Question

Express the solution of the given initial value problem in terms of a convolution integral. \(y^{\mathrm{iv}}-y=g(t) ; \quad y(0)=0, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=0, \quad y^{\prime \prime \prime}(0)=0\)

Short answer

Question: Express the solution of the initial value problem \(y^{(4)}(t) - y(t) = g(t)\), with initial conditions \(y(0) = 0, y'(0) = 0, y''(0) = 0, y'''(0) = 0\), as a convolution integral. Answer: The solution to the given initial value problem can be expressed as a convolution integral \(y(t) = (g*h)(t) = \int_0^t g(\tau)h(t - \tau) d\tau\), where \(g(t)\) is the given function and \(h(t)\) is the inverse Laplace transform of \(\frac{1}{s^4 - 1}\), which can be determined using partial fraction decomposition and inverse Laplace transform techniques.

Step-by-step solution

- Taking Laplace transforms of both sides

First, we're going to take the Laplace transforms of both sides of the given differential equation. \(\mathcal{L}\{y^{(4)}(t) - y(t)\} = \mathcal{L}\{g(t)\}\) Now, using the property of Laplace transforms that \(\mathcal{L}\{y^{(4)}(t)\} = s^4Y(s) - s^3y(0) - s^2y'(0) - sy''(0) - y'''(0)\) and \(\mathcal{L}\{y(t)\} = Y(s)\), we can simplify the equation as: \(s^4Y(s) - Y(s) = G(s)\), where \(G(s) = \mathcal{L}\{g(t)\}\).

- Solving for Y(s)

Next, we solve for \(Y(s)\) by factoring it out of the given equation: \(Y(s)(s^4 - 1) = G(s)\) Now, we can find \(Y(s)\) as: \(Y(s) = \frac{G(s)}{s^4 - 1}\)

- Expressing the solution as convolution integral

To find the solution in terms of the convolution integral, we need to find the inverse Laplace transform of \(Y(s)\): \(y(t) = \mathcal{L}^{-1}\{\frac{G(s)}{s^4 - 1}\}\) Now, using the property of Laplace transforms that the inverse Laplace transform of a product of functions equals the convolution of their inverse Laplace transforms, we have: \(y(t) = \mathcal{L}^{-1}\{G(s)\}*\mathcal{L}^{-1}\{ \frac{1}{s^4 - 1}\}\) Now, let \(h(t) = \mathcal{L}^{-1}\{ \frac{1}{s^4 - 1}\}\), the inverse Laplace of \(\frac{1}{s^4 - 1}\). Then, the solution can be expressed in terms of a convolution integral as: \(y(t) = (g*h)(t) = \int_0^t g(\tau)h(t - \tau) d\tau\) Here, \(g(t)\) is the given function and \(h(t)\) is the inverse Laplace transform of \(\frac{1}{s^4 - 1}\), which can be determined using partial fraction decomposition and inverse Laplace transform techniques.

Key concepts

Laplace Transform

The Laplace Transform is a powerful mathematical tool used to transform a function of time, often a complex differential equation, into a function of a complex variable, usually denoted as \(s\). This transformation makes it easier to analyze and solve differential equations by converting them into algebraic equations.

Key features include:

• It converts differentiation into multiplication by quirk of algebra, which greatly simplifies solving differential equations.
• The Laplace Transform for any function \(f(t)\) is defined as \(\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t)dt\).
• The transform can handle initial value problems by incorporating initial conditions directly into the algebraic manipulations.

These properties allow engineers and scientists to work with linear systems and solve initial value problems efficiently. Through the transformation of the function, the complex differential equation is manipulated, and solutions are found in a simpler form.

Inverse Laplace Transform

The Inverse Laplace Transform is used to convert a function in the \(s\)-domain back to the \(t\)-domain. After solving an algebraic equation obtained from a Laplace Transform, the result is often in the \(s\)-domain. To find the actual time-domain solution, an inverse transform is necessary.

Some essential points include:

• The process is the reverse of the Laplace Transformation and is represented as \(\mathcal{L}^{-1}\{F(s)\} = f(t)\).
• The inverse transform uses tables and properties of Laplace Transforms to find expressions back in time-domain.
• In solving the initial value problem, the inverse Laplace transform is often employed in conjunction with convolution when dealing with complex expressions.

Overall, the inverse provides critical insights into system behaviors, helping uncover the solution of differential equations in their original function form. This transformation is particularly useful in control systems and signal processing.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation accompanied by specific conditions, known as initial conditions, which the solution must satisfy at a particular point (usually \(t=0\)). These conditions are essential, as they provide the necessary information to determine a unique solution to the differential equation.

Main insights include:

• Initial conditions specify the value of the function and its derivatives at the starting point.
• This type of problem frequently appears in engineering and physics when a system's state at a given time is known and future states are needed.
• They enable the practical application of differential equations to real-world phenomena by anchoring them with known values.

For our given problem, using the Laplace Transform helps incorporate these initial conditions directly into the solution process, making the method efficient and practical for solving complex differential equations. By utilizing tools like the convolution integral as seen in the step-by-step solution, you can find specific solutions that adhere to these initial values.