Problem 32 - Differential Equations and Linear Algebra

Question

Let \(y(t)\) be the solution to the initial-value problem \(y^{\prime}+a y=f(t), y(0)=y_{0},\) where \(a\) and \(y_{0}\) are constants. Verify that $$ L[y]=\frac{L[f]}{s+a}+\frac{y_{0}}{s+a} $$ and show that $$ y(t)=y_{0} e^{-a t}+\int_{0}^{t} e^{-a(t-w)} f(w) d w $$

Step-by-Step Solution

Answer

The Laplace transform of the given initial-value problem is \(L[y(t)] = \frac{L[f(t)]}{s+a} + \frac{y_0}{s+a}\). To find \(y(t)\), we take the inverse Laplace transform of \(L[y(t)]\). Using the convolution theorem and properties of inverse Laplace transforms, we get \(y(t) = y_0 e^{-at} + \int_{0}^{t} e^{-a(t-w)}f(w)dw\), which shows that \(y(t)\) satisfies the given equation.

Step 1: Find the Laplace Transform of the IVP

Let's first find the Laplace transform of both sides of the differential equation. The differential equation is given as: \(y'(t) + ay(t) = f(t)\) Taking the Laplace transform, we get: \(L[y'(t)] + aL[y(t)] = L[f(t)]\) Now, remember that the Laplace transform of the derivative of y(t) is given by: \(L[y'(t)] = sL[y(t)] - y(0)\) So we can substitute this expression into our previous equation: \(sL[y(t)] - y(0) + aL[y(t)] = L[f(t)]\) We also know that \(y(0) = y_0\), so we can substitute that as well: \(sL[y(t)] - y_0 + aL[y(t)] = L[f(t)]\) Now, we'll solve for \(L[y(t)]\).

Step 2: Solve for L[y(t)]

To solve for \(L[y(t)]\), we first factor out \(L[y(t)]\) from the left side of the equation: \((s+a)L[y(t)] = L[f(t)] + y_0\) Now, we can divide both sides by (s+a) to isolate \(L[y(t)]\): \(L[y(t)] = \frac{L[f(t)]}{s+a} + \frac{y_0}{s+a}\) This verifies the first part of the problem statement. Now, we'll move on to show that y(t) satisfies the given equation.

Step 3: Find the Inverse Laplace Transform of L[y(t)]

We are given that: \(L[y(t)] = \frac{L[f(t)]}{s+a} + \frac{y_0}{s+a}\) To find y(t), we need to find the inverse Laplace transform of \(L[y(t)]\). We do that by finding the inverse Laplace transform of each term on the right-hand side of the equation: \(y(t) = \mathcal{L}^{-1}\left[\frac{L[f(t)]}{s+a}\right] + \mathcal{L}^{-1}\left[\frac{y_0}{s+a}\right]\) Using the convolution theorem, we can write the first term as: \(\mathcal{L}^{-1}\left[\frac{L[f(t)]}{s+a}\right] = \int_{0}^{t} e^{-a(t-w)}f(w)dw\) For the second term, we can use the fact that the inverse Laplace transform of \(\frac{1}{s+a}\) is \(e^{-at}\): \(\mathcal{L}^{-1}\left[\frac{y_0}{s+a}\right] = y_0 e^{-at}\) Now, we can add both terms to get the desired expression for y(t):

Step 4: Combine Terms to Get y(t)

Combining the results from step 3, we have: \(y(t) = y_0 e^{-at} + \int_{0}^{t} e^{-a(t-w)}f(w)dw\) This shows that y(t) satisfies the given equation, and we are done.

Key Concepts

Initial Value Problem

In mathematics, an Initial Value Problem (IVP) is related to solving differential equations where the solution must satisfy a given initial condition. It's like finding a path on a map that starts at a particular point. You know your starting position, and the differential equation dictates how you can move from there.
For instance, if we consider the equation:

• \( y'(t) + ay(t) = f(t) \)
• with the initial condition \( y(0) = y_0 \)

It implies that at time \( t = 0 \), the value of \( y \) is \( y_0 \). Using the Laplace Transform helps in converting the differential equations into an algebraic form in the frequency domain, which can be easier to manipulate. After simplifying, you can find the transformed solution \( L[y] \) and eventually inverse it back to the time domain. This allows us to understand how the system behaves over time starting from \( y_0 \).

Inverse Laplace Transform

The Inverse Laplace Transform is a crucial concept when dealing with differential equations. Once you've transformed a problem using the Laplace Transform, you often need to revert to understand the time-domain dynamics.
The key idea is to take a function in the Laplace domain and convert it back into a function of time. Mathematically, this is represented as an inverse operation:

• \( y(t) = \mathcal{L}^{-1}[L[y(t)]] \)

In our exercise:

• The term \( \mathcal{L}^{-1}\left[\frac{y_0}{s+a}\right] \) is simply \( y_0 e^{-at} \)
• The complex part \( \mathcal{L}^{-1}\left[\frac{L[f(t)]}{s+a}\right] \) is tackled using the Convolution Theorem, resulting in an integral expressing how past values of \( f(t) \) affect the current state.

Understanding these inversions is essential because they help translate solutions back to the time context from the frequency domain, making them applicable and interpretable in real-world scenarios.

Convolution Theorem

The Convolution Theorem is a powerful tool in the world of mathematical analysis, particularly for solving differential equations using the Laplace Transform.
This theorem connects the Laplace transform of a convolution of two functions with the product of their individual transforms. Essentially, it simplifies operations in the Laplace domain but still expresses results in terms of time-domain concepts:

• If \( F(s) = L[f(t)] \) and \( G(s) = L[g(t)] \), then \( \mathcal{L}^{-1}[F(s)G(s)] = f * g \)
• The convolution \( f * g \) represents an integral that combines the two functions: \( (f * g)(t) = \int_0^t f(t-w)g(w)dw \)

In our context, using the convolution helps us solve parts of the original differential equation by showcasing how the function \( f(t) \) modulates the response \( y(t) \) over time. By integrating the influence of \( f(t-w) \) over past intervals, this theorem provides a full picture of system dynamics and response behaviors.