Chapter 2, Problem 2.9P

To determine

(a)

The inverse Laplace transform $f\left(t\right)$ of the given function.

Answer to Problem 2.9P

Inverse Laplace transform of the function $F\left(s\right)=\frac{5s}{s^2+9}$ is $f\left(t\right)=2\sin 3t$.

Explanation of Solution

Given:

$F\left(s\right)=\frac{5s}{s^2+9}$.

Concept Used:

Laplace transform of function $f\left(t\right)$ is defined as

$L\left(f\left(t\right)\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

Where $f\left(t\right)$ is piecewise continuo function defined for all $t\ge 0$.

$\begin{array}{l}L\left(t^n\right)=\frac{\overline{)n+1}}{s^{n+1}}=\frac{\overline{)n}}{s^{n+1}}\\ L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right);L\left(f\left(t\right)\right)=F\left(s\right)\\ L\left(\sin at\right)=L\left(\frac{e^{iat}-e^{-iat}}{2i}\right)=\frac{a}{s^2+a^2}\\ L\left(\cos at\right)=L\left(\frac{e^{iat}+e^{-iat}}{2}\right)=\frac{s}{s^2+a^2}\\ L\left(t^{n}f\left(t\right)\right)={\left(-1\right)}^n\frac{d^n}{d{s}^n}\left(F\left(s\right)\right);L\left(f\left(t\right)\right)=F\left(s\right)\end{array}$

Inverse Laplace transform of given function $F\left(s\right)$ is given by

$f\left(t\right)=L^{-1}\left(F\left(s\right)\right)$.

Calculation:

Here, we have

$F\left(s\right)=\frac{5s}{s^2+9}$

Thus, the inverse Laplace transform $f\left(t\right)$ of the given function is given as

$\begin{array}{l}f\left(t\right)=L^{-1}\left(F\left(s\right)\right)\\ \Rightarrow f\left(t\right)=L^{-1}\left(\frac{5s}{s^2+9}\right)\\ \Rightarrow f\left(t\right)=5L^{-1}\left(\frac{s}{s^2+3^2}\right)\\ \Rightarrow f\left(t\right)=5\cos 3t\left[\because L\left(\cos at\right)=\frac{s}{s^2+a^2}\right]\end{array}$.

To determine

(b)

The inverse Laplace transform $f\left(t\right)$ of the given function.

Answer to Problem 2.9P

Inverse Laplace transform of the function $F\left(s\right)=\frac{6}{s^2-9}$ is $f\left(t\right)=2\sinh 3t$.

Explanation of Solution

Given:

$F\left(s\right)=\frac{6}{s^2-9}$.

Concept Used:

Laplace transform of function $f\left(t\right)$ is defined as

$L\left(f\left(t\right)\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

Where $f\left(t\right)$ is piecewise continuo function defined for all $t\ge 0$.

$\begin{array}{l}L\left(t^n\right)=\frac{\overline{)n+1}}{s^{n+1}}=\frac{\overline{)n}}{s^{n+1}}\\ L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right);L\left(f\left(t\right)\right)=F\left(s\right)\\ L\left(\sin at\right)=L\left(\frac{e^{iat}-e^{-iat}}{2i}\right)=\frac{a}{s^2+a^2}\\ L\left(\cos at\right)=L\left(\frac{e^{iat}+e^{-iat}}{2}\right)=\frac{s}{s^2+a^2}\\ L\left(\sinh at\right)=L\left(\frac{e^{at}-e^{-at}}{2i}\right)=\frac{a}{s^2-a^2}\\ L\left(\cosh at\right)=L\left(\frac{e^{at}+e^{-at}}{2}\right)=\frac{s}{s^2-a^2}\\ L\left(t^{n}f\left(t\right)\right)={\left(-1\right)}^n\frac{d^n}{d{s}^n}\left(F\left(s\right)\right);L\left(f\left(t\right)\right)=F\left(s\right)\end{array}$

Inverse Laplace transform of given function $F\left(s\right)$ is given by

$f\left(t\right)=L^{-1}\left(F\left(s\right)\right)$.

Calculation:

Here, we have

$F\left(s\right)=\frac{6}{s^2-9}$

Thus, the inverse Laplace transform $f\left(t\right)$ of the given function is given as

$\begin{array}{l}f\left(t\right)=L^{-1}\left(F\left(s\right)\right)\\ \Rightarrow f\left(t\right)=L^{-1}\left(\frac{6}{s^2-9}\right)\\ \Rightarrow f\left(t\right)=2L^{-1}\left(\frac{3}{s^2-3^2}\right)\\ \Rightarrow f\left(t\right)=2\sinh 3t\left[\because L\left(\sinh at\right)=\frac{a}{s^2-a^2}\right]\end{array}$.

