Question

Find the inverse Laplace transform of the given function. $$ \frac{1-2 s}{s^{2}+4 s+5} $$

Short answer

Question: Find the inverse Laplace transform of the given function $$\frac{1-2s}{s^2 + 4s + 5}$$. Answer: The inverse Laplace transform of the given function is $$2e^{-2t}\cos(t) - 3e^{-2t}\sin(t)$$.

Step-by-step solution

Perform Partial Fraction Decomposition

To perform partial fraction decomposition on $$\frac{1-2s}{s^2 + 4s + 5}$$, we first need to identify the factors of the denominator. The denominator $$s^2 + 4s + 5$$ cannot be factored into linear terms, but it can be written in the standard quadratic form as $$(s+2)^2 + 1$$, with $$\alpha = -2$$ and $$\omega = 1$$. Now, let's perform partial fraction decomposition: $$ \frac{1-2s}{(s+2)^2 + 1} = \frac{A(s+2) + B}{(s+2)^2 + 1} $$ For the coefficients A and B, multiply both sides by the denominator to clear fractions: $$1-2s = A(s+2) + B$$ Now, we need to find the values for A and B. Let's solve for B first by setting s equal to 0: $$1 - 2(0) = A(0+2) + B \Rightarrow B + 2A = 1$$ Next, let's determine A. Since there is only one equation left, typically we would differentiate both sides. But in this case, let's substitute $$s = -3$$ and solve for A directly: $$1 - 2(-3) = A(-3+2) + B \Rightarrow 7 = -A + B$$ Now we have two equations: 1. $$B + 2A = 1$$ 2. $$7 = -A + B$$ Now we can solve for A and B: $$A = 2$$ $$B = -3$$ So, our function after partial fraction decomposition is: $$ \frac{1-2s}{(s+2)^2 + 1} = \frac{2(s+2) - 3}{(s+2)^2 + 1} $$

Find the Inverse Laplace Transform

Now that we have the function in the form $$\frac{2(s+2) - 3}{(s+2)^2 + 1}$$, we can find the inverse Laplace transform of each term separately using the Laplace transform pairs: 1. Inverse Laplace transform of $$\frac{2(s+2)}{(s+2)^2 + 1} = 2e^{-2t}\cos(t)$$ 2. Inverse Laplace transform of $$\frac{-3}{(s+2)^2 + 1} = -3e^{-2t}\sin(t)$$ Now we add them together to find the inverse Laplace transform of the original function: $$ L^{-1}\left\{\frac{1-2s}{s^2 + 4s + 5}\right\} = 2e^{-2t}\cos(t) - 3e^{-2t}\sin(t) $$ So, the inverse Laplace transform of the given function is: $$ 2e^{-2t}\cos(t) - 3e^{-2t}\sin(t) $$