Question

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

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

Short answer

The inverse Laplace transform is $\frac{1}{4}{\mathrm{e}}^{-\mathrm{t}/4}$.

Step-by-step solution

Definition of Laplace Transform

Let $\mathbf{f}$be a function defined for $\mathbf{t}\mathbf{\ge }\mathbf{0}$. Then the integral $\mathcal{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}{\mathbf{e}}^{\mathbf{-}\mathbf{xt}}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$ is said to be the Laplace transform of $\mathbf{f}$, provided that the integral converges.

Consider the formula for inverse Laplace given below:

1. $1={\mathcal{L}}^{-1}\left\{\frac{1}{\mathrm{s}}\right\}$
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. $e^{at}={\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\}$
   
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. $\cosh kt={\mathcal{L}}^{-1}\left\{\frac{s}{s^2-k^2}\right\}$
   

Determine the inverse transform

Solve for the inverse Laplace transform.

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

From the inverse transform formulas:

$L^{-1}\left\{\frac{1}{4s+1}\right\}=\frac{1}{4}{e}^{-\frac{t}{4}}$