Question

Use the Laplace transform to solve the initial value problem. \(4 y^{\prime \prime}+4 y^{\prime}+y=3 \sin t+\cos t, \quad y(0)=2, y^{\prime}(0)=-1\)

Short answer

Based on the given step by step solution, create a short answer for the following problem: Initial Value Problem: \(4y'' + 4y' + y = 3\sin{t} + \cos{t}\), with initial conditions \(y(0) = 2\) and \(y'(0) = -1\) Solution: \(y(t) = -7 + 4t + \frac{1}{2}e^{-t}(9\cos t - 4\sin t)\)

Step-by-step solution

Apply the Laplace Transform to the Differential Equation

Apply the Laplace transform to both sides of the given differential equation: \(L\{4y'' + 4y' + y = 3\sin t + \cos t\}\) Using the linearity property of the Laplace transform and the known Laplace transforms for derivatives, sine, and cosine: \(4s^2 Y(s) - 4sy(0) - 4sy'(0) + 4sY(s) - 4y(0) + Y(s) = \frac{3}{s^2+1} + \frac{s}{s^2+1}\) Substitute the initial conditions y(0) = 2 and y'(0) = -1: \(4s^2 Y(s) - 4s(2) - 4s(-1) + 4sY(s) - 4(2) + Y(s) = \frac{3}{s^2+1} + \frac{s}{s^2+1}\)

Solve the Algebraic Equation for Y(s)

Simplify the equation and solve for Y(s): \(4s^2 Y(s) + 4s + 4sY(s) - 8 + Y(s) = \frac{3}{s^2+1} + \frac{s}{s^2+1}\) Factor out Y(s) and combine terms: \(Y(s)(4s^2 + 4s + 1) = 8 + \frac{3}{s^2+1} + \frac{s}{s^2+1}\) Divide both sides by \((4s^2 + 4s + 1)\): \(Y(s) = \frac{8 + \frac{3}{s^2+1} + \frac{s}{s^2+1}}{4s^2 + 4s + 1}\)

Use the Inverse Laplace Transform to Find y(t)

To find y(t), we need to find the inverse Laplace transform of Y(s): \(y(t) = L^{-1}\{Y(s)\}\) First, simplify the expression for Y(s): \(Y(s) = \frac{8(s^2+1) + 3s + s(s^2+1)}{s^2(4+4s+s^2)}\) \(Y(s) = \frac{9s^3+9s^2+8s+8}{s^2(4+4s+s^2)}\) Now we can use partial fraction decomposition to separate Y(s) into simpler terms, so we can use the inverse Laplace transform on each term. \(Y(s) = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{4s^2 + 4s + 1}\) After performing partial fraction decomposition: \(Y(s) = -\frac{7}{s} - \frac{4}{s^2} + \frac{9s+2}{4s^2 + 4s + 1}\) Now we can find the inverse Laplace transforms for each of the three terms, and then add them together to get y(t): \(y(t) = -7L^{-1}\{\frac{1}{s}\} - 4L^{-1}\{\frac{1}{s^2}\} + L^{-1}\{\frac{9s+2}{4s^2 + 4s + 1}\}\) \(y(t) = -7 + 4t + L^{-1}\{\frac{9s+2}{4s^2 + 4s + 1}\}\) To solve for the last term, we can complete the square in the denominator and then use the inverse Laplace transform for the shifted cosine function: \(L^{-1}\{\frac{9s+2}{4s^2 + 4s + 1}\} = L^{-1}\{\frac{9s+2}{(2s+1)^2+1^2}\} = \frac{1}{2}e^{-t}(9\cos t - 4\sin t)\) So the final solution for y(t) is: \(y(t) = -7 + 4t + \frac{1}{2}e^{-t}(9\cos t - 4\sin t)\)

Verify the Solution

We have found the solution to the given initial value problem: \(y(t) = -7 + 4t + \frac{1}{2}e^{-t}(9\cos t - 4\sin t)\) Verify that this solution satisfies the initial conditions: \(y(0) = -7 + 4(0) + \frac{1}{2}e^{0}(9\cos 0 - 4\sin 0) = -7 + 0 + \frac{1}{2}(9 - 0) = 2\) \(y'(t) = 4 - \frac{1}{2}e^{-t}(9\cos t - 4\sin t) + \frac{1}{2}e^{-t}(-9\sin t - 4\cos t)\) \(y'(0) = 4 - \frac{1}{2}e^{0}(9\cos 0 - 4\sin 0) + \frac{1}{2}e^{0}(-9\sin 0 - 4\cos 0) = 4 - 0 - 2 = -1\) Our solution satisfies both initial conditions, so we have successfully solved the initial value problem using the Laplace transform.

Key concepts

Initial Value Problem

An initial value problem is a common type of differential equation paired with specified values at the initial point. Here, the differential equation describes the system's behavior, such as motion or electric circuits, while the initial conditions anchor the solution by specifying the system's state at the outset.
In our problem, we have the differential equation given by:

• \[4 y'' + 4 y' + y = 3 \sin t + \cos t\]
• With initial conditions: \(y(0) = 2\) and \(y'(0) = -1\)

These conditions imply the system's response starts at the position \(y(0) = 2\) with an initial velocity \(y'(0) = -1\). Solving such problems involves finding a function \(y(t)\) that satisfies both the differential equation and the initial conditions.
Utilizing the Laplace transform simplifies the process by converting differential equations, typically complicated and involving calculus, into algebraic equations, which are much easier to solve.

Inverse Laplace Transform

The inverse Laplace Transform plays a vital role in solving initial value problems by transforming a solution from the Laplace domain back into the time domain. This process allows us to interpret mathematical results in real-world contexts.
In our exercise, after finding \(Y(s)\), which is a function in the Laplace domain, we aim to convert it back into \(y(t)\), a time-domain function. The inverse Laplace Transform is not always straightforward but often involves breaking down \(Y(s)\) into simpler components.

• We use properties of the inverse Laplace transform, such as linearity, to process individual terms separately.
• Complex terms might require additional techniques like completing the square or recognizing shifted forms of simpler transforms known from tables.

Our solution process used these techniques to derive the time-domain expression \(y(t) = -7 + 4t + \frac{1}{2}e^{-t}(9\cos t - 4\sin t)\), a function that provides meaningful insights into the system’s behavior over time.

Partial Fraction Decomposition

Partial fraction decomposition is a mathematical technique used to break down complex rational expressions into simpler fractions, which makes applying the inverse Laplace Transform more feasible.
In the exercise, after simplifying \(Y(s)\), partial fraction decomposition allows us to divide \(Y(s)\) into more manageable parts for which the inverse Laplace transform is easier to find. Here’s how it works:

• Split the expression \(Y(s)\) into fractions with simpler denominators, specifically aiming for forms like \(\frac{A}{s}\), \(\frac{B}{s^2}\), or linear transformations.
• Calculate the constants involved in these fractions, often leading to a system of equations that need solving.

By applying partial fraction decomposition, we reduced \(Y(s)\) such that each part could be directly transformed back into the time domain. For instance, terms were transformed using known inverse Laplace forms, which collectively provided the complete solution \(y(t)\).
This step made the inverse transformation manageable and directly linked the abstract results back to practical terms related to our initial value problem.