Problem 42 - Advanced Engineering Mathematics

Question

Use the Laplace transform to solve the given integral equation or integrodifferential equation. $$ f(t)=\cos t+\int_{0}^{t} e^{-\tau} f(t-\tau) d \tau $$

Step-by-Step Solution

Answer

The solution is \( f(t) = \cos t + \sin t \).

Step 1: Identify the components of the equation

The given equation is \( f(t) = \cos t + \int_{0}^{t} e^{-\tau} f(t-\tau) d\tau \). It includes the function \( f(t) \), the integral \( \int_{0}^{t} e^{-\tau} f(t-\tau) d\tau \), and the function \( \cos t \). The goal is to solve for \( f(t) \) using the Laplace transform.

Step 2: Apply the Laplace Transform

Apply the Laplace transform to both sides of the equation to transform it into a simpler algebraic equation:\[ \mathcal{L}\{ f(t) \} = \mathcal{L}\{ \cos t \} + \mathcal{L}\left\{ \int_{0}^{t} e^{-\tau} f(t-\tau) d\tau \right\} \].Using the linearity of the Laplace transform, this becomes:\[ F(s) = \frac{s}{s^2 + 1} + \mathcal{L}\{e^{-t} * f(t)\} \]

Step 3: Laplace Transform of Convolution

Utilize the convolution theorem: the Laplace transform of a convolution \( f(t) * g(t) \) is the product of their Laplace transforms. In this case, \( g(t) = e^{-t} \) with \( \mathcal{L}\{e^{-t}\} = \frac{1}{s+1} \). Hence, \( \mathcal{L}\{e^{-t} * f(t)\} = F(s) \cdot \frac{1}{s+1} \).

Step 4: Solve for F(s)

Substitute the known transforms into the equation from Step 2:\[ F(s) = \frac{s}{s^2 + 1} + F(s) \cdot \frac{1}{s+1} \].Rearrange to:\[ F(s) \left( 1 - \frac{1}{s+1} \right) = \frac{s}{s^2 + 1} \].Simplify to:\[ F(s) \left( \frac{s}{s+1} \right) = \frac{s}{s^2 + 1} \].Then, \[ F(s) = \frac{s(s+1)}{s(s+1)^2 + (s+1)} \].

Step 5: Simplify F(s)

Upon simplification of the denominator, \( s(s+1)^2 + (s+1) \) becomes \((s+1)(s^2 + 1)\). Simplify the Laplace transform to:\[ F(s) = \frac{s+1}{s^2 + 1} \] after cancelation.

Step 6: Inverse Laplace Transform

Take the inverse Laplace transform to find \( f(t) \). Since \( F(s) = \frac{s+1}{s^2 + 1} \), split into two fractions:\[ F(s) = \frac{s}{s^2 + 1} + \frac{1}{s^2 + 1} \].The inverse Laplace transforms are \( \mathcal{L}^{-1}\{ \frac{s}{s^2 + 1} \} = \cos t \) and \( \mathcal{L}^{-1}\{ \frac{1}{s^2 + 1} \} = \sin t \).Thus, \( f(t) = \cos t + \sin t \).

Key Concepts

Integral Equations

Integral equations are fascinating mathematical expressions that involve unknown functions under an integral sign. Essentially, they allow us to express a function in terms of an integral, which often involves itself. Consider this equation:

• It involves both the function we seek, like \( f(t) \),
• And a kernel function, which in our case, is \( e^{-\tau} \).

Here, the integral equation is given by:\[ f(t) = \cos t + \int_{0}^{t} e^{-\tau} f(t-\tau) d\tau \].
The goal is to determine the function \( f(t) \). By using the Laplace transform, which is a powerful tool to convert such integral expressions into algebraic equations, we greatly simplify our problem.

Integrodifferential Equations

Integrodifferential equations are intriguing since they blend the elements of differential and integral equations. In these equations, the unknown function not only satisfies differential relations but is also linked to integral terms. Our original exercise was to solve:

• A blend of integral terms \(\int_{0}^{t} e^{-\tau} f(t-\tau) d\tau \)
• And differential terms represented by \( f(t) \)

The presence of both integral and differential components makes the solution challenging and interesting.
In our case, using the Laplace transform helps to decouple and solve these mixed equations by transforming them into an algebraic form, as shown in the step-by-step solution. We applied the Laplace transform to both the integral and standard terms to simplify the problem.

Convolution Theorem

The convolution theorem is a cornerstone concept when working with integral equations like the one in our exercise. It simplifies the Laplace transform of an integral involving two functions by expressing it as the product of their individual Laplace transforms.

• Consider the convolution of functions \( f(t) \) and \( g(t) = e^{-t} \).
• Using the convolution theorem, we find \( \mathcal{L}\{e^{-t} * f(t)\} = F(s) \cdot \frac{1}{s+1} \).

This translates a convolution integral into a simpler multiplication in the s-domain. This is invaluable as it turns a complex integral problem into a more approachable algebraic troubleshooting task.
In the solution, we could effectively use this theorem to manipulate and solve the Laplace transformed equation.

Inverse Laplace Transform

The inverse Laplace transform is vital when converting solutions back into the time domain. Once we transform a given integral or integrodifferential equation using the Laplace transform technique, we receive an expression in terms of \( F(s) \). However, to fully solve for \( f(t) \), we need to return to the time domain by using its inverse.

• In our exercise, after simplifying to \( F(s) = \frac{s+1}{s^2 + 1} \)
• We separate it: \( F(s) = \frac{s}{s^2 + 1} + \frac{1}{s^2 + 1} \)

We recognize these terms from Laplace tables:

• \( \mathcal{L}^{-1}\{ \frac{s}{s^2 + 1} \} = \cos t \)
• \( \mathcal{L}^{-1}\{ \frac{1}{s^2 + 1} \} = \sin t \)

Hence, the solution in the time domain is \( f(t) = \cos t + \sin t \). The ability to seamless switch between mathematical worlds (transform and inverse) makes the Laplace transform a powerful tool in solving complex problems.