Question

Find the Laplace transform of each of the following functions: (a) \(t^{n}\), (g) \(e^{a t} \cos \omega t .\) (b) \(\cos \omega t\). (h) \(t \sinh k t\). (c) \(\sinh k t\). (i) \(t \cosh k t\). (d) \(\cosh k t\). (j) \(\sqrt{t^{-1}}\). (e) \(t e^{a t}\). (k) \(\sqrt{t}\). (f) \(e^{a t} \sin \omega t\). (l) \((\sin a t) / t .\)

Short answer

Question: Find the Laplace transform of the following functions: (a) \(t^n\) (b) \(\cos{\omega t}\) Answer: (a) The Laplace transform of \(t^n\) is \(\frac{n!}{s^{n+1}}\). (b) The Laplace transform of \(\cos(\omega t)\) is \(\frac{s}{s^2 + \omega^2}\).

Step-by-step solution

(a) Laplace transform of \(t^n\)

To find the Laplace transform of \(t^n\), use the Laplace transform definition: $$F(s) = \int_{0}^{\infty} e^{-st}t^n dt$$ Use integration by parts with \(u = t^n\) and \(dv = e^{-st} dt\). Then \(du = n t^{n-1} dt\) and \(v = -\frac{1}{s} e^{-st}\). Applying integration by parts: $$F(s) = \left[-\frac{1}{s} e^{-st}t^n\right]_{0}^{\infty} - \int_{0}^{\infty}-\frac{n}{s} e^{-st}t^{n-1} dt$$ While the first term evaluates to zero, we get: $$F(s) = \frac{n}{s} \int_{0}^{\infty} e^{-st}t^{n-1} dt$$ We see that the remaining integral is the Laplace transform of \(t^{n-1}\). We can continue this process for n times until we get: $$F(s) = \frac{n!}{s^{n+1}}L[t^0]=\frac{n!}{s^{n+1}}$$ Thus, the Laplace transform of \(t^n\) is \(\frac{n!}{s^{n+1}}\).

(b) Laplace transform of \(\cos{\omega t}\)

To find the Laplace transform of \(\cos{\omega t}\), use the Laplace transform definition: $$F(s) = \int_{0}^{\infty} e^{-st} \cos{\omega t} dt$$ Use integration by parts with \(u = \cos(\omega t)\) and \(dv = e^{-st} dt\). Then \(du = -\omega \sin(\omega t) dt\) and \(v = -\frac{1}{s} e^{-st}\). Applying integration by parts: $$F(s) = \left[-\frac{1}{s} e^{-st}\cos(\omega t)\right]_{0}^{\infty} + \frac{\omega}{s} \int_{0}^{\infty}e^{-st}\sin(\omega t) dt$$ The limit of the first term goes to zero for s > 0. Again, apply integration by parts to the remaining integral: $$F(s) = \frac{\omega}{s} \left[-\frac{1}{s} e^{-st}\sin(\omega t)\right]_{0}^{\infty} - \frac{\omega}{s} \int_{0}^{\infty}-\frac{\omega}{s} e^{-st} \cos(\omega t) dt$$ The first term evaluates to zero. We have: $$F(s) = \frac{\omega^2}{s^2} L[\cos(\omega t)]$$ Comparing both sides, we get: $$L[\cos(\omega t)] = \frac{s}{s^2 + \omega^2}$$ Thus, the Laplace transform of \(\cos(\omega t)\) is \(\frac{s}{s^2 + \omega^2}\). The solutions for the other functions can be found using a similar approach by applying the Laplace transform definition and, in some cases, using integration by parts. The final Laplace transform results are: (c) \(L[\sinh(kt)] = \frac{k}{s^2 - k^2}\) (d) \(L[\cosh(kt)] = \frac{s}{s^2 - k^2}\) (e) \(L[t e^{at}] = \frac{1}{(s-a)^2}\) (f) \(L[e^{at}\sin(\omega t)] = \frac{\omega}{(s - a)^2 + \omega^2}\) (g) \(L[e^{at}\cos(\omega t)] = \frac{s-a}{(s - a)^2 + \omega^2}\) (h) \(L[t \sinh(kt)] = \frac{2ks}{(s^2 - k^2)^2}\) (i) \(L[t \cosh(kt)] = \frac{s^2 + k^2}{(s^2 - k^2)^2}\) (j) \(L[\sqrt{t^{-1}}] = \sqrt{\frac{\pi}{s}}\) (k) \(L[\sqrt{t}] = \frac{\sqrt{\pi}}{2s^{\frac{3}{2}}}\) (l) \(L[\frac{\sin(at)}{t}] = \arctan\frac{a}{s}\)

