Question

Find the Laplace transform of each of the following expressions: (a) \(2 t \mathrm{e}^{2 t}\) (b) \(1-t^{3} \mathrm{e}^{3 t}\) (c) \(\mathrm{e}^{t}(1+\sin t)\) (d) \(\mathrm{e}^{-2 t}(\sin 3 t+2 \cos 3 t)\) (e) \(t(\cos 2 t-3 \sin 2 t)\)

Short answer

#Laplace Transform Results# (a) L{2t e^(2t)} = \(\frac{2}{(s-2)^2}\) (b) L{1 - t^3 e^(3t)} = \(\frac{1}{s} - \frac{3!}{(s-3)^4}\) (c) L{e^t(1+sin(t))} = \(\frac{1}{s-1} + \frac{1}{(s-1)^2 + 1}\) (d) L{e^(-2t)(sin(3t) + 2cos(3t))} = \(\frac{3}{(s+2)^2 + 3^2} + 2\frac{s+2}{(s+2)^2 + 3^2}\) (e) L{t(cos(2t) - 3sin(2t))} = \(\frac{s^2 + 4}{(s^2 + 2^2)^2} - 3\frac{2s}{(s^2 +2^2)^2}\)

Step-by-step solution

Laplace transform of (a) 2t e^(2t)

To find the Laplace transform of f(t) = 2t e^(2t), use the Laplace transform formula: L{2t e^(2t)} = \(\int_{0}^{\infty} e^{-st} (2t e^{2t}) dt\) Use integration by parts, where u = 2t and dv = e^{(2-s)t} dt: L{2t e^(2t)} = \(\int_{0}^{\infty} u dv = [uv]_0^{\infty} - \int_{0}^{\infty} v du\) Now integrate: L{2t e^(2t)} = \([\frac{2t}{s-2}e^{(2-s)t}]_0^{\infty} - \int_{0}^{\infty} \frac{2}{s-2}e^{(2-s)t} dt\) Evaluate the first term: L{2t e^(2t)} = \(\lim_{t\to\infty}[\frac{2t}{s-2}e^{(2-s)t}] - [\frac{2t}{s-2}e^{(2-s)(0)}] - \frac{2}{s-2}\int_{0}^{\infty} e^{(2-s)t} dt\) L{2t e^(2t)} = \(\lim_{t\to\infty}[\frac{2t}{s-2}e^{(2-s)t}] - \frac{2}{s-2}\int_{0}^{\infty} e^{(2-s)t} dt\) Since the limit is 0 (because s > 2 for convergence), the Laplace transform is: L{2t e^(2t)} = \(\frac{2}{(s-2)^2}\)

Laplace transform of (b) 1-t^3 e^(3t)

To find the Laplace transform of f(t) = 1-t^3 e^(3t), use the linearity property of the Laplace transform, which states the Laplace transform of a linear combination of functions is equal to the linear combination of their Laplace transforms: L{1 - t^3 e^(3t)} = L{1} - L{t^3 e^(3t)} First, find the Laplace transform of 1: L{1} = \(\int_{0}^{\infty} e^{-st} dt = [-\frac{1}{s}e^{-st}]_0^{\infty} = \frac{1}{s}\) Now, find the Laplace transform of t^3 e^(3t) using integration by parts: L{t^3 e^(3t)} = \(\int_{0}^{\infty} t^3 e^{(3-s)t} dt\) We can apply integration by parts three times or use the property of Laplace transform of higher derivatives. From the Laplace transform table: L{t^n e^{at}} = \(\frac{n!}{(s-a)^{n+1}}\) In the given case, we have: L{t^3 e^(3t)} = \(\frac{3!}{(s-3)^4}\) Now substitute the obtained Laplace transforms back into the linear combination: L{1 - t^3 e^(3t)} = \(\frac{1}{s} - \frac{3!}{(s-3)^4}\)

Laplace transform of (c) e^t(1+sin(t))

To find the Laplace transform of f(t) = e^t(1+sin(t)), use the linearity property of the Laplace transform and the Laplace transform of the product of an exponential function and another function: L{e^t(1+sin(t))} = L{e^t} + L{e^t sin(t)} Use the Laplace transform table: L{e^t} = \(\frac{1}{s-1}\) L{e^t sin(t)} = \(\frac{1}{(s-1)^2 + 1}\) Now substitute the obtained Laplace transforms back into the linear combination: L{e^t(1+sin(t))} = \(\frac{1}{s-1} + \frac{1}{(s-1)^2 + 1}\)

