MATH 223
Lecture 04
Laplace Transformation
| $ \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*}
• 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*}
• 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*}
• 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*}
• 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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