Question

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

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

Short answer

$\frac{1}{3}\sin t-\frac{1}{6}\sin 2t$

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. role="math" localid="1664018947904" $\mathrm{sinhkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
7. $\mathrm{coshkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   

Use Inverse Transform

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

Here we have to decompose the fraction.

Decomposing

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

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

Now,

$\frac{1}{3}{L}^{-1}\left\{\frac{1}{s^2+1}\right\}-\frac{1}{3}{L}^{-1}\left\{\frac{1}{2}\frac{2}{s^2+4}\right\}=\frac{1}{3}\sin t-\frac{1}{6}\sin 2t$