To determine

(c)

The inverse Laplace transform $f\left(t\right)$ of the given function.

Answer to Problem 2.9P

Inverse Laplace transform of the function $F\left(s\right)=\frac{45}{s{\left(s+3\right)}^2}$ is $f\left(t\right)=5-5\left(3t+1\right)e^{-3t}$.

Explanation of Solution

Given:

$F\left(s\right)=\frac{45}{s{\left(s+3\right)}^2}$.

Concept Used:

Laplace transform of function $f\left(t\right)$ is defined as

$L\left(f\left(t\right)\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

Where $f\left(t\right)$ is piecewise continuo function defined for all $t\ge 0$.

$\begin{array}{l}L\left(t^n\right)=\frac{\overline{)n+1}}{s^{n+1}}=\frac{\overline{)n}}{s^{n+1}}\\ L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right);L\left(f\left(t\right)\right)=F\left(s\right)\\ L\left(t^{n}f\left(t\right)\right)={\left(-1\right)}^n\frac{d^n}{d{s}^n}\left(F\left(s\right)\right);L\left(f\left(t\right)\right)=F\left(s\right)\end{array}$

Inverse Laplace transform of given function $F\left(s\right)$ is given by

$f\left(t\right)=L^{-1}\left(F\left(s\right)\right)$.

Calculation:

Here, we have

$F\left(s\right)=\frac{45}{s{\left(s+3\right)}^2}$

Thus, the inverse Laplace transform $f\left(t\right)$ of the given function is given as

To determine

(d)

The inverse Laplace transform $f\left(t\right)$ of the given function.

Answer to Problem 2.9P

Inverse Laplace transform of the function $F\left(s\right)=\frac{2}{s\left(s^2+4s+13\right)}$ is $f\left(t\right)=\frac{-2}{39}\left[\left\{\frac{-e^{-2t}}{13}\left\{2\sin \left(3t\right)+3\cos \left(3t\right)\right\}\right\}+3\right]$.

Explanation of Solution

Given:

$F\left(s\right)=\frac{2}{s\left(s^2+4s+13\right)}$.

Concept Used:

Laplace transform of function $f\left(t\right)$ is defined as

$L\left(f\left(t\right)\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

Where $f\left(t\right)$ is piecewise continuo function defined for all $t\ge 0$.

$\begin{array}{l}L\left(t^n\right)=\frac{\overline{)n+1}}{s^{n+1}}=\frac{\overline{)n}}{s^{n+1}}\\ L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right);L\left(f\left(t\right)\right)=F\left(s\right)\\ L\left(\sin at\right)=L\left(\frac{e^{iat}-e^{-iat}}{2i}\right)=\frac{a}{s^2+a^2}\\ L\left(\cos at\right)=L\left(\frac{e^{iat}+e^{-iat}}{2}\right)=\frac{s}{s^2+a^2}\\ L\left(t^{n}f\left(t\right)\right)={\left(-1\right)}^n\frac{d^n}{d{s}^n}\left(F\left(s\right)\right);L\left(f\left(t\right)\right)=F\left(s\right)\end{array}$

Inverse Laplace transform of given function $F\left(s\right)$ is given by

$f\left(t\right)=L^{-1}\left(F\left(s\right)\right)$.

Calculation:

Here, we have

$F\left(s\right)=\frac{2}{s\left(s^2+4s+13\right)}$

Thus, the inverse Laplace transform $f\left(t\right)$ of the given function is given as

To determine

(e)

The inverse Laplace transform $f\left(t\right)$ of the given function.

Answer to Problem 2.9P

Inverse Laplace transform of the function $F\left(s\right)=\frac{20}{s\left(s^2+4\right)}$ is $f\left(t\right)=5\left(1-\cos \left(2t\right)\right)$.

Explanation of Solution

Given:

$F\left(s\right)=\frac{20}{s\left(s^2+4\right)}$.

Concept Used:

Laplace transform of function $f\left(t\right)$ is defined as

$L\left(f\left(t\right)\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

Where $f\left(t\right)$ is piecewise continuo function defined for all $t\ge 0$.

