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\title[Matematika 4]
{Matematika 4}

\subtitle{2. vaja}

\author[Borut] % 
{B.~Jur\v{c}i\v{c}~Zlobec\inst{1}}

\institute[Univerza v Ljubljani Slovenia] % 
{
  \inst{1}%
    Univerza v Ljubljani,\\Fakulteta za Elektrotehniko\\ 
    1000 Ljubljana, Tr\v{z}a\v{s}ka 25, Slovenija
}


\date[\today] % 
{Matematika FE, Ljubljana, Slovenija \today}

\subject{Matematika}

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\section{Laplaceova transformacija}
\subsection{Laplaceova transformacija}
\begin{frame}{Formule}
\begin{itemize}
	\item  \formulas{$\displaystyle F(s)=\mathcal{L}(f(t))= 
	\int_0^\infty f(t) e^{-s t}dt$}.
	\item  \formulas{$\displaystyle f(t)=\mathcal{L}^{-1}(F(s))=
	\frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma-i T}^{\gamma+i T} 
	e^{s t}F(s)ds$},\\ kjer je $\gamma > \Re(s)$.
	\item S pomo"cjo Chauchyjevega izreka o residuih:
	\formulas{$\displaystyle f(t)=\sum_k \Res(e^{s t}F(s),s_k)$}.
	\item Ra"cunanje residuov v polu $n$-te stopnje $s_k$:
	\formulas{$\displaystyle \Res(e^{s t}F(s),s_k)=\frac{1}{(n-1)!}
	\lim_{s\to s_k}\frac{d^{n-1}}{ds^{n-1}}
	\left((s-s_k)^ne^{s t}F(s)\right)$}.
	\item Ra"cunanje residuov v polu prve stopnje v $s_k$, "ce je
	$\displaystyle F(s)=\frac{P(s)}{Q(s)}$, 
	\formulas{$\displaystyle \Res(e^{s t}F(s),s_k)=e^{s_k t}\frac{P(s_k)}{Q'(s_k)}$}.
\end{itemize}
\end{frame}

\begin{frame}{Pravila: $a>0$, $b\in\mathbb{R}$}
\begin{enumerate}[I]
\item $\mathcal{L}\left(b_1\,f(t)+b_2\,g(t)\right)=
	b_1\,F(s)+b_2\,G(s)$\label{lin}
\item $\displaystyle\mathcal{L}\left((f*g)(t)\right)=
	F(s)G(s)$\label{konv} 
\item $\displaystyle\mathcal{L}\left(H(t-a)f(t-a)\right)=
	e^{-as}F(s)$\label{pt}
\begin{multicols}{2}
\item $\displaystyle\mathcal{L}\left(e^{b t}f(t)\right)=
	F(s-b)$\label{ps} 
\item $\displaystyle\mathcal{L}\left(f(a\,t)\right)=
	\frac{1}{a}F\left(\frac{s}{a}\right)$\label{at}
\end{multicols}
\item $\displaystyle\mathcal{L}\left(f^{(n)}(t)\right)=
	s^nF(s)-s^{n-1}f(0)-s^{n-2}f'(0)-\dots-f^{(n-1)}(0)$\label{dt} 
\item $\displaystyle\mathcal{L}\left(t^nf(t)\right)=
	(-1)^nF^{(n)}(s)$\label{ds}   
\begin{multicols}{2}
\item $\displaystyle\mathcal{L}\left(\frac{f(t)}{t}\right)=
        \int_s^\infty F(\sigma)d\sigma$\label{is}  
\item $\displaystyle\mathcal{L}\left(\int_0^tf(\tau)d\tau\right)=
        \frac{F(s)}{s}$\label{it}  
\end{multicols}
\begin{multicols}{2}
\item $\displaystyle \lim_{t\searrow0} f(t)=\lim_{s\to\infty}(sF(s))$
\item $\displaystyle \lim_{t\to\infty}f(t)=\lim_{s\to0}(sF(s))$
\end{multicols}
\end{enumerate}
\end{frame}

