Problem 13 - Differential Equations and Linear Algebra

Question

Use the Laplace transform to solve the given initial-value problem. \(y^{\prime \prime}-3 y^{\prime}+2 y=4 e^{3 t}, \quad y(0)=0, \quad y^{\prime}(0)=0\).

Step-by-Step Solution

Answer

The solution to the initial-value problem \(y'' - 3y' + 2y = 4e^{3t}\), with initial conditions \(y(0) = 0\) and \(y'(0) = 0\), is given by \[y(t) = -2e^t + e^{2t} + e^{3t}.\]

Step 1: Apply Laplace transform to the given differential equation.

First, we need to recall that the Laplace transform of the first derivative is given by \(\mathcal{L}\{y'(t)\} = sY(s) - y(0)\) and for the second derivative, we have \[\mathcal{L}\{y''(t)\} = s^2Y(s)- sy(0) - y'(0).\] Now applying the Laplace transform to both sides of the given differential equation, we get \[\mathcal{L}\{y''(t) - 3y'(t) + 2y(t)\} = \mathcal{L}\{4e^{3t}\}.\] Plugging the Laplace transforms of the derivatives, we get \[(s^2Y(s) - sy(0) - y'(0)) - 3(sY(s)-y(0)) + 2Y(s) = \frac{4}{s-3}.\] Substituting the initial conditions, \(y(0) = 0\) and \(y'(0) = 0\), we have \[s^2Y(s) - 3sY(s) + 2Y(s) = \frac{4}{s-3}.\]

Step 2: Solve for Y(s)

Next, we will solve this algebraic equation for Y(s), the Laplace transform of the unknown function y(t). First, factor out Y(s) from the left side of the equation, \[Y(s)(s^2 - 3s + 2) = \frac{4}{s-3}.\] Next, divide both sides of the equation by \((s^2 - 3s + 2)\) to isolate Y(s). \[Y(s) = \frac{4}{(s-3)(s^2 - 3s + 2)}.\] Now, we need to factor the denominator to put it in partial fraction form. The denominator factors as \((s-3)(s-1)(s-2)\). Thus, the equation becomes \[Y(s) = \frac{4}{(s-3)(s-1)(s-2)}.\]

Step 3: Perform partial fraction decomposition

In order to simplify the expression and find the inverse Laplace transform, we should perform partial fraction decomposition. \[Y(s) = \frac{A}{s-1} + \frac{B}{s-2} + \frac{C}{s-3}.\] By clearing the denominators and comparing coefficients of like terms, we find that \(A = -2\), \(B = 1\), and \(C = 1\). Then, the partial fraction decomposition becomes \[Y(s) = -2\left(\frac{1}{s-1}\right) + \left(\frac{1}{s-2}\right) + \left(\frac{1}{s-3}\right).\]

Step 4: Inverse Laplace transform

Finally, we will use the inverse Laplace transform to find the solution y(t). Recall that the inverse Laplace transform of \(\frac{1}{s-a}\) is \(e^{at}\). Applying inverse Laplace transform, we get \[y(t) = \mathcal{L}^{-1}\{Y(s)\} = -2e^{t} + e^{2t} + e^{3t}.\] Thus, the solution to the initial-value problem is \[y(t) = -2e^{t} + e^{2t} + e^{3t}.\]

Key Concepts

Initial-Value Problem

An initial-value problem consists of a differential equation along with specified values, called initial conditions, at a particular point. This information helps in finding a unique solution to the differential equation. For the initial-value problem in the exercise, we have a second-order differential equation given by:

• \( y'' - 3y' + 2y = 4e^{3t} \)

This equation is equipped with two initial conditions:

• \( y(0) = 0 \)
• \( y'(0) = 0 \)

The role of these conditions is crucial. They ensure the solution to the differential equation is not just any solution, but the one that passes through these given points. When we consider the Laplace transform, these initial conditions manifest in the form of constants that appear when transforming derivatives, thereby personalizing the transformed equation to reflect the unique characteristics of our problem.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler, more manageable pieces. In the context of Laplace transforms, it helps us simplify the inverse transformation step by reducing the expression into known transform pairs.

For the exercise's equation, once the Laplace transform is applied, we arrive at an expression:

• \( Y(s) = \frac{4}{(s-3)(s-1)(s-2)} \)

This rational expression can be decomposed into partial fractions, simplifying the analysis. We write it as:

• \( \frac{A}{s-1} + \frac{B}{s-2} + \frac{C}{s-3} \)

By clearing the denominators and equating coefficients, we solve for \( A, B, \text{and } C \) which results in simpler, separate terms. In this exercise, these values ended up being \(-2, 1, \text{and } 1\) respectively. Partial fractions are vital as they directly correspond to inverse Laplace transforms of basic functions, paving the way for a straightforward solution.

Inverse Laplace Transform

The inverse Laplace transform is used to convert a function from the Laplace domain back into the time domain. It is essential for interpreting the results of a Laplace transform, especially when solving differential equations.

Once we have decomposed the expression using partial fraction decomposition:

• \( Y(s) = -2\left(\frac{1}{s-1}\right) + \left(\frac{1}{s-2}\right) + \left(\frac{1}{s-3}\right) \)

We apply the inverse Laplace transform to each term separately. The rule for these transforms is straightforward: the inverse Laplace transform of \( \frac{1}{s-a} \) is an exponential function \( e^{at} \).

Thus applying it yields:

• \( y(t) = -2e^{t} + e^{2t} + e^{3t} \)

This expression gives us the time-domain solution to our initial-value problem. By leveraging the properties of inverse Laplace transforms, we turn complex algebraic expressions in the frequency domain back into interpretable time-based functions.