Inverse laplace transform issue $\frac{2}{4s^2+s}$

Question

I'm stuck trying to perform an inverse laplace transform of $\dfrac{2}{4s^2+s}$. MathCad said that the original is $2 -2e^{-\frac{1}{4}t}$, and I know that this is the right solution, but I can't understand how it was derived. I also understand that the image can be represented as $2\dfrac{1}{4s^2+s}$, and the original might be $2(1-e^{-\frac{1}{4}t})$.

So, can you please explain step-by-step why $$\frac{2}{4s^2+s} \xrightarrow{\text{Inverse Laplace Transform}} 2-2e^{-\frac{1}{4}t}\ ?$$

Answer 1

If $$Y(s)=\frac{2}{4s^2+s}$$ is our job so you can do it as follows:

$$Y(s)=\frac{2}{s(4s+1)}=\frac{2}s-\frac{8}{4s+1}$$ So $$y(x)=\mathcal{L}^{-1}(2/s)-\mathcal{L}^{-1}(8/4s+1)$$

Answer 2

Using partial fractions, we have $$\frac{2}{4s^2 + s} = \frac{2}{s} + \frac{-8}{4s + 1}.$$

As $\mathcal{L}\{kf(t)\} = k\mathcal{L}\{f(t)\}$, and $\mathcal{L}\{1\} = \dfrac{1}{s}$, we see that $\mathcal{L}\{2\} = \dfrac{2}{s}$.

As $\mathcal{L}\{e^{\alpha t}\} = \dfrac{1}{s-a}$, we have $\mathcal{L}\{e^{-\frac{1}{4}t}\} = \dfrac{1}{s+\frac{1}{4}}$, so $\mathcal{L}\{-2e^{-\frac{1}{4}t}\} = \dfrac{-2}{s+\frac{1}{4}} = \dfrac{-8}{4s + 1}$.

Finally as $\mathcal{L}\{f(t) + g(t)\} = \mathcal{L}\{f(t)\} + \mathcal{L}\{g(t)\}$ we have

$$\mathcal{L}\{2 - 2e^{-\frac{1}{4}t}\} = \mathcal{L}\{2\} + \mathcal{L}\{-2e^{-\frac{1}{4}t}\} = \frac{2}{s} + \frac{-8}{4s + 1} = \frac{2}{4s^2 + s}$$

so $\mathcal{L}^{-1}\left\{\dfrac{2}{4s^2+s}\right\} = 2 - 2e^{-\frac{1}{4}t}$.