Question

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+4 y^{\prime}+5 y=26 e^{3 t}, \quad y_{0}=1, y_{0}^{\prime}=5$$

Short answer

y(t) = 26 e^{3t} * u(t) - e^{-2t}(\text{cos}(t) - 2\text{sin}(t)) + e^{3 t} * u(t)

Step-by-step solution

- Take the Laplace Transform of Both Sides

Apply the Laplace transform to each term of the differential equation. Using the properties of Laplace transforms, we get: \(\text{L}\{y''(t)\text}\) + 4\(\text{L}\{y'(t)\}\) + 5\(\text{L}\{y(t)\}\) = \(\text{L}\{26e^{3t}\}\).

- Apply Initial Conditions for Laplace Transform

Substitute the initial conditions into the transformed equation: \(s^2 Y(s) - sy(0) - y'(0) + 4(s Y(s) - y(0)) + 5Y(s) = \frac{26}{s-3}\).

- Simplify the Equation

Substitute \(y(0) = 1\) and \(y'(0) = 5\): \(s^2 Y(s) - s - 5 + 4(s Y(s) - 1) + 5Y(s) = \frac{26}{s-3}\).Combine like terms to get: \((s^2 + 4s + 5)Y(s) - s - 9 = \frac{26}{s-3}\).

- Solve for \(Y(s)\)

Isolate \(Y(s)\): \(Y(s) = \frac{26}{(s-3)(s^2 + 4s + 5)} + \frac{s + 9}{s^2 + 4s + 5}\).

- Perform Partial Fraction Decomposition

Separate the equation into partial fractions. Notice that the second term already has a form that can be inverted using known Laplace transforms: \(Y(s) = \frac{26}{(s-3)((s+2)^2+1)} + \frac{s + 2 + 2}{(s+2)^2 + 1}\).

- Take the Inverse Laplace Transform

Using known inverse Laplace transforms: \(y(t) = 26 e^{3t} * \text{u}(t) - e^{-2t} (\text{cos}(t) - 2\text{sin}(t)) + e^{3t} * \text{u}(t)\).

Key concepts

Laplace Transform

The Laplace Transform is a powerful integral transform used to convert differential equations into algebraic equations. This makes them easier to solve. It transforms a time-domain function, typically denoted as \( f(t) \), into a complex frequency-domain function, typically denoted as \( F(s) \). The basic definition is:
\[ \text{L}\{f(t)\} = \frac{ \int_{0}^{\text{∞}} e^{-st} f(t) dt } \].
Any function that satisfies certain conditions can be transformed using this method. Laplace Transforms are especially useful for linear differential equations with constant coefficients.

Differential Equations

Differential equations involve unknown functions and their derivatives. They are classified based on the type and order of the derivatives involved. In this exercise, we deal with a second-order linear differential equation given by:
\[ y^{\prime \prime} + 4y^{\prime} + 5y = 26e^{3t} \].
Solving differential equations typically requires methods like Laplace Transforms, especially when initial conditions are given.

Initial Conditions

Initial conditions are values given for the function and its derivatives at specific points, usually at \( t = 0 \). For the given problem, the initial conditions are:
- \( y(0) = 1 \)
- \( y^{\prime}(0) = 5 \)
These conditions are substituted into the Laplace-transformed equations to reflect the specific situation. This helps to uniquely determine the solution of the differential equation.

Partial Fraction Decomposition

Partial Fraction Decomposition is a technique used to break down complex rational expressions into simpler fractions. This helps in finding the inverse Laplace Transform. For our equation, the term \( Y(s) \) needs to be decomposed into simpler fractions:
\[ Y(s) = \frac{26}{(s-3)(s^2+4s+5)} + \frac{s+9}{s^2+4s+5} \].
This step transforms the problem into multiple simpler problems, making the inverse Laplace Transform easier to apply.

Inverse Laplace Transform

The Inverse Laplace Transform is used to convert a function back from the frequency domain to the time domain. After performing partial fraction decomposition, we find inverse transforms for each term separately. For example:
- \( \frac{26}{(s-3)((s+2)^2+1)} \) uses the known inverse transform for a shifted exponential and a shifted, damped sinusoidal.
Combining the results, we get:
\[ y(t) = 26e^{3t} - e^{-2t}(\cos(t) - 2\sin(t)) + e^{3t} \].
Each term corresponds to the individual inverse transforms of the decomposed fractions. This final step provides the solution in the original time domain.