# Find the inverse Laplace transform f(t)=L1(F(s)) of the function F(s)=4s-10s^2-4s+20.

## Question:

Find the inverse Laplace transform f(t)=L1(F(s)) of the function F(s)=4s-10s{eq}^2 {/eq}-4s+20.

## Problems involving Inverse Laplace:

This problem involves finding the inverse Laplace transform of the given function. The idea is to use the partial fractions method to separate the given rational polynomial as a sum of two polynomials whose inverse Laplace transform is known.

## Answer and Explanation:

Given the Laplace transform of a function,

{eq}\displaystyle F(s) = \frac{4s - 10}{s^2 - 4s + 20} {/eq}

We need to find the inverse Laplace transform of this function.

{eq}\displaystyle L^{-1} (F(s)) = L^{-1} \Big( \frac{4s-10}{s^2 - 4s + 20} \Big) = L^{-1} \Big( 4 \frac{ s-2}{(s-2)^2 +16} + \frac{-2}{(s-2)^2 +16} \Big) {/eq}

So, we have

{eq}\displaystyle 4 L^{-1} \Big( \frac{s-2}{(s-2)^2 +16} \Big) - 2 L^{-1} \Big( \frac{1}{(s-2)^2 + 16} \Big) {/eq}

Which upon using simple laplace inverse formula we get,

{eq}\displaystyle f(t) = 4e^{2t} \cos (4t) - \frac{1}{2} e^{2t} \sin (4t ) {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2