Question

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

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{s^2+3s}\right\}$

Short answer

$\frac{1}{3}-\frac{1}{3}{e}^{-3t}$

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. $t^n={\mathcal{L}}^{-1}\left\{\frac{n!}{s^{n+1}}\right\},n=1,2,3,\dots$
   
3. $e^{at}={\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\}$
   
4. $\sin kt={\mathcal{L}}^{-1}\left\{\frac{k}{s^2+k^2}\right\}$
   
5. $\cos kt={\mathcal{L}}^{-1}\left\{\frac{s}{s^2+k^2}\right\}$
   
6. $\sinh kt={\mathcal{L}}^{-1}\left\{\frac{k}{s^2-k^2}\right\}$
   
7. $\cosh kt={\mathcal{L}}^{-1}\left\{\frac{s}{s^2-k^2}\right\}$
   

Use Inverse Transform

$L^{-1}\left\{\frac{1}{s^2+3s}\right\}$

Here we have to decompose the fraction, however it's much easier to complete the square in the denominator and realize that it's exponential and hyperbolic functions, but this is in the next chapter.

Decomposing

$\frac{1}{s^2+3s}=\frac{1}{s(s+3)}=\frac{a}{s}+\frac{b}{s+3}\to \frac{1}{s^2+3s}=\frac{1}{3s}-\frac{1}{3(s+3)}$

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

Now, $\frac{1}{3}{L}^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{3}{L}^{-1}\left\{\frac{1}{s-(-3)}\right\}$

$=\frac{1}{3}-\frac{1}{3}{e}^{-3t}$