\begin{frame}{Laplaceova transformacija nekaterih funkcij}
Konstanta $a>0$, medtem ko je $b\in\mathbb{R}$. 
\begin{multicols}{2}
\begin{enumerate}
	\item $\displaystyle\mathcal{L}\left(H(t) \right)=
		\frac{1}{s}$\label{ht}
        \item $\displaystyle\mathcal{L}\left(t^n\right)=
		\frac{n!}{s^{n+1}}$\label{tn}
        \item $\displaystyle\mathcal{L}\left(e^{bt}\right)=
		\frac{1}{s-b}$\label{ebt}
        \item $\displaystyle\mathcal{L}\left(H(t-a)\right)=
		\frac{1}{s}e^{-at}$\label{hta}
        \item $\displaystyle\mathcal{L}\left(\sin(bt)\right)=
		\frac{b}{s^2+b^2}$\label{sin}
	\item $\displaystyle\mathcal{L}\left(\cos(bt)\right)= 
		\frac{s}{s^2+b^2}$\label{cos}
	\item $\displaystyle\mathcal{L}\left(t\sin(bt)\right)= 
		\frac{2bs}{(s^2+b^2)^2}$\label{tsin}
	\item $\displaystyle\mathcal{L}\left(t\cos(at)\right)= 
		\frac{s^2-b^2}{(s^2+b^2)^2}$\label{tcos}
	\item $\displaystyle\mathcal{L}\left(\delta(t)\right)=
		1$\label{deltat}
	\item $\displaystyle\mathcal{L}\left(\delta(t-a)\right)= 
		e^{-as}$\label{deltata}
\end{enumerate}
\end{multicols}
\end{frame}

