Problem 31 - Advanced Engineering Mathematics

Question

Use the Laplace transform to solve the given initial-value problem. $$ \frac{d y}{d t}-y=1, \quad y(0)=0 $$

Step-by-Step Solution

Answer

The solution is \( y(t) = -1 + e^t \).

Step 1: Take the Laplace Transform of the Differential Equation

Apply the Laplace transform to both sides of the equation \( \frac{d y}{d t} - y = 1 \). The properties of the Laplace transform state that \( \mathcal{L}\{ \frac{d y}{d t} \} = sY(s) - y(0) \) and \( \mathcal{L}\{y\} = Y(s) \), where \( Y(s) \) is the Laplace transform of \( y(t) \). Since \( y(0) = 0 \), we have: \[ \mathcal{L}\{ \frac{d y}{d t} \} - \mathcal{L}\{y\} = \mathcal{L}\{1\} \] which simplifies to \[ sY(s) - Y(s) = \frac{1}{s}. \]

Step 2: Solve for Y(s)

Factor \( Y(s) \) out of the left side of the equation: \( (s - 1) Y(s) = \frac{1}{s} \). Solving for \( Y(s) \), we get: \[ Y(s) = \frac{1}{s(s - 1)}. \] This is the expression for \( Y(s) \), the Laplace transform of \( y(t) \).

Step 3: Perform Partial Fraction Decomposition

Since \( Y(s) = \frac{1}{s(s - 1)} \), perform partial fraction decomposition to express this as \( \frac{A}{s} + \frac{B}{s-1} \). Solving for \( A \) and \( B \), we have: \( 1 = A(s - 1) + Bs \). Setting \( s = 0 \) gives \( 1 = -A \), so \( A = -1 \). Setting \( s = 1 \) gives \( 1 = B \), so \( B = 1 \). Thus, \[ Y(s) = \frac{-1}{s} + \frac{1}{s-1}. \]

Step 4: Inverse Laplace Transform

Take the inverse Laplace transform of each term: \( \mathcal{L}^{-1}\{ \frac{-1}{s} \} = -1 \) and \( \mathcal{L}^{-1}\{ \frac{1}{s-1} \} = e^t \). Thus, the inverse Laplace transform of \( Y(s) \) is \( y(t) = -1 + e^t \).

Step 5: Verify the Initial Condition

Verify that the solution satisfies the initial condition \( y(0) = 0 \). Substituting \( t = 0 \) into \( y(t) = -1 + e^t \) gives \( y(0) = -1 + e^0 = -1 + 1 = 0 \). The initial condition is satisfied.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function. They are essential for expressing dynamic processes in various fields such as physics, engineering, and economics. In this exercise, we deal with a first-order linear differential equation, which is one of the simplest types. It is given by \( \frac{d y}{d t} - y = 1 \), where \( y(t) \) represents the function of time. The differential equation describes the relationship between the function \( y(t) \) and its rate of change over time \( \frac{d y}{d t} \).

Solving differential equations like this often involves transforming them into algebraic equations using techniques such as the Laplace Transform. This makes them easier to solve by removing the calculus component and reducing them to simple algebraic manipulation. Many real-world problems modeled by such equations require an initial condition, like \( y(0) = 0 \), to ensure a unique solution.

Partial Fraction Decomposition

Partial fraction decomposition is a method used in algebra to break down complex rational expressions into simpler fractions. This process is particularly useful in the context of Laplace Transforms when solving differential equations. It helps us to simplify the transformed function \( Y(s) \) for easier inversion back to the time domain.

In the example, we have the expression \( Y(s) = \frac{1}{s(s - 1)} \). By expressing \( Y(s) \) as a sum of simpler fractions \( \frac{A}{s} + \frac{B}{s-1} \), we can individually manage these components. To find \( A \) and \( B \), we set up the equation \( 1 = A(s - 1) + Bs \) and solve it by selecting strategic values for \( s \):

• When \( s = 0 \): Results in \( 1 = -A \), so \( A = -1 \).
• When \( s = 1 \): Results in \( 1 = B \), so \( B = 1 \).

This simplification allows us to rewrite the equation, facilitating the inverse Laplace transform in the next steps.

Inverse Laplace Transform

The inverse Laplace transform is a technique used to convert functions from the frequency domain back to the time domain. After finding \( Y(s) \) with the Laplace Transform and simplifying it using partial fraction decomposition, we reverse the process to find the solution \( y(t) \) in its original form. This is crucial for understanding the behavior of the system described by the differential equation.

For the given problem, we have \( Y(s) = \frac{-1}{s} + \frac{1}{s-1} \). Using known inverse Laplace transforms:

• \( \mathcal{L}^{-1}\{ \frac{-1}{s} \} = -1 \)
• \( \mathcal{L}^{-1}\{ \frac{1}{s-1} \} = e^t \)

Thus, the inverse Laplace transform results in the solution \( y(t) = -1 + e^t \).

Finally, we validate the solution by checking that it satisfies the initial condition \( y(0) = 0 \). This step is crucial as it confirms that our solution accurately models the original problem, ensuring a consistent interpretation of the differential equation.