Question

Use the table of Laplace transforms to find the Laplace transforms of the following functions. (a) \(\cosh t \sin t\) (b) \(\sin ^{2} t\) (c) \(\cos ^{2} 2 t\) (d) \(\cosh ^{2} t\) (e) \(t \sinh 2 t\) (f) \(\sin t \cos t\) (g) \(\sin \left(t+\frac{\pi}{4}\right)\) (h) \(\cos 2 t-\cos 3 t\) (i) \(\sin 2 t+\cos 4 t\)

Short answer

Short answer: The Laplace transforms of the given functions are as follows: (a) \(\mathcal{L}\{\cosh t \sin t\} = \frac{1}{2}\left(\frac{1}{(s-1)^2+1^2} - \frac{1}{(s+1)^2+1^2}\right)\) (b) \(\mathcal{L}\{\sin^2 t\} = \frac{1}{2s} - \frac{s}{4(s^2 + 4)}\) (c) \(\mathcal{L}\{\cos^2 2t\} = \frac{1}{2s} + \frac{s}{4(s^2 + 16)}\) (d) \(\mathcal{L}\{\cosh^2 t\} = \frac{s^2+2s+1}{2(s^2+1)(s+1)}\) (e) \(\mathcal{L}\{t\sinh 2t\} = \frac{4}{(s^2-4)^2}\) (f) \(\mathcal{L}\{\sin t \cos t\} = \frac{1}{2s^2+4}\) (g) \(\mathcal{L}\{\sin\left(t+\frac{\pi}{4}\right)\} = \frac{\sqrt{2}s}{s^2+1} - \frac{\sqrt{2}}{s^2+1}\) (h) \(\mathcal{L}\{\cos 2t - \cos 3t\} = \frac{s}{s^2+4} - \frac{s}{s^2+9}\) (i) \(\mathcal{L}\{\sin 2t + \cos 4t\} = \frac{2}{s^2 + 4} + \frac{s}{s^2 + 16}\)

Step-by-step solution

Multiply the function by e^(-st)

Let's multiply the given function by \(e^{-st}\): \(e^{-st}\cosh t \sin t\)

Use the property of Laplace Transform

Using the Laplace Transform's multiplication property: \(\mathcal{L}\{\cosh t \sin t\} = \frac{1}{2}(\mathcal{L}\{e^{t}\sin t\} - \mathcal{L}\{e^{-t}\sin t\})\) Now, let's find the Laplace transform of \(e^{t}\sin t\) and \(e^{-t}\sin t\) using the table. Using the Laplace transform of \(e^{at}\sin(bt)\), we get: \(\mathcal{L}\{e^{t}\sin t\} = \frac{b}{(s-a)^2+b^2}=\frac{1}{(s-1)^2+1^2}\) \(\mathcal{L}\{e^{-1}\sin t\} = \frac{b}{(s-a)^2+b^2}=\frac{1}{(s+1)^2+1^2}\)

Substitute back and simplify

Now, substitute these transforms back into the equation and simplify: \(\mathcal{L}\{\cosh t \sin t\} = \frac{1}{2}\left(\frac{1}{(s-1)^2+1^2} - \frac{1}{(s+1)^2+1^2}\right)\) (b) \(\sin ^{2} t\)

Use the trigonometric identity

To find the Laplace transform of \(\sin^2 t\), we first need to use the trigonometric identity: \(\sin^2 t = \frac{1-\cos(2t)}{2}\) Now, find the Laplace transform of the RHS. \(\mathcal{L}\{\sin^2 t\} = \mathcal{L}\{\frac{1}{2}-\frac{\cos(2t)}{2}\}\)

Use the properties of Laplace transform

Using the Laplace transform properties, we get: \(\mathcal{L}\{\frac{1}{2}\} - \mathcal{L}\{\frac{\cos(2t)}{2}\}\) Now we find the Laplace transforms for each term: \(\mathcal{L}\{\frac{1}{2}\} = \frac{1}{2s}\) \(\mathcal{L}\{\frac{\cos(2t)}{2}\} = \frac{s}{4(s^2 + 2^2)}\)

Combine the terms

Now, combine the transforms: \(\mathcal{L}\{\sin^2 t\} = \frac{1}{2s} - \frac{s}{4(s^2 + 4)}\) (c) \(\cos ^{2} 2 t\)

Use the trigonometric identity

To find the Laplace transform of \(\cos^2 2t\), we first need to use the trigonometric identity: \(\cos^2 2t = \frac{1+\cos(4t)}{2}\) Now, find the Laplace transform of the RHS. \(\mathcal{L}\{\cos^2 2t\} = \mathcal{L}\{\frac{1}{2}+\frac{\cos(4t)}{2}\}\)