$\begin{array}{l}L\left(t^n\right)=\frac{\overline{)n+1}}{s^{n+1}}=\frac{\overline{)n}}{s^{n+1}}\\ L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right);L\left(f\left(t\right)\right)=F\left(s\right)\\ L\left(\sin at\right)=L\left(\frac{e^{iat}-e^{-iat}}{2i}\right)=\frac{a}{s^2+a^2}\\ L\left(\cos at\right)=L\left(\frac{e^{iat}+e^{-iat}}{2}\right)=\frac{s}{s^2+a^2}\\ L\left(t^{n}f\left(t\right)\right)={\left(-1\right)}^n\frac{d^n}{d{s}^n}\left(F\left(s\right)\right);L\left(f\left(t\right)\right)=F\left(s\right)\end{array}$

Inverse Laplace transform of given function $F\left(s\right)$ is given by

$f\left(t\right)=L^{-1}\left(F\left(s\right)\right)$.

Calculation:

Here, we have

$F\left(s\right)=\frac{20}{s\left(s^2+4\right)}$

Thus, the inverse Laplace transform $f\left(t\right)$ of the given function is given as

$\begin{array}{l}f\left(t\right)=L^{-1}\left(F\left(s\right)\right)\\ \Rightarrow f\left(t\right)=L^{-1}\left(\frac{20}{s\left(s^2+4\right)}\right)\\ \Rightarrow f\left(t\right)=L^{-1}\left(\frac{\frac{20}{\left(s^2+4\right)}}{s}\right)={\displaystyle \underset{0}{\overset{t}{\int }}{L}^{-1}\left(\frac{20}{\left(s^2+4\right)}\right)}dt\left(1\right)\\ \Rightarrow L^{-1}\left(\frac{20}{\left(s^2+4\right)}\right)=10L^{-1}\left(\frac{2}{s^2+2^2}\right)=10\sin \left(2t\right)\left[\because L\left(\sin at\right)=\frac{a}{s^2+a^2}\right]\\ \Rightarrow L^{-1}\left(\frac{20}{\left(s^2+4\right)}\right)=10\sin \left(2t\right)\left(2\right)\\ \text{Using }\left(2\right)\text{in }\left(1\right),\text{we get}\\ \Rightarrow f\left(t\right)=L^{-1}\left(\frac{\frac{20}{\left(s^2+4\right)}}{s}\right)\\ \Rightarrow f\left(t\right)={\displaystyle \underset{0}{\overset{t}{\int }}10\sin \left(2t\right)}dt\\ \Rightarrow f\left(t\right)=10\left\{{\left[-\frac{\cos \left(2t\right)}{2}\right]}_0^t\right\}\\ \Rightarrow f\left(t\right)=-5\left\{\left[\cos \left(2t\right)-1\right]\right\}\\ \Rightarrow f\left(t\right)=5\left(1-\cos \left(2t\right)\right)\end{array}$.

To determine

(f)

The inverse Laplace transform $f\left(t\right)$ of the given function.

Answer to Problem 2.9P

Inverse Laplace transform of the function $F\left(s\right)=\frac{20s}{{\left(s^2+4\right)}^2}$ is $f\left(t\right)=5t\sin 2t$.

Explanation of Solution

Given:

$F\left(s\right)=\frac{20s}{{\left(s^2+4\right)}^2}$.

Concept Used:

Laplace transform of function $f\left(t\right)$ is defined as

$L\left(f\left(t\right)\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

Where $f\left(t\right)$ is piecewise continuo function defined for all $t\ge 0$.

$\begin{array}{l}L\left(t^n\right)=\frac{\overline{)n+1}}{s^{n+1}}=\frac{\overline{)n}}{s^{n+1}}\\ L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right);L\left(f\left(t\right)\right)=F\left(s\right)\\ L\left(\sin at\right)=L\left(\frac{e^{iat}-e^{-iat}}{2i}\right)=\frac{a}{s^2+a^2}\\ L\left(\cos at\right)=L\left(\frac{e^{iat}+e^{-iat}}{2}\right)=\frac{s}{s^2+a^2}\\ L\left(t^{n}f\left(t\right)\right)={\left(-1\right)}^n\frac{d^n}{d{s}^n}\left(F\left(s\right)\right);L\left(f\left(t\right)\right)=F\left(s\right)\end{array}$

Inverse Laplace transform of given function $F\left(s\right)$ is given by

$f\left(t\right)=L^{-1}\left(F\left(s\right)\right)$

Using convolution theorem

$L^{-1}\left(G\left(s\right)H\left(s\right)\right)={\displaystyle \underset{0}{\overset{t}{\int }}f\left(\tau \right)}g\left(t-\tau \right)d\tau$.

Calculation:

Here, we have

$F\left(s\right)=\frac{20s}{{\left(s^2+4\right)}^2}$

Thus, the inverse Laplace transform $f\left(t\right)$ of the given function is given as