Question

Use the Laplace transforms to solve each of the initial-value. \(y^{\prime \prime}+9 y=36 e^{-3 t}\) \(y(0)=2, \quad y^{\prime}(0)=3\)

Short answer

The solution to the initial-value problem \(y^{\prime \prime}+9 y=36 e^{-3 t}\), with initial conditions \(y(0)=2\) and \(y^{\prime}(0)=3\), using Laplace transforms is given by: \(y(t) = 2 - e^{-3t}(11\cos3t + 2\sin3t)\).

Step-by-step solution

Apply Laplace Transform to both sides

First, we will apply the Laplace transform to both sides of the given differential equation. \( \mathcal{L} \{y^{\prime \prime} + 9y\} = \mathcal{L} \{36e^{-3t}\} \) With the properties of the Laplace transform, we have: \( \mathcal{L} \{y^{\prime \prime}\} + \mathcal{L} \{9y\} = \frac{36}{s+3} \)

Simplify the transformed equation and solve for Y(s)

Using the Laplace transform formulas, we can expand and simplify the transformed equation: \((s^2Y(s) - sy(0) - y^{\prime}(0)) + 9Y(s) = \frac{36}{s+3}\) Now, plug in the initial conditions y(0) = 2 and y'(0) = 3. \((s^2Y(s) - 2s - 3) + 9Y(s) = \frac{36}{s+3}\) Combine terms with Y(s) and solve for Y(s). \(Y(s)(s^2 + 9) = \frac{36}{s+3} + 2s + 3\) Now, let's divide by \((s^2 + 9)\) to isolate Y(s). \(Y(s) = \frac{36}{(s+3)(s^2 + 9)} + \frac{2s}{s^2 + 9} + \frac{3}{s^2 + 9}\)

Apply inverse Laplace Transform

Now, we will apply inverse Laplace transform to find y(t). \(y(t) = \mathcal{L}^{-1} \{\frac{36}{(s+3)(s^2 + 9)} + \frac{2s}{s^2 + 9} + \frac{3}{s^2 + 9}\}\) To make this easier, we'll use the algebraic technique of partial fraction decomposition. Using partial fractions, we get: \(Y(s) = 2 - \frac{11s - 9}{(s+3)(s^2 + 9)}\) Now, we can apply inverse Laplace transform on individual terms: \(y(t) = 2\mathcal{L}^{-1}\{1\} - \mathcal{L}^{-1} \{\frac{11s - 9}{(s+3)(s^2 + 9)}\}\) By using inverse Laplace transform properties and formulas, we obtain: \(y(t) = 2 - e^{-3t}(11\cos3t + 2\sin3t)\)

Final solution

The final solution to the given initial-value problem, using Laplace transforms, is: \(y(t) = 2 - e^{-3t}(11\cos3t + 2\sin3t)\)

Key concepts

Initial Value Problems

Initial value problems are a class of differential equations where the solution is required to satisfy given initial conditions. In our problem, we are required to solve the differential equation \[ y^{\prime \prime} + 9y = 36 e^{-3t} \]with the initial conditions \( y(0) = 2 \) and \( y^{\prime}(0) = 3 \).

• An initial value problem provides specific values for the function and its derivatives at a starting point, typically at \( t = 0 \).
• These conditions help us find a unique solution among the many possible solutions of a differential equation.
• Solving initial value problems often involves finding the values of arbitrary constants in the general solution.

In this case, using Laplace transforms simplifies the process as it turns differential equations into algebraic equations and allows us to directly substitute these initial conditions, leading to a straightforward computation of the specific solution.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify complex rational expressions, especially when working with Laplace transforms. This method involves breaking down a complex fraction into a sum of simpler fractions, making it easier to perform operations such as the inverse Laplace transform.

• This technique is crucial when dealing with the fraction \( \frac{36}{(s+3)(s^2 + 9)} \).
• By expressing the fraction as a sum of simpler terms, individual inverse Laplace transforms can be applied easily to each term.

In our exercise, partial fraction decomposition allowed us to express \[ Y(s) = 2 - \frac{11s - 9}{(s+3)(s^2 + 9)} \]resulting in terms that are more manageable when applying inverse Laplace transformations. By doing this, we turned a complicated problem into simpler parts, aiding us in reaching the exact solution for \( y(t) \).

Inverse Laplace Transform

The inverse Laplace transform is used to convert expressions from the Laplace domain back into the time domain. For the initial-value problem, once we calculated the transform \[ Y(s) = 2 - \frac{11s - 9}{(s+3)(s^2 + 9)} \],we needed to apply the inverse Laplace transform to find \( y(t) \).

• The primary goal here is to revert \( Y(s) \) into the original time-domain function \( y(t) \).
• This involves using known inverse Laplace transforms of standard functions such as exponentials and trigonometric expressions.

For instance, the terms resulting from partial fraction decomposition were simplified using inverse transform tables:- The term \( \mathcal{L}^{-1}\{ \frac{1}{s+3} \} \) relates to \( e^{-3t} \).- Trigonometric terms required application of inverse transformations linked to cosine and sine functions.By efficiently applying these rules, we derived that \[ y(t) = 2 - e^{-3t}(11\cos3t + 2\sin3t). \]

Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They play a vital role in various fields, modeling processes that change continuously. In this problem, the differential equation \[ y^{\prime \prime} + 9y = 36 e^{-3t} \]is a second-order linear differential equation with constant coefficients.

• Such equations frequently arise in physics and engineering, describing phenomena like mechanical vibrations and electrical circuits.
• The goal is to solve for \( y(t) \), representing the system's behavior over time.

Solving the equation traditionally involves integrating to reduce the order and applying initial conditions to compute constants. However, using Laplace transforms simplifies this process by converting differential equations into algebraic equations in the Laplace domain.Ultimately, this method led us to the solution \[ y(t) = 2 - e^{-3t}(11\cos3t + 2\sin3t), \]where \( y(t) \) satisfies both the differential equation and the initial conditions.