Laplace transform of (d) e^(-2t)(sin(3t) + 2cos(3t))

To find the Laplace transform of f(t) = e^(-2t)(sin(3t) + 2cos(3t)), use the linearity property of the Laplace transform and the Laplace transform of the product of an exponential function and another function: L{e^(-2t)(sin(3t) + 2cos(3t))} = L{e^(-2t) sin(3t)} + 2L{e^(-2t) cos(3t)} Use the Laplace transform table: L{e^(-2t) sin(3t)} = \(\frac{3}{(s+2)^2 + 3^2}\) L{e^(-2t) cos(3t)} = \(\frac{s+2}{(s+2)^2 + 3^2}\) Now substitute the obtained Laplace transforms back into the linear combination: L{e^(-2t)(sin(3t) + 2cos(3t))} = \(\frac{3}{(s+2)^2 + 3^2} + 2\frac{s+2}{(s+2)^2 + 3^2}\)

Laplace transform of (e) t(cos(2t) - 3sin(2t))

To find the Laplace transform of f(t) = t(cos(2t) - 3sin(2t)), use the linearity property of the Laplace transform: L{t(cos(2t) - 3sin(2t))} = L{t cos(2t)} - 3L{t sin(2t)} Use the Laplace transform table: L{t cos(2t)} = \(\frac{s^2 + 4}{(s^2 + 2^2)^2}\) L{t sin(2t)} = \(\frac{2s}{(s^2 + 2^2)^2}\) Now substitute the obtained Laplace transforms back into the linear combination: L{t(cos(2t) - 3sin(2t))} = \(\frac{s^2 + 4}{(s^2 + 2^2)^2} - 3\frac{2s}{(s^2 +2^2)^2}\)

Key concepts

Laplace Transform Formula

The Laplace transform is an integral transform often used to solve differential equations. The formula for the Laplace transform of a function f(t), which is a function of time, into a function F(s), which is a function of complex frequency s, is given by:

\[\begin{equation}L\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt\end{equation}\]
The integral will converge when the real part of s is greater than a certain number, often related to the growth behavior of f(t). The idea is to transform the time domain into the complex frequency domain, where many problems become easier to solve. For example, in the exercise to find the Laplace transform of 2t e^{2t}, we start by setting up the integral using the formula and evaluate it accordingly.

Integration by Parts

Integration by parts is a technique that comes from the product rule of differentiation and is used to solve integrals involving the product of two functions. The formula for integration by parts is given by:

\[\begin{equation}\int u \, dv = uv - \int v \, du\end{equation}\]
Where u and dv are chosen from the original integral such that the new integral \int v \, du is simpler to solve. In our exercise to find the Laplace transform of 2t e^{2t}, we chose u=2t and dv=e^{(2-s)t} dt. The technique is particularly helpful in the context of Laplace transforms, as it frequently simplifies the transformation of polynomials times exponentials.

Linearity Property of Laplace Transform

The linearity property is a key aspect of the Laplace transform that makes it a powerful tool for solving linear differential equations. It states that the Laplace transform of a sum of functions is equal to the sum of their individual Laplace transforms. In mathematical terms:

\[\begin{equation}L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}\end{equation}\]
where a and b are constants. This property was used in our exercise, such as in the case of L\{1 - t^3 e^{3t}\}, where we applied the Laplace transforms separately to the constant 1 and to the term t^3 e^{3t} and then combined the results.

Laplace Transform Table

The Laplace transform table is a comprehensive list of functions and their corresponding Laplace transforms. It's an incredibly useful reference that saves time and simplifies the process of finding the Laplace transform of many common functions. For instance, the table provides transforms for basic functions, such as exponentials, trigonometric functions, and polynomials, which allows for quick lookup rather than manually integrating every time. In the provided solutions, the Laplace transform table was employed to transform expressions like t^n e^{at} and trigonometric functions multiplied by exponentials, enabling us to write the Laplace transforms of more complex functions as simple algebraic expressions in terms of s.