\begin{frame}{Poi"s"ci Laplaceovo transformacijo funkcije}
\formulas{$\displaystyle f(t)=H(t)$}.
\begin{itemize}
	\item $\displaystyle\int_{0}^{\infty}H(t)
	e^{-s t}\,dt=\int_{0}^{\infty}
        e^{-s t}\,dt \rightarrow$
	\item Ker mora biti 
	$\displaystyle\lim_{t\to\infty}e^{-s t}=0$, je
	$\Re(s)>0.\rightarrow$  
	\item $\displaystyle
	\left.\frac{1}{s}e^{-st}\right|_0^\infty=\frac{1}{s}$. 
	\item \formulas{$\displaystyle\mathcal{L}\left(H(t)\right)=
	\frac{1}{s}$}, $\Re(s)>0$.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci Laplaceovo transformacijo funkcije}
\formulas{$\displaystyle f(t)=e^{t}$}.
\begin{itemize}
	\item $\displaystyle\int_{0}^{\infty}
	e^t e^{-s t}\,dt=\int_0^\infty e^{(1-s)t}\,dt\rightarrow$
	\item Ker mora biti 
	$\displaystyle\lim_{t\to\infty}e^{(1-s)t}=0$, je
	$\Re(s)>1.\rightarrow$  
	\item $\displaystyle
	\left.\frac{1}{1-s}e^{(1-s)t}\right|_0^\infty=\frac{1}{s-1}$. 
	\item \formulas{$\displaystyle\mathcal{L}\left(e^t\right)=
	\frac{1}{s-1}$}, $\Re(s)>1$.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci Laplaceovo transformacijo funkcije}
\formulas{$\displaystyle f(t)=e^{i\omega t}$}.
\begin{itemize}
	\item $\displaystyle\int_{0}^{\infty}
	e^{i\omega t} e^{-s t}\,dt=
	\int_0^\infty e^{(i\omega-s)t}\,dt\rightarrow$
	\item Ker mora biti 
	$\displaystyle\lim_{t\to\infty}e^{(i\omega-s)t}=0$, je
	$\Re(s)>0.\rightarrow$  
	\item $\displaystyle
	\left.\frac{1}{i
        \omega-s}e^{(i\omega-s)t}\right|_0^\infty=\frac{1}{i\omega-s}$. 
	\item \formulas{$\displaystyle\mathcal{L}\left(e^t\right)=
	\frac{1}{s-i\omega}$}, $\Re(s)>0$.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci Laplaceovo transformacijo funkcije}
\formulas{$\displaystyle f(t)=\cos(\omega t)$}.
\begin{itemize}
	\item $\displaystyle\cos(\omega t)=
	\frac{e^{i\omega t}+e^{-i\omega t}}{2}\rightarrow$
	\begin{multicols}{2}
	\item $\displaystyle\mathcal{L}\left(e^{i\omega
	t}\right)=\frac{1}{s-i\omega}\rightarrow$
	\item $\displaystyle\mathcal{L}\left(e^{-i\omega
        t}\right)=\frac{1}{s+i\omega}\rightarrow$
	\end{multicols}
	\item $\displaystyle\mathcal{L}\left(\cos(\omega t)\right)=
	\frac{1}{2}\left(\frac{1}{s-i\omega}+\frac{1}{s+i\omega}\right)
	=\frac{s}{s^2+\omega^2}$.\\[5mm]
	\item \formulas{$\displaystyle\mathcal{L}\left(\cos(\omega t)\right)=
	\frac{s}{s^2+\omega^2}$}, $\Re(s)>0$.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}\left(f(t)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle f(t)=e^{-\lambda t}\cos(\omega t)$}.
\begin{itemize}
\item Uporabimo (\ref{cos})$\rightarrow$(\ref{ps}).
\item $\displaystyle\mathcal{L}\left(\cos(\omega t)\right)=
	\frac{s}{s^2+\omega^2}\rightarrow$
\item $\displaystyle\mathcal{L}\left(e^{-\lambda t}\cos(\omega t)\right)=
	\frac{s+\lambda}{(s+\lambda)^2+\omega^2}$.\\[5mm]
\item \formulas{$\displaystyle\mathcal{L}
	\left(e^{-\lambda t}\cos(\omega t)\right)=
	\frac{s+\lambda}{(s+\lambda)^2+\omega^2}$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}\left(f(t)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle f(t)=t^2e^{-\lambda t}$}.
\begin{itemize}
\item Lahko uporabimo (\ref{ds}) ali pa (\ref{tn})$\rightarrow$(\ref{ps}).
\item $\displaystyle\mathcal{L}\left(t^2\right)=
	\frac{2}{s^3}\rightarrow$
\item $\displaystyle\mathcal{L}\left(t^2e^{-\lambda t} \right)=
	\frac{2}{(s+\lambda)^3}$.\\[5mm]
\item \formulas{$\displaystyle\mathcal{L}
	\left(t^2e^{-\lambda t}\right)=
	\frac{2}{(s+\lambda)^3}$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}\left(f(t)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle f(t)=t^2\sin(\omega t)$}.
\begin{itemize}
\item Uporabimo (\ref{ds}).
\item $\displaystyle\mathcal{L}\left(\sin(\omega t)\right)=
	\frac{\omega}{s^2+\omega^2}\rightarrow$
\item $\displaystyle\mathcal{L}\left(t^2\sin(\omega t) \right)=
	\left(\frac{\omega}{s^2+\omega^2}\right)^{(2)}=
	-\frac{2\omega\left(\omega ^2-3s^2\right)}
	{\left(s^2+\omega^2\right)^3}$\\[5mm]
	\item \formulas{$\displaystyle\mathcal{L}
	\left(t^2\sin(\omega t)\right)=
	-\frac{2\omega\left(\omega^2-3s^2\right)}
	{\left(s^2+\omega^2\right)^3}$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}\left(f(t)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle 
f(t)=\sin(t-\frac{\pi}{3})$}.
\begin{itemize}
\item Ne moremo uporabiti (\ref{pt}), ker je
$\displaystyle f(t)=H(t)\sin(t-\frac{\pi}{3})$.
\item Uporabimo adicijski izrek 
	$\displaystyle\sin t\cos\frac{\pi}{3}-
	\sin\frac{\pi}{3}\cos t\rightarrow$.
\item Uporabimo (\ref{lin})$\rightarrow$ (\ref{sin},
	\ref{cos})$\rightarrow$.\\[5mm] 
\item \formulas{$\displaystyle\mathcal{L}\left(f(t)\right)=
\frac{1-\sqrt{3} s}{2 s^2+2}$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}\left(f(t)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle 
f(t)=\sin^2t$}.
\begin{itemize}
\item Uporabimo formulo 
	$\displaystyle\sin^2t=\frac{1}{2}\left(1-\cos(2t)\right)\rightarrow$.
\item Uporabimo (\ref{lin})$\rightarrow$ (\ref{cos})$\rightarrow$.\\[5mm] 
\item \formulas{$\displaystyle\mathcal{L}\left(f(t)\right)=
\frac{2}{s^3+4 s}$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}\left(f(t)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle 
f(t)=\begin{cases}1&0\le t \le 1\\0& drugod\end{cases}$}.
\begin{itemize}
\item Lahko zapi"semo $f(t)=H(t)-H(t-1)$.
\item Uporabimo (\ref{ht}) $\rightarrow$ (\ref{pt}).\\[5mm] 
\item \formulas{$\displaystyle\mathcal{L}\left(f(t)\right)=
\frac{1}{s}\left(1-e^{-t}\right)$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}\left(f(t)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle 
f(t)=\int_0^t\frac{1-e^{-\tau}}{\tau}d\tau$}.
\begin{itemize}
\item Uporabimo (\ref{it})$\rightarrow$(\ref{is})
	$\rightarrow$(\ref{tn},\ref{ebt}).
\item $\displaystyle F(s)=\frac{1}{s}\mathcal{L}\left(\frac{1-e^{-\tau}}{
	\tau}\right)=\rightarrow$ 
\item $\displaystyle =\frac{1}{s}\int_s^\infty\left(\frac{1}{\sigma}-
	\frac{1}{\sigma+1}\right)d\sigma=\rightarrow$
\item $\displaystyle =\int_s^\infty\frac{d\sigma}{\sigma(\sigma+1)}=
	\left.\ln\frac{\sigma}{\sigma+1}\right|_s^\infty=
	-\frac{1}{s}\ln\frac{s}{s+1}$
\item
\formulas{$\displaystyle\mathcal{L}\left(
	\int_0^t\frac{1-e^{-\tau}}{\tau}d\tau\right)=
	\frac{1}{s}\ln\frac{s+1}{s}$}, 
$\Re(s)>0$.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}\left(f(t)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle 
f(t)=\int_0^te^{2\tau}\cos(t-\tau)$}.
\begin{itemize}
\item Uporabimo
(\ref{konv})$\rightarrow$(\ref{ebt},\ref{cos})$\rightarrow$.
\item $\displaystyle F(s)=\mathcal{L}\left(e^{2t}\right)
	\mathcal{L}\left(\cos{t}\right)=\rightarrow$
\item $\displaystyle=\frac{1}{s-2}\frac{s}{s^2+1}=
	\frac{s}{(s-2)(s^2+1)}$.
\item \formulas{$\displaystyle\mathcal{L}\left(\int_0^t e^{2\tau}\cos(t-\tau)\right)=
\frac{s}{(s-2)(s^2+1)}$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}\left(f(t)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle 
f(t)=\begin{cases}
	\sin t&\frac{\pi}{2}\le t \\0& drugod
	\end{cases}$}.
\begin{itemize}
\item Lahko zapi"semo
$f(t)=H(t-\frac{\pi}{2})\sin\left((t-\frac{\pi}{2})+\frac{\pi}{2}\right)$.
\item Uporabimo (\ref{pt}) $\rightarrow$ adicijski izrek $\rightarrow$ 
(\ref{cos})$\rightarrow$\\[5mm]
\item \formulas{$\displaystyle\mathcal{L}\left(f(t)\right)=
	e^{-\frac{\pi}{2}s}\frac{s}{s^2+1}$}.
\end{itemize}
\end{frame}


