Problem 21 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The solution is \( y(t) = (t + 2)e^{-4t} \).

Step 1: Take the Laplace Transform of the Differential Equation

Apply the Laplace transform to both sides of the equation. Recall the Laplace transform of a derivative, \( \mathcal{L}\{y'\} = sY(s) - y(0) \). So, we have:\[\mathcal{L}\{y' + 4y\} = \mathcal{L}\{e^{-4t}\}\]This becomes:\[sY(s) - y(0) + 4Y(s) = \frac{1}{s+4}\]Plug in \( y(0) = 2 \):\[sY(s) - 2 + 4Y(s) = \frac{1}{s+4}\]

Step 2: Solve for Y(s)

Rearrange the equation to isolate \( Y(s) \):\[(s + 4)Y(s) - 2 = \frac{1}{s + 4}\]Add 2 to both sides:\[(s + 4)Y(s) = \frac{1}{s + 4} + 2\]Simplify the right side:\[Y(s) = \frac{1}{(s + 4)^2} + \frac{2}{s + 4}\]

Step 3: Take the Inverse Laplace Transform

Use known inverse transforms to find \( y(t) \). The inverse Laplace transform of \( \frac{1}{(s + 4)^2} \) is \( t e^{-4t} \), and the inverse of \( \frac{2}{s + 4} \) is \( 2e^{-4t} \):\[y(t) = t e^{-4t} + 2e^{-4t}\]Thus, the solution is:\[y(t) = (t + 2)e^{-4t}\]

Step 4: Verify the Solution with Initial Condition

Check if the solution satisfies \( y(0) = 2 \). Substitute \( t = 0 \) into the solution:\[y(0) = (0 + 2)e^{0} = 2\]This confirms the initial condition is satisfied.

Key Concepts

Initial Value Problem

In mathematics, an initial value problem (IVP) is a type of differential equation that comes with specific values known as initial conditions. These are crucial for finding a unique solution. The given problem involves a differential equation along with the initial condition \(y(0) = 2\). This implies that at time \(t = 0\), the value of \(y\) is known to be 2.

The purpose of stating an initial condition is to determine a specific solution among infinitely many possible solutions of a differential equation. For a first-order differential equation like \(y^{\prime} + 4y = e^{-4t}\), one initial condition is sufficient to find the unique solution. In practical terms, this approach helps in understanding how an initial state evolves over time.

When approaching an IVP, here are the steps we generally follow:

• Take the mathematical equation given as a differential equation.
• Identify the initial conditions.
• Use the conditions to determine a unique solution.

Differential Equations

Differential equations are equations that involve functions and their derivatives. They are essential in modeling situations where change is being studied, such as motion across time, population growth, or electrical currents.

In the equation \(y^{\prime} + 4y = e^{-4t}\), we can see it's a linear first-order differential equation. Here, \(y^{\prime}\) represents the derivative of \(y\) with respect to \(t\). The term \(4y\) and the forcing function \(e^{-4t}\) give the equation specific behaviors and complexity.

Linear differential equations such as this have particular properties:

• They can often be solved using advanced techniques like the Laplace Transform.
• They superpose nicely, meaning scaled solutions add together nicely.
• The solutions are continuous and smooth.

Understanding these key points of differential equations allows us to navigate through solutions smoothly and accurately.

Inverse Laplace Transform

The inverse Laplace transform is a technique to retrieve a function of time from its Laplace transform representation. Once a mathematical problem is solved in the Laplace domain, this transform helps convert it back into the time domain.

In our solution, we first found \(Y(s)\) in the Laplace domain. We needed to use known inverse Laplace transforms to find \(y(t)\). The important inverse transforms used include:

• The inverse of \(\frac{1}{(s + 4)^2}\) translates to \(t e^{-4t}\) in the time domain.
• The inverse of \(\frac{2}{s + 4}\) gives \(2e^{-4t}\) in the time domain.

These transforms allow us to express the solution to the differential equation in a form that is meaningful and useful for interpreting real-world scenarios. The ability to toggle between the frequency domain (Laplace) and time domain enhances our problem-solving arsenal.