Laplace Transform of cosh3(2t): A Step-by-Step Guide

In this article, we will explore the process of finding the Laplace transform of the function cosh3(2t). The Laplace transform has numerous applications in engineering and physics, particularly in solving differential equations and analyzing systems. Let's break down the process into manageable steps.

Step 1: Express cosh(2t)

The hyperbolic cosine function, denoted cosh(x), is defined as:

cosh(2t) frac{e^{2t} e^{-2t}}{2}

Step 2: Use the Binomial Theorem

We need to express cosh3(2t) in a form that is easier to transform. We start by expanding it using the binomial theorem:

cosh3(2t) left(frac{e^{2t} e^{-2t}}{2}right)^3

Expanding this, we get:

cosh3(2t) frac{1}{8} (e^{6t} 3e^{4t} 3e^{2t} 1 3e^{-2t} 3e^{-4t} e^{-6t})

Step 3: Apply the Laplace Transform

The Laplace transform, denoted by L{f(t)}, is defined as:

L{f(t)} ∫0∞ e-st f(t) dt

The Laplace transform for exponential functions is given by:

L{eat} frac{1}{s - a} (for s > a)

Step 4: Compute the Laplace Transforms

Using the above definition, we can find the Laplace transforms of each term:

L{e6t} frac{1}{s - 6}

L{e4t} frac{1}{s - 4}

L{e2t} frac{1}{s - 2}

L{e-2t} frac{1}{s 2}

L{e-4t} frac{1}{s 4}

L{e-6t} frac{1}{s 6}

Step 5: Combine the Results

We now combine these results to find the Laplace transform of cosh3(2t):

L{cosh3(2t)} frac{1}{8} left[ frac{1}{s - 6} frac{3}{s - 4} frac{3}{s - 2} frac{1}{s} frac{3}{s 2} frac{3}{s 4} frac{1}{s 6} right]

Final Result

The Laplace transform of cosh3(2t) is thus:

L{cosh3(2t)} frac{1}{8} left[ frac{1}{s - 6} frac{3}{s - 4} frac{3}{s - 2} frac{1}{s} frac{3}{s 2} frac{3}{s 4} frac{1}{s 6} right]

This expression can be further simplified, but this is the complete transformation.

Alternative Approach

An alternative approach is to use the known Laplace transform of cosh(2t) and properties of the Laplace transform. Recall that:

L{cosh(2t)} frac{s}{s^2 - 4}

Since we are dealing with cosh3(2t), we can express it as:

cosh3(2t) frac{1}{8} (e^{-6t} 3e^{-4t} 3e^{-2t} 1 3e^{2t} 3e^{4t} e^{6t})

The Laplace transform of each term is:

L{e^{-6t}} frac{1}{s 6}

L{e^{-4t}} frac{1}{s 4}

L{e^{-2t}} frac{1}{s 2}

L{1} frac{1}{s}

L{e^{2t}} frac{1}{s - 2}

L{e^{4t}} frac{1}{s - 4}

L{e^{6t}} frac{1}{s - 6}

Substituting these into the original expression:

L{cosh3(2t)} frac{1}{8} left( frac{1}{s 6} frac{3}{s 4} frac{3}{s 2} frac{1}{s} frac{3}{s - 2} frac{3}{s - 4} frac{1}{s - 6} right)

The result can be further simplified to:

L{cosh3(2t)} frac{1}{8} left( frac{1}{s 6} frac{3}{s 4} frac{3}{s 2} frac{1}{s} frac{3}{s - 2} frac{3}{s - 4} frac{1}{s - 6} right)

frac{s(s^2 - 28)}{(s^4 - 40s^2 144)}

Conclusion

Thus, we have derived the Laplace transform of cosh3(2t) using both the direct approach and the alternative method, providing a comprehensive understanding of the process.

Final Answer