\subsection{Inverzna Laplaceova transormacija}

\begin{frame}{Poi"s"ci $\mathcal{L}^{-1}\left(F(s)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle 
F(s)=\frac{s}{s^2+2s+2}$}.
\begin{itemize}
\item Lahko zapi"semo $\displaystyle
	F(s)=\frac{s}{(s+1)^2+1}=
	\frac{s+1}{(s+1)^2+1}-\frac{1}{(s+1)^2+1}\rightarrow$
\item (\ref{ps}) $\rightarrow$ (\ref{sin},
\ref{cos})$\rightarrow$\\[5mm]
\item \formulas{$\displaystyle\mathcal{L}^{-1}\left(F(s)\right)=
	e^{-t}\left(\cos t- \sin t\right)$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}^{-1}\left(F(s)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle 
F(s)=\frac{s^2-4}{s^3+2s^2-3s}$}.
\begin{itemize}
\item Lahko zapi"semo 
$\displaystyle F(s)=\frac{s^2-4}{s(s+3)(s-1)}\rightarrow$.
\item Razcepimo na parcialne ulomke $\rightarrow\displaystyle 
	F(s)=\frac{4}{3s}+\frac{5}{12(s+3)}-\frac{3}{4 (s-1)}$.
\item Uporabimo (\ref{lin})$\rightarrow$(\ref{ht})$\rightarrow$(\ref{ps})$\rightarrow$\\[5mm]
\item \formulas{$\displaystyle\mathcal{L}^{-1}\left(F(s)\right)=
	\frac{5 e^{-3 t}}{12}-\frac{3
   e^t}{4}+\frac{4}{3}$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}^{-1}\left(F(s)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle 
F(s)=\frac{s^2+4}{s^4+2s^3+2s^2}$}.
\begin{itemize}
\item Lahko zapi"semo 
$\displaystyle F(s)=\frac{s^2+4}{s^2(s^2+2s+2)}\rightarrow$.
\item Razcepimo na parcialne ulomke 
	$\rightarrow\displaystyle 
	\frac{2 s+3}{s^2+2s+2}+
	\frac{2}{s^2}-\frac{2}{s}=
	\frac{2(s+1)+1}{(s+1)^2+1}+
	\frac{2}{s^2}-\frac{2}{s}$.
\item Uporabimo (\ref{lin})$\rightarrow$(\ref{tn},\ref{sin},\ref{cos})$\rightarrow$(\ref{ps})
	$\rightarrow$\\[5mm]
\item \formulas{$\displaystyle\mathcal{L}^{-1}\left(F(s)\right)=
2 (t-1)+e^{-t} (\sin t+2 \cos t)$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}^{-1}\left(F(s)\right)$ z uporabo
pravil.}
\formulas{$\displaystyle F(s)=\frac{s}{(s^2+1)^2}$}.
\begin{itemize}
\item Uporabimo
(\ref{konv})$\rightarrow$(\ref{sin},\ref{cos})$\rightarrow$.
\item $\displaystyle\mathcal{L}^{-1}\left(\frac{1}{s^2+1}\right)
\mathcal{L}^{-1}\left(\frac{s}{s^2+1}\right)=
\int_0^t\sin\tau\cos(t-\tau)\,d\tau\rightarrow$
\item $\displaystyle\cos t\int_0^t\cos\tau\sin\tau\,d\tau+\sin
t\int_0^t\sin^2\tau\,d=\tau\rightarrow$
\item $\displaystyle\frac{1}{2}t\sin t$\\[5mm] 
\item \formulas{$\displaystyle
	\mathcal{L}^{-1}\left(\frac{s}{(s^2+1)^2}\right)=\frac{1}{2}t\sin t$}
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}^{-1}\left(F(s)\right)$ z
uporabo pravil.}
\formulas{$\displaystyle 
F(s)=\frac{s^2+4}{s^6+2s^4+s^2}$}.
\begin{itemize}
\item Lahko zapi"semo 
$\displaystyle F(s)=\frac{s^2+4}{s^2(s^2+1)^2}\rightarrow$.
\item Razcepimo na parcialne ulomke 
	$\rightarrow\displaystyle 
	\frac{4}{s^2}-\frac{4}{s^2+1}-\frac{3}{\left(s^2+1\right)^2}$.
\item Uporabimo
(\ref{lin})$\rightarrow$(\ref{tn},\ref{sin},\ref{cos})$\rightarrow$(\ref{konv})
	$\rightarrow$

