Question

27. In Problems 1-30 use appropriate algebra and Theorem 7.2.1 to find the given inverse Laplace transform.

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{2s-4}{\left(s^2+s\right)\left(s^2+1\right)}\right\}$

Short answer

$\mathrm{cost}+3\mathrm{sint}-4+3{\mathrm{e}}^{-\mathrm{t}}$

Step-by-step solution

Definition of inverse Transform

THEOREM 7.2.1 Some Inverse Transforms

1. $1={\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}$
   
2. ${\mathrm{t}}^{\mathrm{n}}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{n}!}{{\mathrm{s}}^{\mathrm{n}+1}}\right\},\mathrm{n}=1,2,3,\dots$
   
3. ${\mathrm{e}}^{\mathrm{at}}={\mathcal{L}}^{-1}\left\{\frac{1}{\mathrm{s}-\mathrm{a}}\right\}$
   
4. $\mathrm{sinkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
5. $\mathrm{coskt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
6. $\mathrm{sinhkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   
7. $\mathrm{coshkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   

Use Inverse Transform

$L^{-1}\left\{\frac{2s-4}{\left(s^2+s\right)\left(s^2+1\right)}\right\}=L^{-1}\left\{\frac{2s-4}{s(s+1)\left(s^2+1\right)}\right\}$

Here we have to decompose the fraction.

Decomposing

$\frac{2s-4}{s(s+1)\left(s^2+1\right)}=\frac{a}{s}+\frac{b}{s+1}+\frac{cs+d}{s^2+1}\to \frac{2s-4}{s(s+1)\left(s^2+1\right)}=\frac{s+3}{\left(s^2+1\right)}-\frac{4}{s}+\frac{3}{s+1}$

The trick is to algebraically manipulate the expression to make it look like something we know

Now,

${\mathrm{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2+1}\right\}+3{\mathrm{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^2+1}\right\}-4{\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}}\right\}+3{\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}-(-1)}\right\}=\mathrm{cost}+3\mathrm{sint}-4+3{\mathrm{e}}^{-\mathrm{t}}$