Taking the inverse laplace of this?

[Marshillboy]

How do I take the inverse laplace transform of something that looks like this? It's part of a larger piecewise-defined second order differential equation, but this is the part I'm stuck on.

(-4s-1)/(4s^2 + s + 4)

Completing the square doesn't work for the bottom, so I figure I need to separate the whole thing into two separate fractions. I still can't figure out how to take the inverse laplace of either resulting fraction, however.

 

[algebrat]

You can simplify the numerator by breaking across the difference, using the rule $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$.

For the denominator, just looking at it I can't see why you can't complete the square, if I'm wrong, please show us why.

Also, if you come to something in your steps that you're not sure how to take the inverse Laplace of, show us, we might be able to give an idea.

 

[Marshillboy]

I see.

So, doing so would give me

$\frac{-s}{(s+1/8)^{2}+63/64}+\frac{-1/4}{(s+1/8)^{2}+63/64}$

The second fraction would be simple to use with the $e^{at} sin(bt)$ identity since it's just a coeffecient on top, but what about the first?

 

[hunt_mat]

This is just standard look up tables, look it up on wikipedia.

 

[blue_raver22]

The inverse Laplace transform is a mathematical operation that allows us to transform a function from the Laplace domain back to the time domain. In order to take the inverse Laplace transform of the given expression, we need to use a combination of techniques such as partial fractions and inverse Laplace transform tables.

First, we can simplify the given expression by factoring the denominator as (4s+1)(s+4). This allows us to rewrite the expression as (-4s-1)/(4s+1)(s+4). Next, we can use partial fractions to separate this expression into two fractions, one with a denominator of 4s+1 and the other with a denominator of s+4.

Once we have these two fractions, we can use the inverse Laplace transform tables to find the corresponding time domain functions. The inverse Laplace transform of 1/(4s+1) is e^(-1/4t) and the inverse Laplace transform of 1/(s+4) is e^(-4t). We can then combine these two functions to get the final solution for the inverse Laplace transform of the given expression.

In summary, taking the inverse Laplace transform of a complex expression like the one provided requires a combination of techniques and the use of inverse Laplace transform tables. It may be helpful to consult a math textbook or seek assistance from a mathematician to ensure accurate and efficient calculations.