Question

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}-4 y=4 e^{2 t}, \quad y_{0}=0, y_{0}^{\prime}=1$$

Short answer

The solution to the differential equation is \(y(t) = \frac{1}{2}e^{2t} - \frac{1}{2}e^{-2t} + t e^{2t}\).

Step-by-step solution

- Apply the Laplace Transform

Take the Laplace transform of both sides of the differential equation. Recall that the Laplace transform of a derivative is given by \(\text{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)\), \(\text{L}\{y(t)\} = Y(s)\), and \(\text{L}\{e^{2t}\} = \frac{1}{s-2}\).

- Substitute Initial Conditions

Substitute the initial conditions \(y(0) = 0\) and \(y'(0) = 1\) into the transformed equation. This gives: \((s^2Y(s) - sy(0) - y'(0)) - 4Y(s) = 4 \text{L}\{e^{2t}})\).

- Simplify the Transformed Equation

Simplify the equation using the initial conditions: \(s^2Y(s) - s \times 0 - 1 - 4Y(s) = 4 \times \frac{1}{s-2}\), which simplifies to \(s^2Y(s) - 1 - 4Y(s) = \frac{4}{s-2}\).

- Solve for Y(s)

Combine terms and solve for \(Y(s)\): \(Y(s)(s^2 - 4) = \frac{4}{s-2} + 1\). Rearrange to obtain: \(Y(s) = \frac{4}{(s-2)(s^2-4)} + \frac{1}{s^2-4}\).

- Partial Fraction Decomposition

Perform partial fraction decomposition on \(Y(s)\): \(\frac{4}{(s-2)(s^2-4)} = \frac{4}{(s-2)(s-2)(s+2)}\). Decompose into simpler fractions and solve for the constants.

- Inverse Laplace Transform

Apply the inverse Laplace transform to each term individually. Use known transforms such as \(L^{-1}\{ \frac{1}{s-a} \} = e^{at} \) and \( L^{-1}\{ \frac{1}{s^2-a^2} \} = \frac{\text{sinh}(at)}{a}\). Combine the inverse transforms to find \(y(t)\).

Key concepts

Differential Equations

Differential equations are mathematical equations involving derivatives of an unknown function. They can describe a wide range of phenomena such as movement, growth, and decay.
In this exercise, we deal with a second-order linear differential equation: \( y'' - 4y = 4e^{2t} \). Here,

• \( y'' \) is the second derivative of \( y(t) \)
• \( y \) is the original function.

Our goal is to find \( y(t) \), given certain initial conditions (we'll address those next).

Initial Conditions

Initial conditions are values given for the function and its derivatives at specific points. These are critical because they allow us to find a unique solution to the differential equation.
In this problem, the initial conditions are:\( y(0) = 0 \) and \( y'(0) = 1 \).
When we take the Laplace transform of a differential equation, these conditions help us simplify the transformed equation. Specifically, they will be used to replace terms involving \( y(0) \) and \( y'(0) \) as we see in Step 2, making our equation more solvable.

Laplace Transform Properties

The Laplace Transform is a powerful tool for solving differential equations. It changes functions from the time domain into the complex frequency domain. The main properties we use include:

• \( \text{L}\big\{ y'(t) \big\} = sY(s) - y(0) \)
• \( \text{L}\big\{ y''(t) \big\} = s^2Y(s) - sy(0) - y'(0) \)
• \( \text{L}\big\{ e^{at} \big\} = \frac{1}{s-a} \)

These transformations allow us to convert our differential equation into an algebraic form that is easier to handle.When solving for \( Y(s) \), which represents our function in the frequency domain, we can then manipulate it algebraically to solve for \( y(t) \) in the time domain.

Inverse Laplace Transform

Once we have an algebraic expression for \( Y(s) \), we need to convert it back to the time domain using the Inverse Laplace Transform. This often involves partial fraction decomposition to simplify the expression.
Important transforms include:

• \( L^{-1}\big\{\frac{1}{s-a}\big\} = e^{at} \)
• \( L^{-1}\big\{\frac{1}{s^2-a^2}\big\} = \frac{\text{sinh}(at)}{a} \)

By applying these inverse transforms to each term of \( Y(s) \), we get our final solution in terms of \( y(t) \).