Lecture 4 MATH 223 (Lt Col Tahmina Sultana)

 

 MATH 223

Lecture 04 

Laplace Transformation

    ♦ Some Important Results:

$\displaystyle F(t)$	$\displaystyle \mathcal{L}\{F(t)\}= f(s)$
$\displaystyle t$	$\displaystyle \frac{1}{s^2}$
$\displaystyle t^2$	$\displaystyle \frac{2}{s^3}$
$\displaystyle t^2$	$\displaystyle \frac{6}{s^4}$
$\displaystyle t^n$	$\displaystyle \frac{n!}{s^{n + 1}}$
$\displaystyle e^{at}$	$\displaystyle \frac{1}{s - a}$
$\displaystyle \sin at$	$\displaystyle \frac{a}{s^2 + a^2}$
$\displaystyle \cos at$	$\displaystyle \frac{s }{s^2 + a^2}$
$\displaystyle \sinh at$	$\displaystyle \frac{a}{s^2 - a^2}$
$\displaystyle \cosh at$	$\displaystyle \frac{s}{s^2 - a^2}$

    ♦ Linear Properties of Laplace Transformation:

  $$\begin{align*} & \mathcal{L}\{F_1(t)\} = f_1 (s) = \int^\infty_0 e^{- st} . F_1(t) ~dt \\ & \mathcal{L}\{F_2(t)\} = f_2 (s) = \int^\infty_0 e^{- st} . F_2 (t) ~dt \\ \end{align*}$$  

Then if, $C_1$ and $C_2$ are any constants 

  $$\begin{align*} \therefore \mathcal{L}\{C_1~F_1(t) + C_2~F_2 (t) \} = C_1~f_1 (s) + C_2~f_2 (s) \end{align*}$$  

   • Proof:

  $$\begin{align*} & \mathcal{L}\{C_1 ~F_1 (t) + C_2 ~ F_2(t)\}\\ = & \int^\infty_0 e^{- st} .\{ C_1 F_1(t) + C_2 F_2 (t)\} ~dt\\ = & \int^\infty_0 e^{- st} ~ C_1 F_1(t) ~dt + \int^\infty_0 e^{- st} C_2 F_2(t) ~dt \\ = & C_1 \int^\infty_0 e^{- st} ~ F_1(t) ~dt + C_2 \int^\infty_0 F_2(t) ~dt \\ = & C_1 \mathcal{L}\{F_1(t)\} + C_2 \mathcal{L}\{F_2(t)\} \\ = & C_1 f_1(s) + C_2f_2(s) \end{align*}$$  

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   • Example: Find $\mathcal{L}\{\sin 4t\}$ 

Solution:   

 We know that,  $$\mathcal{L}\{\sin at\} = \frac{a}{s^2 + a^2}$$  

  $$\begin{align*} \therefore & \mathcal{L}\{\sin 4t\} \\ = & \frac{4}{s^2 + 4^2} \\ = & \frac{4}{s^2 + 16 } \end{align*}$$  

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    • Example: Find Laplace Transform of  $\mathcal{L}\{4e^{5t} + 6 t^2 - 3 \sin 4t + 2 \cos 2t\}$  

Solution:  $$\begin{align*} & \mathcal{L}\{4e^{5t} + 6 t^2 - 3 \sin 4t + 2 \cos 2t\} \\ = & \mathcal{L}\{4 e^{5t}\} + \mathcal{L}\{6 t^3\} - \mathcal{L}\{3 \sin 4t\} + \mathcal{L}\{2 \cos 2t\} \\ = & 4 \mathcal{L}\{e^{5t}\} + 6 \mathcal{L}\{t^3\} - 3 \mathcal{L}\{\sin 4t\} + \mathcal{L}\{2 \cos 2t\} \\ = & 4 \frac{1}{s - 5} + 6 \frac{3!}{s^{3 + 1}} - 3 \frac{4}{3 + 16} + 2 \frac{s}{s^2 + 2^2} \end{align*}$$  

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   • Example: Find Laplace Transform of  $\mathcal{L}\{e^{3t} t^2\}$  

Solution:  $$\begin{align*} & e^{3t} = e^{at} \\ & ~~ a = 3 \\ & ~~\\ & F(t) = t^2 \\ \textrm{we know,}& ~~\\ & f(s) = \mathcal{L}\{F(t)\}\\ & ~~~~~~~= \mathcal{L}\{t^2\} \\ & ~~~~~~~= \frac{2}{s^3} \\ & ~~~\\ \textrm{here, } & f(s - a) = f (s - 3) = \mathcal{L}\{e^{3t ~ t^2} \}\\ & f (s) = \frac{2}{s^3} \\ & f(s - 3 ) = \frac{2}{(s - 3)^3} \\ \therefore & \mathcal{L}\{e^{3t}~ t^2\} = \frac{2}{(s - 3)^3} \end{align*}$$  

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    • Example: Find  $\mathcal{L}\{e^{- 2t. \sin 4t}\}$  

Solution:  $$\begin{align*} & e^{at } = e^{ - 2t} \\ \therefore ~ & a = - 2 \\ & ~~~ \\ & \mathcal{L}\{e^{- 2t. \sin 4t}\}= f(s + 2 ) \\ \textrm{And,} & F(t) = \sin 4t \\ & ~~\\ & f(s) = \mathcal{L}\{\sin 4t\} \\ & ~~~~~~~~~= \frac{4}{s^2 + 16} \\ & ~~~\\ \therefore & f(s + 2) = \frac{4}{(s + 2 )^2 + 16} \end{align*}$$                                                 

   • Home Work:  $\mathcal{L}\{e^{4t}~ \cosh 5t\}$  

 

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