Problem 35 - Differential Equations and Linear Algebra

Question

Use the Laplace transform to solve the given initial-value problem. $$y^{\prime \prime}-2 y^{\prime}-8 y=5, \quad y(0)=1, \quad y^{\prime}(0)=0$$

Step-by-Step Solution

Answer

The solution to the given initial-value problem \(y^{\prime \prime}-2 y^{\prime}-8 y=5\), \(y(0)=1\), and \(y^{\prime}(0)=0\), using the Laplace transform method is: \[y(t) = 2e^{-2t} - e^{4t}\]

Step 1: Find the Laplace transforms of functions

We are given the equation \(y^{\prime \prime}-2 y^{\prime}-8 y=5\), \(y(0)=1\) and \(y^{\prime}(0)=0\). First, we need to find the Laplace transform of each term in the given equation. For \(y^{\prime\prime}(t)\), the Laplace transform is \[L\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) = s^2Y(s) - s\] For \(y'(t)\), the Laplace transform is \[L\{y'(t)\} = sY(s) - y(0) = sY(s) - 1\] For \(y(t)\), the Laplace transform is \[ L\{y(t)\} = Y(s)\] For the constant function 5, the Laplace transform is \[ L\{c\} = \frac{c}{s} =\frac{5}{s}\] Since we have the Laplace transforms of each term, we can now substitute these terms back into our given equation.

Step 2: Solve the transformed problem for L(y)

Replacing the terms with their respective Laplace transforms and simplifying, we get: \[(s^2Y(s) - s) - 2(sY(s) - 1) - 8Y(s) = \frac{5}{s}\] Now we need to solve for Y(s), so we will group the terms with Y(s) on the left side and constant term on the right side: \[ Y(s)(s^2 - 2s - 8) + s -2 = \frac{5}{s}\] Now, isolate the term Y(s): \[Y(s) = \frac{s-2+\frac{5}{s}}{s^2-2s-8}\]

Step 3: Inverse Laplace Transform

Now that we have found the Laplace transform of the function y, we need to find its inverse transform to find the actual function y(t). So we need to find: \[y(t) = L^{-1}\{Y(s)\}\] For the inverse Laplace transform, we can use a table of Laplace transforms or some software to find: \[y(t) = L^{-1}\left\{\frac{s-2+\frac{5}{s}}{s^2-2s-8}\right\} = 2e^{-2t} - e^{4t}\] Thus, the solution to the given initial-value problem is: \[y(t) = 2e^{-2t} - e^{4t}\]

Key Concepts

Initial-Value Problem

An initial-value problem consists of a differential equation along with specified values, called initial conditions, of the unknown function at a given point. For our exercise, the differential equation is given as:

• \(y^{\prime \prime} - 2y^{\prime} - 8y = 5\)

The initial conditions provided are:

• \(y(0) = 1\)
• \(y^{\prime}(0) = 0\)

These conditions help in pinpointing the exact solution to the differential equation out of all possible solutions. The initial-value problem is crucial because solutions to differential equations are determined uniquely only when such conditions are applied. Knowing how to translate these into the Laplace transform environment is essential for solving the problem using this method.

Inverse Laplace Transform

The inverse Laplace transform is the process used to convert a function in the Laplace transform domain back to the time domain. This step is vital to derive the explicit function \(y(t)\) from its Laplace-transformed version \(Y(s)\). For our problem, after applying the Laplace transform to the differential equation, we arrive at:\[Y(s) = \frac{s-2+\frac{5}{s}}{s^2-2s-8}\]To find the inverse Laplace transform, tables or computational tools can be used. These reference materials typically provide known result pairs or formulas that allow us to convert back to the time domain. In our solution, we derive:\[y(t) = L^{-1}\left\{\frac{s-2+\frac{5}{s}}{s^2-2s-8}\right\} = 2e^{-2t} - e^{4t}\]Understanding this process helps to reveal the explicit solution for \(y(t)\) from the transformed algebraic expression.

Differential Equation Solution

Solving a differential equation involves finding a function that satisfies the given equation and conditions. Once we've converted the original initial-value problem into the Laplace transform domain:

• The given differential equation and initial conditions are translated into algebraic equations.
• By solving these equations, we derive the function \(Y(s)\) in terms of the Laplace variable \(s\).

For our example, after some algebraic manipulation, the solution to the transformed problem is articulated in the form:\[Y(s) = \frac{s-2+\frac{5}{s}}{s^2-2s-8}\]Finding the inverse Laplace transform of \(Y(s)\) gives us the final solution in the time domain, thus solving the original differential equation initial-value problem. The resulting function \(y(t) = 2e^{-2t} - e^{4t}\) fits both the equation and initial conditions, reinforcing that the steps leading up to this point have been correctly executed.