\item $\displaystyle\mathcal{L}\left(\int_0^t
\sin(\tau)\sin(t-\tau)d\tau\right)=\frac{1}{s^2+1}\frac{1}{s^2+1}\rightarrow$
\\[5mm]
\item \formulas{$\displaystyle\mathcal{L}^{-1}\left(F(s)\right)=
\frac{1}{2} (8 t-11 \sin (t)+3 t
   \cos (t)) $}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci $\mathcal{L}^{-1}\left(F(s)\right)$ s pomo"cjo
residuov.}
\formulas{$\displaystyle 
F(s)=\frac{s^2+4}{s^6+2s^4+s^2}$}.
\begin{itemize}
\item Lahko zapi"semo 
$F(s)=\frac{s^2+4}{s^2(s^2+1)^2}\rightarrow$.
\item Singularne to"cke $0$, $i$ in $-i$ so poli druge stopnje.
\item $\displaystyle\Res(e^{s t}F(s),i)=\lim_{s\to i}
\left(e^{s t}(s-i)^2 F(s)\right)'=\rightarrow$
\item $\left.\left( e^{s t}\frac{s^2+1}{s^2(s+i)^2}\right)'\right|_{s=i}=
-\frac{1}{8} e^{i t}(-6 t-22 i)$.
\item $\Res\left(e^{s t},0\right)=4t$, $\Res\left(e^{s t},-i\right)=
	-\frac{1}{8} e^{-i t} (-6 t+22 i)$. 
\item Vsota residuov je $\frac{1}{2} (8 t-11 \sin (t)+3 t\cos (t))$.
\\[5mm]
\item \formulas{$\displaystyle\mathcal{L}^{-1}\left(F(s)\right)=
\frac{1}{2} (8 t-11 \sin (t)+3 t
   \cos (t)) $}.
\end{itemize}
\end{frame}

\subsection[Resevanje diferencialnih in integralskih enacb]
{Re\v{s}evanje diferencialnih in integralskih 
ena\v{c}b s pomo\v{c}jo Laplaceove transformacije}
\begin{frame}{Re"si diferencialno ena"cbo}
\formulas{$\dot{x}(t)+x(t)=e^{-t}$, $x(0)=1$}.
\begin{itemize}
	\item $\displaystyle s\left(\mathcal{L}_t[x(t)](s)\right)+
	\mathcal{L}_t[x(t)](s)-x(0)=\frac{1}{s+1}$,
	\item $\displaystyle\mathcal{L}_t[x(t)](s)=
	\frac{s x(0)+x(0)+1}{(s+1)^2}$,
	\item \formulas{$\displaystyle x(t)\to e^{-t}(t+1)$}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si diferencialno ena"cbo}
\formulas{$\ddot{x}(t)+\dot{x}(t)=te^{-t}$, $x(0)=0$, $\dot{x}(0)=0$}.
\begin{itemize}
        \item $\displaystyle s^2\left(\mathcal{L}_t[x(t)](s)\right)
	+s\left(\mathcal{L}_t[x(t)](s)\right)=\frac{1}{(s+1)^2}$,
        \item $\displaystyle\mathcal{L}_t[x(t)](s)=
	\frac{1}{s(s+1)^3}$,
        \item \formulas{$\displaystyle x(t)=1-\frac{1}{2} e^{-t} 
	(t(t+2)+2)$}.
\end{itemize}

