Problem 29 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem up to the evaluation of a convolution integral. \(y^{\prime}-a y=f(t), \quad y(0)=\alpha,\) where \(a\) and \(\alpha\) are constants.

Step-by-Step Solution

Answer

To solve the given initial-value problem \(y^{\prime}-a y=f(t), \quad y(0)=\alpha,\) we first find the Laplace transform of the ODE and obtain \(Y(s) = \frac{F(s) + \alpha a}{s+a}\). We then apply the convolution theorem to find the inverse Laplace transform of Y(s) as \(y(t) = \mathcal{L}^{-1}\{F(s)} * e^{-at} +\alpha e^{-at}\). Finally, the convolution integral for y(t) is expressed as: \(y(t) = \int_{0}^{t} (f(\tau)e^{-a(t-\tau)}) d\tau +\alpha e^{-at}\).

Step 1: 1. Laplace Transform of ODE

Take the Laplace transform of the given ODE with respect to t: \(\mathcal{L}\{y'(t) - ay(t) = f(t)\}\) Using linearity of Laplace Transform and first derivative rule, we get: \(sY(s)-\alpha a - aY(s) = F(s)\)

Step 2: 2. Solve for Y(s)

Rearrange and solve for Y(s): \((s+a)Y(s) = F(s) + \alpha a\) \(Y(s) = \frac{F(s) + \alpha a}{s+a}\)

Step 3: 3. Apply Convolution Theorem

We now apply convolution theorem to find the inverse Laplace transform of Y(s). From the convolution theorem, we know that: \(y(t) = \mathcal{L}^{-1}\{Y(s)\} = \mathcal{L}^{-1}\{\frac{F(s)}{s+a}\} + \mathcal{L}^{-1}\{\frac{\alpha a}{s+a}\}\) To get the convolution integral, we need to find the inverse Laplace transforms of the two terms: \(y(t) = \mathcal{L}^{-1}\{F(s)} * e^{-at} +\alpha e^{-at}\)

Step 4: 4. Convolution Integral

Finally, we can write the convolution integral for the function y(t) as: \(y(t) = \int_{0}^{t} (f(\tau)e^{-a(t-\tau)}) d\tau +\alpha e^{-at}\) Here, y(t) is expressed as a convolution integral and we have completed our solution up to the evaluation of the convolution integral.

Key Concepts

Laplace Transform

The Laplace Transform is a powerful tool used to solve differential equations. It transforms a time-domain function into a complex frequency-domain representation. This transformation simplifies the operations involving differential equations by converting derivatives into simple algebraic terms.
Applying the Laplace Transform to a given problem helps in transferring difficult calculus operations into manageable algebra.
For example, to find the transform of the derivative, like in the problem with the equation:

• Given: \( y'(t) - a y(t) = f(t) \)
• Take the Laplace transform: \( \mathcal{L}\{y'(t) - ay(t)\} = sY(s) - \alpha - aY(s) = F(s) \)

The linearity of the Laplace Transform allows it to break down components separately. This makes solving for \( Y(s) \) straightforward once the transform is applied. The goal is to isolate \( Y(s) \) as this expression represents the transformed version of the solution.

Initial-Value Problem

An initial-value problem is a type of differential equation that comes with specified values at the start, typically focusing on conditions such as \( y(0) = \alpha \). These initial conditions ensure a unique solution to the differential equation.In many physical and mathematical scenarios, initial-value problems represent situations where the state at the beginning is known and influences the system's evolution. In our given problem, \( y(0) = \alpha \) serves as the initial value. This condition plays a critical role in forming equations we solve using the Laplace transform.
Specifically, when performing the Laplace transform on derivatives, initial values like \( \alpha \) appear in the transformation process. For instance, in the term \( sY(s) - \alpha \), the constant \( \alpha \) enters as a part of the solution manipulation, ensuring the final solution respects the initial condition.

Convolution Theorem

The Convolution Theorem is a critical concept when dealing with Laplace transforms and inverse transforms. It helps in finding solutions to differential equations by transforming products in the frequency domain into convolution operations in the time domain.This theorem is particularly useful in the final step of finding the inverse Laplace transform. The expression \( Y(s) = \frac{F(s) + \alpha a}{s+a} \) contains terms which when inverse-transformed require convolution. The transformation looks similar to this:

• \( y(t) = \mathcal{L}^{-1}\{\frac{F(s)}{s+a}\} + \mathcal{L}^{-1}\{\frac{\alpha a}{s+a}\} \)
• Which through convolution becomes:
• \( y(t) = \int_{0}^{t} (f(\tau)e^{-a(t-\tau)}) d\tau + \alpha e^{-at} \)

The convolution integral systematically represents the way each element from functions \( f(t) \) and \( e^{-at} \) blend together over time to produce \( y(t) \). This integral captures how initial conditions propagate through the system alongside external inputs.