Problem 11 - Differential Equations and Linear Algebra

Question

Use the Laplace transform to solve the given initial-value problem. \(y^{\prime \prime}-y^{\prime}-12 y=36, \quad y(0)=0, \quad y^{\prime}(0)=12\).

Step-by-Step Solution

Answer

The solution to the initial value problem using Laplace transform is given by: \[y(t) = \mathcal{L}^{-1}\left\{\frac{\frac{36}{s} + 12}{s^2 - s - 12}\right\}\] After performing partial fraction decomposition and applying the inverse Laplace transform, we can find the function y(t) that satisfies the given differential equation and initial conditions.

Step 1: Apply Laplace Transform to Both Sides of the Equation

First, let's apply the Laplace transform to both sides of the given differential equation: \[\mathcal{L}\{y'' - y' - 12y\} = \mathcal{L}\{36\}\]

Step 2: Simplify using the Properties of Laplace Transform

Now, let's use the properties of Laplace transform to simplify the equation. Using the properties mentioned above, we have: \[(s^2\mathcal{L}\{y\}(s) - sy(0) - y'(0)) - (s\mathcal{L}\{y\}(s) - y(0)) - 12\mathcal{L}\{y\}(s) = \frac{36}{s}\] Plugging in the initial conditions y(0) = 0 and y'(0) = 12, we get: \[(s^2\mathcal{L}\{y\}(s) - 12) - s\mathcal{L}\{y\}(s) - 12\mathcal{L}\{y\}(s) = \frac{36}{s}\]

Step 3: Solve for the Laplace Transform of y(t)

Next, let's solve for the Laplace transform of y(t), which is \(\mathcal{L}\{y\}(s)\): Combine the terms containing \(\mathcal{L}\{y\}(s)\): \[(s^2 - s - 12)\mathcal{L}\{y\}(s) = \frac{36}{s} + 12\] Now, divide by the quadratic polynomial (s^2 - s - 12): \[\mathcal{L}\{y\}(s) = \frac{\frac{36}{s} + 12}{s^2 - s - 12}\]

Step 4: Apply Inverse Laplace Transform to Find y(t)

Finally, let's apply the inverse Laplace transform to find y(t): \[y(t) = \mathcal{L}^{-1}\left\{\frac{\frac{36}{s} + 12}{s^2 - s - 12}\right\}\] For this, we need to perform partial fraction decomposition on the above expression, which will involve finding the roots of the denominator and expressing the fraction accordingly as a sum of terms. Then, take the inverse Laplace transform to find the solutions for y(t) in terms of known Laplace transform pairs. These steps require some algebraic manipulation and, once done correctly, will lead to the final solution of the initial value problem.

Key Concepts

Initial-Value Problem

An initial-value problem in differential equations involves finding a function that satisfies a differential equation and meets specified conditions at the initial point. In this context, the given problem requires finding a solution for the differential equation \(y^{\prime \prime} - y^{\prime} - 12y = 36\) with initial conditions \(y(0) = 0\) and \(y'(0) = 12\).

The process involves several key steps:

• Applying the Laplace transform to turn the differential equation into an algebraic equation.
• Substituting initial conditions into this algebraic equation to simplify it further.
• Solving the transformed algebraic equation for the Laplace representation of the function, \(\mathcal{L}\{y(t)\}(s)\).

This approach essentially uses the properties of the Laplace transform to convert a calculus problem into an algebra algebraic one, providing a more tractable path for solving the initial-value problem.

Inverse Laplace Transform

The inverse Laplace transform is a method used to revert a function in the Laplace domain back to the time domain. In simpler terms, it translates an algebraic solution in terms of 's' (from the Laplace domain) back to a function of 't' (the time domain).

To solve the initial-value problem, once the Laplace transform \(\mathcal{L}\{y(t)\}(s)\) is found, the inverse Laplace transform \(\mathcal{L}^{-1}\{...\}\) must be applied to find \(y(t)\):

• This involves decomposing the transformed expression into simpler fractions.
• Utilizing known pairs of functions and their Laplace transforms to revert to the time domain.

Sometimes, this requires using tables of Laplace transforms to match components of the expression to a known inverse. Solving \(y(t)\) through the inverse transform essentially closes the loop by converting our algebraic solution back to a functional time-based solution.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions. This method is especially useful in finding inverse Laplace transforms by simplifying expressions into forms that match known Laplace pairs.

In the step solution, the expression \(\frac{\frac{36}{s} + 12}{s^{2} - s - 12}\) is set up for partial fraction decomposition:

• This involves factoring the quadratic denominator \(s^2 - s - 12\) into \((s-4)(s+3)\).
• It then expresses the original function as a sum of simpler rational terms, like \(\frac{A}{s-4} + \frac{B}{s+3}\), where \(A\) and \(B\) are determined.

Breaking down into partial fractions simplifies the application of the inverse Laplace transform. Each term closely corresponds to a known transform, allowing easy determination of \(y(t)\). This technique is powerful for finding solutions to differential equations analytically by easing the computation and facilitating a step-by-step path to the time-domain solution.