Problem 68 - Advanced Engineering Mathematics

Question

In Problems, use the Laplace transform to solve the given initial-value problem. $$ y^{\prime \prime}-5 y^{\prime}+6 y=\mathcal{u}(t-1), \quad y(0)=0, y^{\prime}(0)=1 $$

Step-by-Step Solution

Answer

The solution is \( y(t) = \frac{1}{6}u(t-1) - \frac{1}{2}u(t-1)e^{2(t-1)} + \frac{1}{3}u(t-1)e^{3(t-1)} \).

Step 1: Take the Laplace Transform

Apply the Laplace transform to both sides of the equation to convert the differential equation into an algebraic equation. Remember that \( ext{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) \) and \( ext{L}\{y'(t)\} = sY(s) - y(0) \). For the given initial conditions, \( y(0)=0 \) and \( y'(0)=1 \), the transforms are: \[ L\{y'' - 5y' + 6y\} = (s^2Y(s) - 1) - 5(sY(s)) + 6Y(s) \] Therefore, the equation becomes \[ (s^2 - 5s + 6)Y(s) = \frac{e^{-s}}{s} \] due to the Laplace transform of the unit step function \( ext{L}\{u(t-1)\} = \frac{e^{-s}}{s} \).

Step 2: Solve for Y(s)

Rearrange the algebraic equation from Step 1 to solve for \( Y(s) \): \[ Y(s) = \frac{e^{-s}}{s(s^2 - 5s + 6)} \] Next, factor the quadratic \( s^2 - 5s + 6 \) which simplifies to \( (s-2)(s-3) \). Thus, \[ Y(s) = \frac{e^{-s}}{s(s-2)(s-3)} \].

Step 3: Apply Partial Fraction Decomposition

Decompose \( \frac{1}{s(s-2)(s-3)} \) into partial fractions to make inverse transformation easier: \[ \frac{1}{s(s-2)(s-3)} = \frac{A}{s} + \frac{B}{s-2} + \frac{C}{s-3} \] Solve for the coefficients \( A, B, \) and \( C \) by multiplying through by the denominator and equating coefficients. This results in: \[ 1 = A(s-2)(s-3) + Bs(s-3) + Cs(s-2) \] Solve this system to find \( A=1/6, B=-1/2, \) and \( C=1/3 \).

Step 4: Find the Inverse Laplace Transform

The inverse Laplace transform is given by \[ y(t) = ext{L}^{-1}\{ \frac{1}{6s} - \frac{e^{-s}}{2s-3} + \frac{e^{-s}}{3s-2}\} \] Using the shifting property and linearity of Laplace Transforms:\( ext{L}^{-1}\{ \frac{1}{s} \} = 1 \), \( ext{L}^{-1}\{ \frac{1}{s-2} \} = e^{2t} \), and \( ext{L}^{-1}\{ \frac{1}{s-3} \} = e^{3t} \),applying the time-shifting property gives the final solution: \[ y(t) = \frac{1}{6}u(t-1) - \frac{1}{2}u(t-1)e^{2(t-1)} + \frac{1}{3}u(t-1)e^{3(t-1)} \].

Key Concepts

Initial-Value Problem

An initial-value problem involves solving a differential equation with given conditions at the start point, typically time zero. These conditions, called initial conditions, make the problem unique and ensure one specific solution. In our exercise, the initial conditions provided are:

• \( y(0) = 0 \)
• \( y'(0) = 1 \)

These conditions mean that when time \( t = 0 \), the function \( y(t) \) is zero, and its first derivative is one. This setup is essential as it enables us to handle functions in dynamic systems precisely, ensuring coherence from the start. Without these initial conditions, the differential equation would have infinitely many solutions.

Differential Equation

A differential equation comprises derivatives that describe how a function changes. It's a mathematical equation meant to represent a physical problem. In the provided exercise, we have:\[ y'' - 5y' + 6y = \mathcal{u}(t-1) \]

• \( y'' \) is the second derivative, representing acceleration or the rate of change of a rate of change.
• \( y' \) is the first derivative, representing velocity or the rate of change of the function.
• \( y \) is the original function itself, representing a position or direct value.

By considering each term, the differential equation outlines a system where different aspects of change sum up to match the external influence on the system, expressed by the unit step function \( \mathcal{u}(t-1) \). Differential equations help us unravel complex real-world processes by breaking everything down to laws of motion, finance models, or any dynamic system's governing behaviors.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler parts, making them easier to work with, especially when considering inverse Laplace transforms. In the solution:\[ \frac{1}{s(s-2)(s-3)} \]is decomposed as:\[ \frac{A}{s} + \frac{B}{s-2} + \frac{C}{s-3} \]This decomposition makes the problem more manageable. Finding the constants \( A \), \( B \), and \( C \) involves setting the combined complex fraction equal to 1 and solving for each coefficient by simplifying the equation and equating equivalent terms. Through this process, these fractions symbolize simpler inverse transformations—turning complex tasks into straightforward combinations. In practical applications, this step is crucial for breaking down tasks and simplifying subsequent calculations.

Inverse Laplace Transform

The inverse Laplace transform is a technique for converting a function in the frequency domain back to the time domain. This step is vital to obtain the original function after manipulation with Laplace transforms. The solution applies:

• Shifting property: Helps to deal with transformations of shifted functions.
• Linearity property: Allows direct application and superposition of results.

By using properties like \( \text{L}^{-1}\{ \frac{1}{s} \} = 1 \), \( \text{L}^{-1}\{ \frac{1}{s-2} \} = e^{2t} \), and employing the shifting property, we translate the frequency representation back to its time function:\[ y(t) = \frac{1}{6}u(t-1) - \frac{1}{2}u(t-1)e^{2(t-1)} + \frac{1}{3}u(t-1)e^{3(t-1)} \]This process closes the mathematical loop, returning the transformed problem to a state we can interpret meaningfully, thus resolving the initial dynamical problem.