Problem 13 - Differential Equations and Linear Algebra

Question

Determine \(L[f * g]\) $$f(t)=e^{t}, \quad g(t)=t e^{2 t}$$

Step-by-Step Solution

Answer

The Laplace transform of the convolution of the given functions is: $$L[f * g] = \frac{1}{(s - 1)(s - 2)^2} \qquad (s > 2)$$

Step 1: Recall the convolution theorem for Laplace transforms

The convolution theorem states that the Laplace transform of the convolution of two functions f(t) and g(t) is the product of their individual Laplace transforms, i.e., $$L[f * g] = L[f(t)] \cdot L[g(t)]$$

Step 2: Compute the Laplace transform of f(t) and g(t)

We know that \(f(t) = e^t\) and \(g(t) = te^{2t}\). We need to find their Laplace transforms, denoted by \(F(s)\) and \(G(s)\) respectively. For \(f(t) = e^t\), the Laplace transform is: $$F(s) = L[f(t)] = \int_0^\infty e^{-st}e^t dt = \int_0^\infty e^{(1-s)t} dt$$ The integral converges when \(s > 1\), and the result is: $$F(s) = \frac{1}{s - 1} \qquad (s > 1)$$ For \(g(t) = te^{2t}\), the Laplace transform is: $$G(s) = L[g(t)] = \int_0^\infty e^{-st}te^{2t} dt = \int_0^\infty te^{(2-s)t} dt$$ Here, we can use integration by parts. Letting \(u=t\) and \(dv=e^{(2-s)t} dt\), we have \(du=dt\) and \(v=\frac{1}{2-s}e^{(2-s)t}\). Applying integration by parts, the integral converges when \(s > 2\), and the result is: $$G(s) = \frac{1}{(s-2)^2} \qquad (s > 2)$$

Step 3: Apply the convolution theorem

Now that we have the Laplace transforms of f(t) and g(t), we can apply the convolution theorem to find the Laplace transform of the convolution of f(t) and g(t): $$L[f * g] = F(s) \cdot G(s) = \frac{1}{s - 1} \cdot \frac{1}{(s - 2)^2} \qquad (s > 2)$$ So, the Laplace transform of the convolution of the given functions is: $$L[f * g] = \frac{1}{(s - 1)(s - 2)^2} \qquad (s > 2)$$

Key Concepts

Laplace Transform

The Laplace Transform is a fundamental mathematical technique used to convert functions from the time domain into the s-domain (complex frequency domain). It is highly useful for solving differential equations and analyzing linear time-invariant systems.
This transform helps simplify complex functions by converting differential equations, which can be tough to solve, into algebraic equations, which are easier to handle.

• The basic formula for the Laplace Transform of a function \( f(t) \) is given by:\[ L[f(t)] = \int_0^\infty e^{-st} f(t) \, dt \]where \( s \) is a complex number.
• The Laplace Transform can only be applied to certain types of functions, specifically those that are piecewise continuous and of exponential order.
• In the problem, we computed the Laplace transforms of \( f(t) = e^t \) and \( g(t) = te^{2t} \).

Understanding the Laplace Transform is key to solving many engineering and physics problems, especially where initial conditions are involved.

Integration by Parts

Integration by parts is a technique used in calculus to integrate products of functions. This method is a straightforward application of the product rule for differentiation, which can simplify integrals that might not otherwise be solvable using standard techniques.
The formula for integration by parts is:

• \[ \int u \, dv = uv - \int v \, du \]
• Here, \( u \) and \( dv \) are parts of the integrand chosen such that the resulting \( \int v \, du \) is simpler to integrate than the original \( \int u \, dv \).
• In our problem, this method was applied to find the Laplace Transform of \( g(t) = te^{2t} \), where \( u = t \) and \( dv = e^{(2-s)t} dt \).

Integration by parts is particularly useful when dealing with products of polynomial and exponential functions, as seen in the problem. It requires practice to choose \( u \) and \( dv \) effectively, but once mastered, it can greatly simplify complex integrals.

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are used to describe various physical phenomena such as heat, motion, electricity, and waves.
Differential equations can be broadly categorized into:

• Ordinary Differential Equations (ODEs): Equations containing functions of one independent variable and their derivatives.
• Partial Differential Equations (PDEs): Equations involving partial derivatives of functions of multiple independent variables.

In the given problem, solving differential equations would ultimately require knowing their Laplace transforms or using techniques like convolution theorem, which we did when finding \( L[f * g] \). This showcases how solving differential equations often involves intermediate techniques like Laplace transforms and integration by parts.
Understanding differential equations is crucial for modeling and analyzing real-world systems, and the Laplace Transform offers a powerful tool in this analysis by translating the problems to algebraic ones, making them much more tractable.