Use the properties of Laplace transform

Using the Laplace transform properties, we get: \(\mathcal{L}\{\frac{1}{2}\} + \mathcal{L}\{\frac{\cos(4t)}{2}\}\) Now we find the Laplace transforms for each term: \(\mathcal{L}\{\frac{1}{2}\} = \frac{1}{2s}\) \(\mathcal{L}\{\frac{\cos(4t)}{2}\} = \frac{s}{4(s^2 + 4^2)}\)

Combine the terms

Now, combine the transforms: \(\mathcal{L}\{\cos^2 2t\} = \frac{1}{2s} + \frac{s}{4(s^2 + 16)}\) I will now show the Laplace transforms for the remaining functions (d) to (i) in less detail. (d) \(\cosh ^{2} t\) \(\mathcal{L}\{\cosh^2 t\} = \frac{s^2+2s+1}{2(s^2+1)(s+1)}\) (e) \(t \sinh 2 t\) \(\mathcal{L}\{t\sinh 2t\} = \frac{4}{(s^2-4)^2}\) (f) \(\sin t \cos t\) \(\mathcal{L}\{\sin t \cos t\} = \frac{1}{2s^2+4}\) (g) \(\sin \left(t+\frac{\pi}{4}\right)\) \(\mathcal{L}\{\sin\left(t+\frac{\pi}{4}\right)\} = \frac{\sqrt{2}s}{s^2+1} - \frac{\sqrt{2}}{s^2+1}\) (h) \(\cos 2 t-\cos 3 t\) \(\mathcal{L}\{\cos 2t - \cos 3t\} = \frac{s}{s^2+4} - \frac{s}{s^2+9}\) (i) \(\sin 2 t+\cos 4 t\) \(\mathcal{L}\{\sin 2t + \cos 4t\} = \frac{2}{s^2 + 4} + \frac{s}{s^2 + 16}\)

Key concepts

Laplace Transform Properties

Understanding the properties of the Laplace transform is crucial in solving differential equations and analyzing systems in the engineering and scientific disciplines. One of these key properties is linearity, which states that the Laplace transform of a sum is the sum of the Laplace transforms, and scaling by a constant simply scales the transform. For example, if you have a function like \(3f(t) + 5g(t)\), its Laplace transform would be \(3\mathcal{L}\{f(t)\} + 5\mathcal{L}\{g(t)\}\).

Another important property is the frequency shifting or modulation property. This property is particularly useful when dealing with exponentials multiplied by functions, such as \(e^{at}f(t)\). The effect is to shift the Laplace transform's argument by \(a\), directly applied in the provided solution. Furthermore, the differentiation and integration properties help with transforming derivatives and integrals of functions, streamlining the process of solving differential equations.

Using these properties allows students to break down complex functions into simpler parts that are easier to transform. Moreover, they make the use of Laplace transforms practical for a wide variety of problems, extending their application far beyond just solving textbook exercises.

Trigonometric Identities

Trigonometric identities are relationships between trigonometric functions that are true for all values of the variable for which the functions are defined. They are used to simplify expressions and solve equations that involve trigonometric functions. For instance, the identity \(\sin^2 t + \cos^2 t = 1\) is fundamental in trigonometry and appears frequently in various forms, such as in the solutions for \(\sin^2 t\) and \(\cos^2 t\), where it was transformed into \(\frac{1}{2} - \frac{1}{2}\cos(2t)\) and \(\frac{1}{2} + \frac{1}{2}\cos(4t)\), respectively.

These identities are particularly useful when combined with Laplace transforms. Applying the identities helps to simplify the functions before taking their Laplace transform, making the process more manageable and leading to more straightforward solutions. When students understand these identities, they equip themselves with powerful tools to tackle a vast array of problems in calculus, physics, and engineering.

Cosh and Sinh Functions

The hyperbolic functions cosh (hyperbolic cosine) and sinh (hyperbolic sine) are analogs of the trigonometric functions, but for a hyperbola rather than a circle. The definitions are as follows: \(\cosh x = \frac{e^{x} + e^{-x}}{2}\) and \(\sinh x = \frac{e^{x} - e^{-x}}{2}\).

These functions share many properties with their trigonometric counterparts. For example, they also have identities similar to the trigonometric ones, such as \(\cosh^2 x - \sinh^2 x = 1\), which is analogous to the Pythagorean identity in trigonometry.

In the context of Laplace transforms, the hyperbolic sine and cosine can be expressed as a combination of exponential functions, thus simplifying the process of finding their transforms. As shown in the step-by-step solution, \(\cosh t \sin t\) was transformed using the exponential definitions of \(\cosh\) and \(\sinh\), resulting in a manageable expression to apply Laplace transforms. Recognizing how to manipulate and transform these hyperbolic functions is vital for students working in areas such as control theory, signal processing, and the solutions of differential equations where these functions frequently occur.