Question

Use the Laplace transforms to solve each of the initial-value. \(y^{\prime \prime}+2 y^{\prime}+5 y=0\) \(y(0)=2, \quad y^{\prime}(0)=4\)

Short answer

The solution to the given initial-value problem is: \[y(t) = 2e^{-t} + e^{-t} \sin(2t)\]

Step-by-step solution

Calculate the Laplace transforms of the given equation.

First, let's find the Laplace transforms of \(y^{\prime \prime}+2 y^{\prime}+5 y=0\). We know that the Laplace transform of \(y(t)\) is denoted by \(Y(s)\), and the Laplace transform of \(y'(t)\) is \(sY(s) - y(0)\), and for \(y''(t)\) is \(s^2 Y(s) - sy(0) - y'(0)\). Using these transforms, the Laplace transform of the given equation is: \(s^2Y(s) - 2s -4 +2(sY(s) - 2))+ 5Y(s)=0\)

Solve for the Laplace transform of the solution.

Next, let's solve the equation for \(Y(s)\). We need to isolate \(Y(s)\) on one side of the equation. \(s^2Y(s) + 2sY(s) + 5Y(s) = 2s + 4\) \[(s^2+2s+5)Y(s) = 2s+4\] Now, divide by \((s^2+2s+5)\) to get \(Y(s)\): \[Y(s) = \frac{2s+4}{s^2+2s+5}\]

Calculate the inverse Laplace transform of the solution.

Now we need to find the inverse Laplace transform of \(Y(s)\) to get the solution in terms of \(y(t)\). Once we do a partial fraction decomposition, we have: \[Y(s)=\frac{2s+4}{s^2+2s+5}=\frac{2}{s+1}+\frac{2}{(s+1)^2+4}\] The inverse Laplace transform of each term is as follows: - \(y_1(t) = \mathcal{L}^{-1}\left \{ \frac{2}{s+1} \right \}(t) = 2e^{-t}\) - \(y_2(t) = \mathcal{L}^{-1}\left \{ \frac{2}{(s+1)^2+4} \right \}(t) = e^{-t} \sin(2t)\) The principle of linearity states that the inverse Laplace transform of a sum is equal to the sum of the Laplace transforms of its elements. Therefore, the solution to the given differential equation is: \(y(t) = y_1(t) + y_2(t) = 2e^{-t} + e^{-t} \sin(2t)\) Our final solution to the initial-value problem is: \[y(t) = 2e^{-t} + e^{-t} \sin(2t)\]

Key concepts

Inverse Laplace Transform

The inverse Laplace transform is crucial when solving differential equations, particularly when working with initial-value problems. It essentially allows us to convert a function in the frequency domain back into the time domain. This is important because many operations are simpler in the Laplace-transformed, or frequency, form. The function we transform back to is denoted as \(y(t)\), which gives us an explicit representation of the system's behavior over time.
Starting from a transformed equation \(Y(s)\), the goal is to figure out \(y(t)\). This involves looking up known inverse Laplace transforms and applying properties like linearity. In our solution:

• We computed \(Y(s) = \frac{2s+4}{s^2+2s+5}\).
• This was decomposed and inversely transformed.

These steps enabled us to find the time-dependent solution \(y(t) = 2e^{-t} + e^{-t} \sin(2t)\). The inverse Laplace transform provides a powerful tool to extract the time-based behavior from a frequency-based model.

Initial-Value Problem

An initial-value problem is a differential equation coupled with specific conditions, known as initial conditions, provided for the solution at an initial point. For example, you might be given conditions like \(y(0) = 2\) and \(y'(0) = 4\). These conditions are crucial because they determine the unique solution to a differential equation from a potentially infinite family of solutions.
Including these values gives us a clear snapshot of the system's state at the initial time point, helping us predict future or past states accurately. In solving our problem, these initial conditions translate into constraints when applying Laplace transforms.
The conditions influence the transformed equation and ensure that when we eventually solve for \(y(t)\), the solution matches these predefined conditions. Hence, the final solution precisely follows the problem's initial setup.

Differential Equations

Differential equations involve equations that relate a function to its derivatives. They are crucial in modeling real-world phenomena such as heat transfer, motion, and financial markets. The differential equation we analyzed is \(y^{\prime\prime}+2y^{\prime}+5y=0\). This is a second-order linear homogeneous differential equation with constant coefficients.

• Second-order indicates the highest derivative is \(y''\).
• Homogeneous means there is no term independent of \(y\) or its derivatives.

The Laplace transform helps by converting this into an easier algebraic form. This results in a simpler setup in which expectations derived from initial conditions can be linearly applied. Eventually solving this algebraic equation helps retrieve the original function that describes the time-based behavior of the system.

Partial Fraction Decomposition

Partial fraction decomposition is a technique to express a complex rational expression as a sum of simpler fractions, which is immensely useful in finding inverse Laplace transforms. It makes it easier to apply known transforms.
To solve the equation \(Y(s) = \frac{2s+4}{s^2+2s+5}\), we need to simplify it using partial fraction decomposition. This involves breaking this expression into parts that match patterns whose inverse transforms are readily available in many Laplace transform tables. In our exercise:

• The term \(\frac{2s+4}{s^2+2s+5}\) was expressed as \(\frac{2}{s+1} + \frac{2}{(s+1)^2+4}\).
• Each piece corresponds to a standard Laplace transform, such as \(e^{-t}\) and \(e^{-t} \sin(2t)\).

This breakdown makes it feasible to apply the inverse Laplace transform to each term individually, ensuring an easy composition into the solution for \(y(t)\). This process highlights how powerful partial fraction decomposition can be for solving initial-value problems involving Laplace transforms.