\end{frame}

\begin{frame}{Re"si diferencialno ena"cbo}
\formulas{$\ddot{x}(t)+x(t)=0$, $x(0)=0$, $\dot{x}(0)=1$}.
\begin{itemize}
\item $\displaystyle s^2
   \mathcal{L}_t[x(t)](s)+\mathcal{
   L}_t[x(t)](s)=s x(0)+\dot{x}(0)$
\item $\displaystyle\mathcal{L}_t[x(t)](s)=
   \frac{1}{s^2+1}$
\item \formulas{$\displaystyle x(t)=\sin t $}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si diferencialno ena"cbo}
\formulas{$\ddot{x}(t)+x(t)=\delta(t)$, $x(0)=0$, $\dot{x}(0)=0$}.
\begin{itemize}
\item $\displaystyle s^2
   \mathcal{L}_t[x(t)](s)+\mathcal{
   L}_t[x(t)](s)=1+s x(0)+\dot{x}(0)$,
\item $\displaystyle\mathcal{L}_t[x(t)](s)=
   \frac{1}{s^2+1}$,
\item \formulas{$\displaystyle x(t)=\sin t $}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si diferencialno ena"cbo}
\formulas{$\ddot{x}(t)+x(t)=\sin{t}$, $x(0)=0$, $\dot{x}(0)=0$}.
\begin{itemize}
	\item $\displaystyle s^2
	\mathcal{L}_t[x(t)](s))+\mathcal{L}_t[x(t)](s)=
	s x(0)+\dot{x}(0)+\frac{1}{s^2+1}$,
	\item $\displaystyle\mathcal{L}_t[x(t)](s)=
   	\frac{1}{(s^2+1)^2}$, 
	\item\formulas{$x(t)=\frac{1}{2} (\sin (t)-t \cos (t))$}.
\end{itemize}
\end{frame}
\begin{frame}{Graf funkcije $x(t)$}
	\begin{center}
		\includegraphics[scale=1]{funkcija-2-4.pdf}
	\end{center}
\end{frame}

\begin{frame}{Re"si diferencialno ena"cbo}
\formulas{$\ddot{x}(t)+x(t)=\sin{(2 t)}$, $x(0)=0$, $\dot{x}(0)=0$}.
\begin{itemize}
\item $\displaystyle s^2
   	\mathcal{L}_t[x(t)](s)+
	\mathcal{L}_t[x(t)](s)-s x(0)-\dot{x}(0)=
	\frac{2}{s^2+4}$,
\item $\displaystyle\mathcal{L}_t[x(t)](s)=
   \frac{2}{\left(s^2+1\right)
   \left(s^2+4\right)}$, 	
\item \formulas{$\displaystyle x(t)=\frac{2}{3} \sin{t}-\frac{1}{3}\sin{(2t)}$}.
\end{itemize}
\end{frame}

\begin{frame}{Graf funkcije $x(t)$}
\begin{center}
\includegraphics[scale=1]{funkcija-2-5.pdf}
\end{center}
\end{frame}


\begin{frame}{Re"si diferencialno ena"cbo}
\formulas{$\ddot{x}(t)+4\dot{x}(t)+4x(t)=0$, 
$x(0)=1$, $\dot{x}(0)=4$}.
\begin{itemize}
	\item $\displaystyle
	s^2\left(\mathcal{L}_t[x(t)](s)\right)+
	4\left(\mathcal{L}_t[x(t)](s)\right)+
	4\left(s\left(\mathcal{L}_t[x(t)](s)\right)-1\right)
	-s-4=0	
	$ 
	\item $\displaystyle
	\mathcal{L}_t[x(t)](s)=\frac{s+8}{(s+2)^2}
	$ 
	\item \formulas{$\displaystyle 
	x(t)=e^{-2 t}(6 t+1)	
	$}
\end{itemize}
\end{frame}

\begin{frame}{Re"si diferencialno ena"cbo}
\formulas{$\ddot{x}(t)+4\dot{x}(t)+4x(t)=0$, 
$x(0)=0$, $\dot{x}(0)=1$}.
\begin{itemize}
\item $s^2 \mathcal{L}_t[x(t)](s)+4
   \mathcal{L}_t[x(t)](s)+4 s
   \mathcal{L}_t[x(t)](s)=4 x(0)
   +s x(0)+\dot{x}(0)$, 
\item $\displaystyle\mathcal{L}_t[x(t)](s)= \frac{1}{(s+2)^2}$, 
\item \formulas{$\displaystyle x(t)=e^{-2 t} t$}.
\end{itemize}
\end{frame}
\begin{frame}
	\begin{center}
		\includegraphics{funkcija-2-3.pdf}
	\end{center}
\end{frame}


