Introduction to Finding the Laplace Transform of ( f(t) 2t cdot 4e^{2t} cos(t) )

The Laplace transform is a powerful tool in engineering and mathematics, particularly for solving linear differential equations. In this article, we will walk through the process of finding the Laplace transform of the function ( f(t) 2t cdot 4e^{2t} cos(t) ) using the complex exponentials method and the shifting theorem. We will also demonstrate how to use the multiplication by ( t ) rule.

Method 1: Using Complex Exponentials

To find the Laplace transform of ( f(t) 2t cdot 4e^{2t} cos(t) ), we can convert the cosine term into complex exponentials using the Euler's formula:

[ f(t) 2t cdot 4e^{2t} cos(t) 8te^{2t} left( frac{e^{it} e^{-it}}{2} right) ]

Simplifying, we get:

[ f(t) 4te^{2t} (e^{it} e^{-it}) ]

Now, we can find the Laplace transform of ( e^{2t} e^{it} ) and ( e^{2t} e^{-it} ) separately. Using the properties of Laplace transforms:

[ mathcal{L}{e^{2t} e^{it}} frac{s-2i}{(s-2)^2 1} quad text{and} quad mathcal{L}{e^{2t} e^{-it}} frac{s 2i}{(s-2)^2 1} ]

Adding these two transforms together, we get:

[ mathcal{L}{4te^{2t} (e^{it} e^{-it})} 4 left( frac{s-2i}{(s-2)^2 1} frac{s 2i}{(s-2)^2 1} right) cdot frac{t}{1} ]

Further simplification gives:

[ mathcal{L}{4te^{2t} (e^{it} e^{-it})} frac{8(s-2i s 2i)}{(s-2)^2 1} frac{16s}{(s-2)^2 1} ]

Method 2: Using the Shifting Theorem and Multiplication by ( t )

We can also use the shifting theorem and multiplication by ( t ) rule to find the Laplace transform of ( f(t) ).

[ f(t) 4 left( te^{2t} cos(t) 2e^{2t} cos(t) right) ]

Using the shifting theorem for ( e^{2t} cos(t) ):

[ mathcal{L}{e^{2t} cos(t)} frac{s-2}{(s-2)^2 1} ]

Applying the shifting theorem and the multiplication by ( t ) rule:

[ mathcal{L}{te^{2t} cos(t)} -frac{d}{ds} left( frac{s-2}{(s-2)^2 1} right) ]

[ mathcal{L}{2e^{2t} cos(t)} frac{2(s-2)}{(s-2)^2 1} ]

Combining these results, we get:

[ mathcal{L}{4te^{2t} cos(t) 8e^{2t} cos(t)} 4 left( -frac{d}{ds} left( frac{s-2}{(s-2)^2 1} right) frac{2(s-2)}{(s-2)^2 1} right) ]

After calculating the derivative and simplification, we obtain:

[ mathcal{L}{4te^{2t} cos(t) 8e^{2t} cos(t)} frac{8s-32}{(s-2)^2 1} ]

Conclusion

In this article, we discussed two methods for finding the Laplace transform of ( f(t) 2t cdot 4e^{2t} cos(t) ). By using complex exponentials and the shifting theorem combined with the multiplication by ( t ) rule, we arrived at the final Laplace transform. Understanding these techniques is crucial for solving differential equations and analyzing linear systems.