Question

Use the Laplace transform to solve the given initial value problem. $$ y^{\prime \prime}+2 y^{\prime}+y=4 e^{-t} ; \quad y(0)=2, \quad y^{\prime}(0)=-1 $$

Short answer

Based on the step by step solution above, answer the following question: **Question:** Find the solution to the initial value problem \(y^{\prime \prime}+2y^{\prime}+y=4e^{-t}\), given the initial conditions \(y(0)=2\) and \(y^{\prime}(0)=-1\). **Answer:** The solution to the given initial value problem is \(y(t) = 4e^{-t} - 3te^{-t}\).

Step-by-step solution

Take Laplace transforms of both sides of the ODE

The given ODE is $$ y^{\prime \prime}+2 y^{\prime}+y=4 e^{-t} $$ Applying the Laplace transform to both sides of the equation, we get: $$ \mathcal{L}\{y^{\prime \prime}\} + 2\mathcal{L}\{y^{\prime}\} + \mathcal{L}\{y\} = 4\mathcal{L}\{e^{-t}\} $$

Apply the Laplace transform properties for derivatives

Using the properties of Laplace transform for the first and second derivatives, we can write the equation as follows: $$ s^2Y(s) - sy(0) - y'(0) + 2(sY(s) - y(0)) + Y(s) = 4 \frac{1}{s+1} $$ Substitute the given initial conditions \(y(0)=2\) and \(y^{\prime}(0)=-1\).

Substitute initial conditions and simplify the equation

Substituting the initial conditions into the transformed equation, we get: $$ s^2Y(s) - 2s + 1 + 2(sY(s) - 2) + Y(s) = 4 \frac{1}{s+1} $$ Now, combine the terms to obtain the following equation: $$ (s^2 + 2s + 1) Y(s) = 4 \frac{1}{s+1} + 3s - 3 $$

Solve for Y(s)

Now, let's solve for \(Y(s)\): $$ Y(s) = \frac{4 \frac{1}{s+1} + 3s - 3}{s^2 + 2s + 1} $$

Apply the inverse Laplace transform to find y(t)

To find y(t), we need to apply the inverse Laplace transform to Y(s). To do this, we need to first express Y(s) as the sum of partial fraction decompositions. $$ Y(s) = \frac{4}{s+1} + \frac{3s -3}{s^2 + 2s + 1} $$ Factor the denominator of the second term: $$ Y(s) = \frac{4}{s+1} + \frac{3s -3}{(s+1)^2} $$ Now, apply the inverse Laplace transform to each term: $$ y(t) = 4e^{-t} - 3te^{-t} $$ The solution to the given initial value problem is: $$ y(t) = 4e^{-t} - 3te^{-t} $$

Key concepts

Initial Value Problem

An initial value problem involves finding a specific solution to a differential equation that satisfies given initial conditions. It typically consists of an ordinary differential equation (ODE), such as \(y'' + 2y' + y = 4e^{-t}\), along with specific starting values for the function and its derivatives.
In this problem, the initial conditions are \(y(0) = 2\) and \(y'(0) = -1\). These conditions are crucial because they allow us to pinpoint a unique solution to the differential equation.
Solving for these, usually through methods like the Laplace Transform, helps in determining the behavior of the system at any future time \(t\).

• The initial values state where your solution starts.
• These values are integrated during the process to find the exact solution.
• For our example, the initial value conditions set the context of the solution related to time \(t = 0\).

Differential Equations

Differential equations are equations that involve derivatives of a function. They're essential tools in modeling situations where change is constant, as in physics, engineering, and other sciences.
The given differential equation in the problem is a second-order linear differential equation, \(y'' + 2y' + y = 4e^{-t}\).
This can be challenging to solve using traditional methods for some students, which is why we employ the Laplace Transform.

• They describe how a quantity changes over time.
• The order of the differential equation indicates the highest derivative present.
• Linear differential equations like ours often have constant coefficients, making them suitable for techniques like the Laplace Transform.

Inverse Laplace Transform

The inverse Laplace transform is used to convert a function from its Laplace transform form back into its original time domain. In our situation, we have an equation \(Y(s)\) that represents the Laplace transform of \(y(t)\).
The task here is to convert \(Y(s) = \frac{4}{s+1} + \frac{3s-3}{(s+1)^2}\) back to \(y(t)\).
Doing so involves using known results of Laplace transforms to identify corresponding time-domain functions.

• The inverse transform is crucial for interpreting the results in a real-world scenario.
• It typically requires techniques such as partial fraction decomposition to simplify the function.
• Once inversion is complete, you're left with a function fully describing the system's behavior over time.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions, making them easier to work with. In our example, we apply partial fraction decomposition to \(Y(s)\) to prepare for the inverse Laplace transform.
The expression \(Y(s) = \frac{4}{s+1} + \frac{3s-3}{(s+1)^2}\) is split into fractions that are easier to interpret.
This step is crucial for applying the inverse Laplace transform because it aligns with the known transform pairs.

• It simplifies the process of finding inverse transforms.
• Helps in managing and solving complex fractions.
• Aligns complex denominators to standard forms for easier manipulation.