s-2} $. Simplify to $ (s^2 + s - 5)Y(s) - (s + 2) = \frac{1}{s-2} $.

Solve for Y(s)

Isolate $ Y(s) $: $ (s^2 + s - 5)Y(s) = \frac{1}{s-2} + s + 2 $, $ Y(s) = \frac{1}{(s-2)(s^2 + s - 5)} + \frac{s + 2}{s^2 + s - 5} $.

Perform Partial Fraction Decomposition

Decompose each term if necessary: $\frac{1}{(s-2)(s^2 + s - 5)} $ and $\frac{s + 2}{s^2 + s - 5} $, and simplify.

Take the Inverse Laplace Transform

Apply the inverse Laplace transform to each term to get back to the time domain. This gives expressions involving known inverse transforms.

Write the Final Solution

Combine all parts after inversion to express $ y(t) $ in terms of time: \ y(t) = \text{(solution)} \.

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Key Concepts

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

Differential Equations

To solve the given differential equation using the Laplace transform, we first need to understand what a differential equation is. A differential equation involves functions and their derivatives. In this case, the differential equation is a second-order one, as it includes $y''$. $ \frac{dy}{dt}$ represents the first derivative and $ \frac{d^2y}{dt^2}$ is the second derivative of the function $y(t)$. Differential equations are used to describe various phenomena such as motion, heat, or fluid dynamics.
By applying the Laplace transform, we convert the differential equation from a time domain to a s-domain, which makes complex differential equations easier to solve.

Initial Conditions

Initial conditions are necessary to solve differential equations completely. They give us the values of the function and its derivatives at a specific point, typically when $t=0$. For our equation, the initial conditions are $y(0) = 1$ and $ y'(0) = 2$.
When applying the Laplace transform, initial conditions help to determine the constants in the transformed equation. This is because the Laplace transform of derivatives includes initial conditions.

Inverse Laplace Transform

After converting the differential equation to the s-domain and solving for $Y(s)$, we need to change it back to the time domain. This is achieved using the inverse Laplace transform. The inverse Laplace transform takes a function in the s-domain and converts it back to the time domain.
To find the inverse Laplace transform, often we look up standard Laplace transform pairs or use partial fraction decomposition to simplify the term. This step is critical as it provides us with the final solution $y(t)$ of the original differential equation.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to simplify complex rational expressions in Laplace transform problems. When we have the transformed equation, $ Y(s) \, $ it might contain a complex fraction.
To handle this, we decompose it into simpler fractions that are easier to invert. For example, breaking down a fraction like $ \frac{1}{(s-2)(s^2+s-5)}$ into simpler fractions helps in finding their inverse Laplace transforms using known standard forms. This step facilitates solving the differential equation by making inverse transforms straightforward.