Question

Discuss how Theorem 7.4.1 can be used to find

${\mathbf{L}}^{\mathbf{-}\mathbf{1}}\left\{\mathbf{ln}\frac{\mathbf{s}\mathbf{-}\mathbf{3}}{\mathbf{x}\mathbf{+}\mathbf{1}}\right\}$

Short answer

The required function for${\mathcal{L}}^{-1}\left\{\ln \frac{s-3}{x+1}\right\}$by using the theorem derivatives of transforms is.$f\left(t\right)=-\frac{1}{t}\left[e^{3t}-e^{-t}\right]$

Step-by-step solution

Define the theorem derivatives of transforms:

If$\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}\mathbf{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}$and$\mathit{n}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{.}\mathbf{.}\mathbf{.}$then$\mathbf{L}\left\{{\mathbf{t}}^{\mathbf{n}}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\right\}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{n}}\frac{{\mathbf{d}}^{\mathbf{n}}}{{\mathbf{ds}}^{\mathbf{\pi }}}\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{.}$

Evaluate the function using the theorem:

Consider the theorem,$\mathcal{L}\left\{t^{n}f\left(t\right)\right\}=(-1{)}^n\frac{d^n}{d{s}^{\pi }}F\left(s\right).$

Substitute,$n=1$

$\mathcal{L}\left\{tf\right(t\left)\right\}=(-1)·\frac{\text{d}}{\text{d}s}F\left(s\right)$

Assume,$F\left(s\right)=\ln \frac{s-3}{s+1}$

$f\left(t\right)=\frac{1}{t}{\mathcal{L}}^{-1}\left\{-\frac{\text{d}}{\text{d}s}\left(\ln \frac{s-3}{s+1}\right)\right\}$

$=-\frac{1}{t}{\mathcal{L}}^{-1}\left\{\frac{\text{d}}{\text{d}s}\left[\ln \right(s-3)-\ln (s+1\left)\right]\right\}\left(\ln a-\ln b=\ln \frac{a}{b}\right)$

Evaluate,

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

$=-\frac{1}{t}\left[{\mathcal{L}}^{-1}\left\{\frac{1}{s-3}\right\}-{\mathcal{L}}^{-1}\left\{\frac{1}{s-(-1)}\right\}\right]\left({\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}\right)$

$=-\frac{1}{t}\left[e^{3t}-e^{-t}\right]$

Therefore, the required solution for the given function ${\mathcal{L}}^{-1}\left\{\ln \frac{s-3}{x+1}\right\}$by using the theorem is $f\left(t\right)=-\frac{1}{t}\left[e^{3t}-e^{-t}\right]$.