\begin{frame}{Re"si diferencialno ena"cbo}
\formulas{$\ddot{x}(t)+\dot{x}(t)+4x(t)=0$, 
$x(0)=0$, $\dot{x}(0)=1$}.
\begin{itemize}
\item $\displaystyle s^2 \mathcal{L}_t[x(t)](s)+s
   \mathcal{L}_t[x(t)](s)+4
   \mathcal{L}_t[x(t)](s)=s
   x(0)+\dot{x}(0)+x(0)$, 
\item $\displaystyle\mathcal{L}_t[x(t)](s)\to
   	\frac{1}{s^2+s+4}$, 
\item \formulas{$\displaystyle x(t)=\frac{2}{\sqrt{15}} e^{-t/2} \sin
   \left(\frac{\sqrt{15}
   t}{2}\right)$.}
\end{itemize}
\end{frame}
\begin{frame}{Graf funkcije $x(t)$}
\begin{center}
	\includegraphics[scale=1]{funkcija-2-2.pdf}
\end{center}
\end{frame}

\begin{frame}{Re"si diferencialno ena"cbo}
\formulas{$\ddot{x}(t)+4x(t)=\sin{(2 t)}$, 
$x(0)=0$, $\dot{x}(0)=1$}.
\begin{itemize}
\item $\displaystyle s^2 \left(\mathcal{L}_t[x(t)](s)\right)+4
   \left(\mathcal{L}_t[x(t)](s)\right)-s
   x(0)-\dot{x}(0)=\frac{2}{s^2+4}$,
\item $\displaystyle\mathcal{L}_t[x(t)](s)=
   \frac{s^2+6}{\left(s^2+4\right)^2}$,
\item \formulas{$\displaystyle x(t)=\frac{1}{8} (5 \sin (2 t)-2 t \cos (2 t))$} 
\end{itemize}
\end{frame}

\begin{frame}{Re"si diferencialno ena"cbo
$\ddot{x}(t)+4x(t)=f(t)$, kjer je} 
\formulas{$x(0)=0$, $\dot{x}(0)=0$ in  
$f(t)\begin{cases}1,&0\le t \le 1\\-1,& 1 < t \le 2\\
0,& \text{drugod} \end{cases}$}.
\begin{itemize}
	\item $\displaystyle\left(s^2+4\right)
   	\left(\mathcal{L}_t[x(t)](s)\right)=s\, x(0)+\dot{x}(0)+
	\frac{1}{s}\left(1-2e^{-s}+e^{-2 s}\right)$, 
	\item $\displaystyle\mathcal{L}_t[x(t)](s)=
	\frac{e^{-2 s}\left(e^s-1\right)^2}{s\left(s^2+4\right)}$,
	\item \formulas{$x(t)=\frac{1}{2}\left(\theta (t-2) \sin^2(2-t)+
	\theta (t-1) (\cos (2-2t)-1)+\sin^2(t)\right)$.}
\end{itemize}
\end{frame}

\begin{frame}{Graf funkcije $x(t)$}
\begin{center}
	\includegraphics[scale=1]{funkcija-2-1.pdf}
\end{center}
\end{frame}

\begin{frame}{Re"si sistem diferencialnih ena"cb}
\formulas{\begin{align*}
&\dot{x}(t)=y(t)-x(t)+z(t), & x(0)=0,\\
&\dot{y}(t)=x(t)-y(t),&  y(0)=0,\\
&\dot{z}(t)=-z(t), &  z(0)=1.
\end{align*}}

\begin{itemize}
\item 
\begin{align*}
   s\mathcal{L}_t[x(t)](s)&=
	-\mathcal{L}_t[x(t)](s)+
	\mathcal{L}_t[y(t)](s)+
	\mathcal{L}_t[z(t)](s)\\s
   \mathcal{L}_t[y(t)](s)&=
	\mathcal{L}_t[x(t)](s)-\mathcal{L}_t[y(t)](s)\\
	s\mathcal{L}_t[z(t)](s)&=
	-\mathcal{L}_t[z(t)](s)+1
\end{align*}
\item $x(t)=\frac{1}{2}
   e^{-2 t} \left(e^{2 t}-1\right),
   y(t)=\frac{1}{2} e^{-2 t}
   \left(e^t-1\right)^2,
   z(t)=e^{-t}$
\end{itemize}
\end{frame}