Key concepts

Integration by Parts

Integration by parts is a fundamental technique in calculus used for finding integrals of products of functions. It's especially useful when dealing with problems involving Laplace transforms. The main idea involves breaking an integral into simpler parts, making it easier to solve. This process typically involves:

• Choosing one function to differentiate (usually the simpler one), called \(u\).
• Choosing another function to integrate, called \(dv\).

The integration by parts formula is expressed as:\[\int u\, dv = uv - \int v\, du\]Here, you have a trade-off: by taking a derivative of \(u\) to form \(du\) and integrating \(dv\) to form \(v\), you potentially simplify the integration problem.
A classic situation where this is applied is with the Laplace transform of \(t^n \cdot e^{-st}\), as demonstrated in the original problem. When \(u = t^n\) and \(dv = e^{-st}dt\), the integration by parts aids in decomposing the original complex integral into simpler ones.

Differential Equations

Differential equations form the cornerstone of modeling in sciences and engineering. These equations involve derivatives, which represent rates of change. Solving them gives insights into dynamic processes. The Laplace transform is a powerful tool here.

• It transforms a differential equation into an algebraic equation.
• This transformation simplifies the solving process.

The transform method is particularly useful for linear systems subject to initial conditions. Applying it allows us to convert complex time-domain differential equations into simpler s-domain equations.
Consider a simple differential equation: \(y' + y = 0\), where \(y\) is a function of time.Using the Laplace transform, you can convert this into an equation in terms of \(s\), simplifying the process significantly. For higher-order differential equations, the Laplace transform can be even more advantageous, representing derivatives as simple algebraic expressions.

Cosine and Sine Functions

Cosine and sine functions are fundamental in trigonometry and appear frequently in physical phenomena, especially those involving oscillations. When dealing with Laplace transforms, these functions have simple transforms, representing their oscillatory nature.

• Cosine functions: \(\cos(\omega t)\)
• Sine functions: \(\sin(\omega t)\)

Their respective Laplace transforms are:

• \(L[\cos(\omega t)] = \frac{s}{s^2 + \omega^2}\)
• \(L[\sin(\omega t)] = \frac{\omega}{s^2 + \omega^2}\)

These transforms simplify initial value problems in calculus, turning complex oscillatory behavior into simpler expressions in the s-domain. Such transformations are crucial in engineering and physics, where wave-like behaviors are common.

Exponential Functions

Exponential functions, of the form \(e^{at}\), are vital in various fields due to their growth and decay properties. In the context of Laplace transforms, they often simplify the handling of differential equations.

• Exponential functions have a straightforward Laplace transform: \(L[e^{at}] = \frac{1}{s - a}\).
• This transform captures the essence of exponential growth or decay in a system.

When combining exponentials with trigonometric functions, like \(e^{at} \cos(\omega t)\) or \(e^{at} \sin(\omega t)\), their Laplace transforms capture the oscillations and growth simultaneously.
Understanding these transforms is essential for tackling problems in control systems and electrical engineering, where exponential behaviors frequently arise. Notably, exponential functions in Laplace transforms allow us to analyze systems' stability by examining their poles and zeros.