Question

In Problems 3 - 24 fill in the blanks or answer true or false.

14.${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{3s-1}\right\}$ =…….

Short answer

$L^{-1}\left\{\frac{1}{3s-1}\right\}=\frac{1}{3}{e}^{\frac{t}{3}}$

Step-by-step solution

Consider the function

The function is,

$\begin{array}{l}{L}^{-1}\left\{\frac{1}{3s-1}\right\}\\ \Rightarrow L^{-1}\left\{\frac{1}{3s-1}\right\}=L^{-1}\left\{\frac{1}{3}·\frac{1}{s-\frac{1}{3}}\right\}\end{array}$

Compute the Laplace inverse

Consider the function,

$\begin{array}{l}{L}^{-1}\left\{\frac{1}{3s-1}\right\}\\ \mathrm{For} \mathrm{function} f\left(t\right) \mathrm{and} \mathrm{constant} a:L^{-1}\left\{a·f\left(t\right)\right\}=a·L^{-1}\left\{f\left(t\right)\right\}\\ L^{-1}\left\{\frac{1}{3}·\frac{1}{s-\frac{1}{3}}\right\}=\frac{1}{3}{L}^{-1}\left\{\frac{1}{s-\frac{1}{3}}\right\}\\ L^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}\\ SubstitutethevalueandfindtheLaplaceinversion\\ \Rightarrow L^{-1}\left\{\frac{1}{s-\frac{1}{3}}\right\}=e^{\frac{t}{3}}\end{array}$

Here,

$\begin{array}{l}\frac{1}{3}{L}^{-1}\left\{\frac{1}{3s-1}\right\}=\frac{1}{3}{e}^{\frac{t}{3}}\\ \Rightarrow L^{-1}\left\{\frac{1}{3s-1}\right\}=\frac{1}{3}{e}^{\frac{t}{3}}\\ Therefore,theLaplaceinverse{L}^{-1}\left\{\frac{1}{3s-1}\right\}is\\ \therefore L^{-1}\left\{\frac{1}{3s-1}\right\}=\frac{1}{3}{e}^{\frac{t}{3}}\end{array}$