\begin{frame}{Re"si integralsko ena"cbo}
\formulas{$x(t)=t^2+\int_0^tx(\tau)d\tau$.}
\begin{itemize}
	\item $\displaystyle\mathcal{L}_t[x(t)](s)=
	\frac{\mathcal{L}_t[x(t)](s)}{s}+\frac{2}{s^3}$,
	\item $\displaystyle \mathcal{L}_t[x(t)](s)=
	\frac{2}{(s-1)s^2}$, 
	\item \formulas{$\displaystyle x(t)\to 2
 	\left(-t+e^t-1\right)$}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si integralsko ena"cbo}
\formulas{$x(t)=t+2-2\cos(t)-\int_0^t(t-\tau)x(\tau)d\tau$.}
\begin{itemize}
        \item $\displaystyle 
	\mathcal{L}_t[x(t)](s)=-\frac{\mathcal{L}_t[x(t)](s)}
	{s^2}-\frac{2s}{s^2+1}+\frac{1}{s^2}+\frac{2}{s}$,
        \item $\displaystyle \mathcal{L}_t[x(t)](s)=
	\frac{s^2+2s+1}{\left(s^2+1\right)^2}$,
        \item \formulas{$\displaystyle x(t)=t \sin (t)+\sin (t)$}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si integralsko ena"cbo}
\formulas{$x(t)=t-\int_0^tx(\tau)\cos(t-\tau)d\tau$.}
\begin{itemize}
	\item $\displaystyle 
	\mathcal{L}_t[x(t)](s)=\frac{1}{s^2}
	-\frac{s \left(\mathcal{L}_t[x(t)](s)\right)}{s^2+1}$
	\item $\displaystyle\mathcal{L}_t[x(t)](s)=
	\frac{s^2+1}{s^2\left(s^2+s+1\right)}$ 
	\item \formulas{$\displaystyle 
	x(t)=-1+t+\frac{e^{-t/2}}{\sqrt{3}}
   	\left(\sqrt{3} \cos\left(\frac{\sqrt{3}t}{2}\right)-
	\sin\left(\frac{\sqrt{3}t}{2}\right)\right)$}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si integro-diferencialno ena"cbo}
\formulas{$\dot{x}(t)=t+\int_0^tx(\tau)\cos(t-\tau)d\tau$, $x(0)=1$}
\begin{itemize}
	\item $\displaystyle
	s\left(\mathcal{L}_t[x(t)](s)\right)-1=
	\frac{s\left(\mathcal{L}_t[x(t)](s)\right)}{s^2+1}+
	\frac{1}{s^2}
	$
	\item $\displaystyle
	\mathcal{L}_t[x(t)](s)=
	\frac{\left(s^2+1\right)^2}{s^5}
	$ 
	\item \formulas{$\displaystyle 
	x(t)=\frac{t^4}{24}+t^2+1$}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si integro-diferencialno ena"cbo}
\formulas{$\displaystyle\ddot{x}(t)+\int_0^t(x(\tau)+\ddot{x}(\tau))\sin(t-\tau)d\tau=
2\cos(t)$,\\[5mm] $x(0)=0$, $\dot{x}(0)=0$}.
\begin{itemize}
	\item $\displaystyle s^2 \left(\mathcal{L}_t[x(t)](s)\right)+
	\mathcal{L}_t[x(t)](s)=\frac{2 s}{s^2+1}$,
	\item $\displaystyle \mathcal{L}_t[x(t)](s)=
	\frac{2s}{\left(s^2+1\right)^2}$, 
	\item \formulas{$\displaystyle x(t)=t \sin (t)$}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si sistem integralskih ena"cb}
\formulas{$x(t)=t+\int_0^ty(\tau)d\tau$, $y(t)=1+\int_0^tx(\tau)d\tau$}
\begin{itemize}
	\item $\displaystyle 
	\mathcal{L}_t[x(t)](s)=
   	\frac{\mathcal{L}_t[y(t)](s)}{s}+\frac{1}{s^2}$\\
	$\displaystyle\mathcal{L}_t[y(t)](s)=\frac{\mathcal{L}_t[x(t)](s)}{s}+
	\frac{1}{s}
	$
	\item $\displaystyle
	\mathcal{L}_t[x(t)](s)=\frac{2}{s^2-1},\quad 
	\mathcal{L}_t[y(t)](s)=\frac{2s}{s^2-1}-\frac{1}{s}
	$ 
	\item \formulas{$\displaystyle   
	x(t)= e^{-t}\left(e^{2t}-1\right),\quad 
	y(t)=e^{-t}+e^t-1
	$} 
\end{itemize}
\end{frame}
\end{document}

