Problem 8 - Advanced Engineering Mathematics

Question

Use the Laplace transform to solve the given differential equation subject to the indicated initial conditions. $$ y^{\prime \prime}-2 y^{\prime}=1+\delta(t-2), \quad y(0)=0, y^{\prime}(0)=1 $$

Step-by-Step Solution

Answer

The solution is \( y(t) = 1 - e^{2t} + u(t-2)(\frac{1}{2} - \frac{1}{2}e^{2(t-2)}) \).

Step 1: Take the Laplace Transform of the Differential Equation

Apply the Laplace Transform to both sides of the equation: \[ L\{y'' - 2y'\} = L\{1 + \delta(t-2)\}. \] Using properties of the Laplace Transform:\[ L\{y''\} = s^2Y(s) - sy(0) - y'(0) = s^2Y(s) - 1, \]\[ L\{2y'\} = 2(sY(s) - y(0)) = 2sY(s), \]\[ L\{1\} = \frac{1}{s}, \]and \[L\{\delta(t-2)\} = e^{-2s}.\] Thus the transformed equation is:\[ s^2Y(s) - 1 - 2sY(s) = \frac{1}{s} + e^{-2s}. \]

Step 2: Simplify and Solve for Y(s)

Rearrange the equation from Step 1:\[ (s^2 - 2s)Y(s) = \frac{1}{s} + e^{-2s} + 1. \]Solve for \(Y(s)\):\[ Y(s) = \frac{1/s + e^{-2s} + 1}{s^2 - 2s}. \] Simplify further:\[ Y(s) = \frac{1}{s(s-2)} + \frac{e^{-2s}}{s(s-2)} + \frac{1}{s^2 - 2s}. \]

Step 3: Perform Partial Fraction Decomposition

Decompose \( \frac{1}{s(s-2)} \):\[ \frac{1}{s(s-2)} = \frac{A}{s} + \frac{B}{(s-2)}. \]Solving gives \( A = \frac{1}{2} \), \( B = -\frac{1}{2} \).Thus,\[ \frac{1}{s(s-2)} = \frac{1}{2s} - \frac{1}{2(s-2)}. \]

Step 4: Inverse Laplace Transform

Apply the inverse Laplace Transform:\[ y(t) = \frac{1}{2}L^{-1}\{\frac{1}{s}\} - \frac{1}{2}L^{-1}\{\frac{1}{s-2}\} + L^{-1}\{\frac{e^{-2s}}{s(s-2)}\} + L^{-1}\{\frac{1}{s-2}\}. \]Use known transforms:\[ L^{-1}\{\frac{1}{s}\} = 1, \quad L^{-1}\{\frac{1}{s-2}\} = e^{2t}, \]and the shift theorem:\[ L^{-1}\{\frac{e^{-2s}}{s(s-2)}\} = u(t-2)\left(\frac{1}{2} - \frac{1}{2}e^{2(t-2)}\right). \]Finally:\[ y(t) = \frac{1}{2} - \frac{1}{2}e^{2t} + u(t-2)(\frac{1}{2} - \frac{1}{2}e^{2(t-2)}) + (1 - e^{2t}). \]

Step 5: Final Simplification

Combine terms from the last step:\[ y(t) = 1 - e^{2t} + u(t-2)(\frac{1}{2} - \frac{1}{2}e^{2(t-2)}). \]This is the solution to the original differential equation under the given initial conditions.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function to its derivatives. In other words, they describe the relationship between changing quantities.
These equations are essential in modeling various physical phenomena such as heat transfer, sound waves, and even financial markets.
In our exercise, the differential equation is \(y'' - 2y' = 1 + \delta(t-2)\), which involves second and first derivatives.

• \(y''\) represents the second derivative of function \(y\) with respect to time \(t\), indicating acceleration or curvature.
• \(y'\) represents the first derivative, signifying velocity or slope.

The goal is often to find the function \(y(t)\) that satisfies the equation, addressing how \(y\) changes over time.

Initial Value Problem

An initial value problem for differential equations involves finding a solution that additionally satisfies specific conditions at the starting point (initial time).
For our problem, the initial conditions given are \(y(0)=0\) and \(y'(0)=1\). These specify the state of \(y\) and its rate of change at time \(t=0\).

• The condition \(y(0)=0\) means that at the initial time \(t=0\), the value of the function \(y\) is 0.
• \(y'(0)=1\) indicates that the rate of change of \(y\) at \(t=0\) is 1, essentially informing us about the initial velocity.

The initial conditions are crucial because they help determine the unique solution to the differential equation. Without these, multiple solutions could fit the equation.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify complex rational expressions by breaking them into simpler terms. This is especially useful in inverse Laplace transform calculations.
To solve the exercise, we decompose \(\frac{1}{s(s-2)}\) into simpler fractions:

• We write it as \(\frac{A}{s} + \frac{B}{s-2}\).
• Determine the constants \(A\) and \(B\) by comparing coefficients or substituting convenient values for \(s\) to solve the system.

In this case, \(A = \frac{1}{2}\) and \(B = -\frac{1}{2}\), thus:\(\frac{1}{s(s-2)} = \frac{1}{2s} - \frac{1}{2(s-2)}.\)This decomposition allows us to easily find the inverse Laplace transform of the expression, as each separate term corresponds to a known transform.

Inverse Laplace Transform

The inverse Laplace transform converts a function from the Laplace domain back to the time domain. It's the reverse process of the Laplace transform and allows us to reconstruct the original function from its transformed counterpart.
In our solution:

• We have several terms like \(\frac{1}{2}L^{-1}\{\frac{1}{s}\}\) that correspond to inverse transforms.
• \(L^{-1}\{\frac{1}{s}\} = 1\) represents a constant solution in the time domain.
• \(L^{-1}\{\frac{1}{s-2}\} = e^{2t}\), indicating an exponential growth.

The shift theorem applies where we have terms like \(e^{-2s}\). This results in shifting the resulting function in time, such as modifying the argument to \(e^{2(t-2)}\) with the unit step function \(u(t-2)\). By utilizing these known transforms, we reconstruct \(y(t)\), providing the precise solution to the initial problem.