Introduction

The inverse Laplace transform is a crucial concept in engineering and mathematics, particularly in solving differential equations and analyzing systems in the time domain. This article provides a detailed step-by-step guide on how to solve the inverse Laplace transform of the given expression frac{s}{s^2 - 4^3}. The process involves recognizing the form, using known transform properties, and applying the convolution theorem when necessary.

Solving the Inverse Laplace Transform

Step 1: Recognize the Form

The given expression frac{s}{s^2 - 4^3} can be compared with the standard form for which the inverse Laplace transform is known. The form frac{s}{s^2 - a^2} is particularly significant, as it relates to the Laplace transform of trigonometric functions.

Step 2: Use the Inverse Transform Formula

The inverse Laplace transform of frac{s}{s^2 - a^2} is given by:

( mathcal{L}^{-1} left{ frac{s}{s^2 - a^2} right} frac{1}{a} sinh(at) ) for ( a eq 0 )

However, for our specific case, we have:

( mathcal{L}^{-1} left{ frac{s}{s^2 - 4^3} right} frac{1}{3 - 1} t^{3-1} sin(2t) frac{1}{2} t^2 sin(2t) )

Here, we identify (a 2) and (n 3), applying the formula as given.

Step 3: Final Result

The final result for the inverse Laplace transform is:

( mathcal{L}^{-1} left{ frac{s}{s^2 - 4^3} right} frac{1}{2} t^2 sin(2t) )

This result is derived directly from the properties of the Laplace transform and the specific form of the given function.

Alternative Approach using Convolution Theorem

Another method to solve the inverse Laplace transform involves the convolution theorem. This theorem states that the inverse Laplace transform of a product of two functions is the convolution of their inverse Laplace transforms.

Step 1: Breakdown into Product of Functions

The expression frac{s}{s^2 - 4^3} can be decomposed into two functions, ( F(s) ) and ( G(s) ), such that:

( F(s) frac{1}{s^2 - 4} )

( G(s) frac{s}{s^2 - 4^2} )

Recognizing ( F(s) ) and ( G(s) ), we can find their respective inverse Laplace transforms:

( mathcal{L}^{-1} left{ frac{1}{s^2 - 4} right} frac{1}{2} sinh(2t) )

( mathcal{L}^{-1} left{ frac{s}{s^2 - 4^2} right} cosh(2t) )

Step 2: Apply the Convolution Theorem

The convolution integral is given by:

( (f * g)(t) int_0^t f(tau) g(t - tau) , dtau )

Here, the inverse Laplace transforms of ( F(s) ) and ( G(s) ) are:

( f(t) frac{1}{2} sinh(2t) )

( g(t) cosh(2t) )

Therefore, the convolution integral becomes:

( (f * g)(t) int_0^t frac{1}{2} sinh(2tau) cosh(2(t - tau)) , dtau )

The integral can be solved using standard integration techniques and properties of hyperbolic functions. After solving, the result is:

( mathcal{L}^{-1} left{ frac{s}{s^2 - 4^3} right} frac{1}{16} t^2 cos(2t) - frac{1}{32} t sin(2t) )

This solution approach further solidifies the understanding of the inverse Laplace transform and its application.

Conclusion

In conclusion, we have demonstrated how to solve the inverse Laplace transform of frac{s}{s^2 - 4^3} using both the standard transform properties and the convolution theorem. Each approach provides valuable insights and enhances our understanding of this essential mathematical technique.