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\QTR{small}{Matematick\U{e1} anal\U{fd}za III online - Laplaceova transform\U{e1}cia - Aplik\U{e1}cie Laplaceovej transform\U{e1}cie\dotfill \thepage }}
%}
\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}
\begin{document}
\author{A. U. Thor}
\title{Lab Report}
\date{The Date }
\maketitle
\begin{abstract}
A Laboratory report created with Scientific Notebook
\end{abstract}
\section{Laplaceova transform\'{a}cia}
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\textbf{%
%TCIMACRO{\hyperref{Obsah}{}{}{mcindex.tex}}%
%BeginExpansion
\msihyperref{Obsah}{}{}{mcindex.tex}%
%EndExpansion
} & \textbf{%
%TCIMACRO{\hyperref{Obsah kapitoly}{}{}{K4.tex}}%
%BeginExpansion
\msihyperref{Obsah kapitoly}{}{}{K4.tex}%
%EndExpansion
} & \textbf{%
%TCIMACRO{\hyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{K45.tex}}%
%BeginExpansion
\msihyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{K45.tex}%
%EndExpansion
} & \textbf{%
%TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{Ot4.tex}}%
%BeginExpansion
\msihyperref{Ot\'{a}zky}{}{}{Ot4.tex}%
%EndExpansion
} & \textbf{%
%TCIMACRO{\hyperref{Cvi\v{c}enia}{}{}{Cv4.tex}}%
%BeginExpansion
\msihyperref{Cvi\v{c}enia}{}{}{Cv4.tex}%
%EndExpansion
} & \textbf{%
%TCIMACRO{\hyperref{Index}{}{}{Ind.tex}}%
%BeginExpansion
\msihyperref{Index}{}{}{Ind.tex}%
%EndExpansion
} \\ \hline
\end{tabular}
\end{center}
\section{Aplik\'{a}cie Laplaceovej transform\'{a}cie}
\subsection{Rie\v{s}enie line\'{a}rnych diferenci\'{a}lnych rovn\'{\i}c s
kon{}\v{s}tantn\'{y}mi koeficientami.}
Ak $a_{k}\in \mathbf{C},\,k=0,1,\dots ,n;\,a_{n}\neq 0,\,b_{k}\in \mathbf{C}%
,\,k=0,1,\dots ,n-1$ s\'{u} dan\'{e} komplexn\'{e} kon\v{s}tanty a $f:%
\mathbf{R\longrightarrow C}$ je po \v{c}astiach spojit\'{a} funkcia, m\^{o}%
\v{z}me sformulova\v{t} \emph{za\v{c}iato\v{c}n\'{u} \'{u}lohu\label{1}} - Z%
\'{U} ako probl\'{e}m n\'{a}js\v{t} rie\v{s}enie oby\v{c}ajnej diferenci\'{a}%
lnej rovnice
\begin{equation}
a_{n}x^{\left( n\right) }\left( t\right) +a_{n-1}x^{\left( n-1\right)
}\left( t\right) +\dots +a_{1}x^{\prime }\left( t\right) +a_{0}x\left(
t\right) =f\left( t\right) \tag{(1)}
\end{equation}%
pre $t>0,$ ktor\'{e} bude sp\'{l}\v{n}a\v{t} podmienky
\begin{equation}
x\left( 0+\right) =b_{0},x^{\prime }\left( 0+\right) =b_{1},\dots ,x^{\left(
n-1\right) }\left( 0+\right) =b_{n-1} \tag{(2)}
\end{equation}%
Z te\'{o}rie oby\v{c}ajn\'{y}ch diferenci\'{a}lnych rovn\'{\i}c vieme, \v{z}%
e Z\'{U} (1), (2) m\'{a} jedin\'{e} rie\v{s}enie. T\'{u}to \'{u}lohu mo\v{z}%
no rie\v{s}i\v{t} aj pomocou Laplaceovej transform\'{a}cie, ak budeme
predpoklada\v{t}, \v{z}e funkcia $f\left( t\right) $ je origin\'{a}l. Potom
aj rie\v{s}enie - funkcia $x:\mathbf{R\longrightarrow C},$ ktor\'{u} roz\v{s}%
\'{\i}rime nulou pre $t<0$ so svojimi deriv\'{a}ciami do r\'{a}du $n$ s\'{u}
origin\'{a}ly. Toto plynie z faktu, \v{z}e
\begin{equation*}
x^{\left( j\right) }\left( t\right) =\sum_{k=1}^{n}\left[ b_{k}+\int_{0}^{t}%
\frac{W_{k}\left( \tau \right) }{W\left( \tau \right) }d\tau \right] \varphi
_{k}\left( t\right) ,\,j=0,1,\dots ,n-1
\end{equation*}%
kde $\left\{ \varphi _{1},\dots ,\varphi _{n}\right\} $ je b\'{a}za vektorov%
\'{e}ho priestoru rie\v{s}en\'{\i} rovnice (1) s nulovou pravou stranou:
\begin{equation*}
a_{n}x^{\left( n\right) }\left( t\right) +a_{n-1}x^{\left( n-1\right)
}\left( t\right) +\dots +a_{1}x^{\prime }\left( t\right) +a_{0}x\left(
t\right) =0
\end{equation*}%
a
\begin{equation*}
W\left( t\right) =\left(
\begin{tabular}{cccc}
$\varphi _{1}\left( t\right) $ & $\varphi _{2}\left( t\right) $ & $\dots $ &
$\varphi _{n}\left( t\right) $ \\
$\varphi _{1}^{\prime }\left( t\right) $ & $\varphi _{2}^{\prime }\left(
t\right) $ & $\dots $ & $\varphi _{n}^{\prime }\left( t\right) $ \\
$\vdots $ & $\vdots $ & $\ddots $ & $\vdots $ \\
$\varphi _{1}^{\left( n-1\right) }\left( t\right) $ & $\varphi _{2}^{\left(
n-1\right) }\left( t\right) $ & $\dots $ & $\varphi _{n}^{\left( n-1\right)
}\left( t\right) $%
\end{tabular}%
\right)
\end{equation*}%
\begin{equation*}
W_{k}\left( t\right) =\left(
\begin{tabular}{cccccc}
$\varphi _{1}\left( t\right) $ & $\varphi _{2}\left( t\right) $ & $\dots $ &
$0$ & $\dots $ & $\varphi _{n}\left( t\right) $ \\
$\varphi _{1}^{\prime }\left( t\right) $ & $\varphi _{2}^{\prime }\left(
t\right) $ & $\dots $ & $0$ & $\dots $ & $\varphi _{n}^{\prime }\left(
t\right) $ \\
$\vdots $ & $\vdots $ & $\vdots $ & $\vdots $ & $\ddots $ & $\vdots $ \\
$\varphi _{1}^{\left( n-1\right) }\left( t\right) $ & $\varphi _{2}^{\left(
n-1\right) }\left( t\right) $ & $\dots $ & $f\left( t\right) $ & $\dots $ & $%
\varphi _{n}^{\left( n-1\right) }\left( t\right) $%
\end{tabular}%
\right) ,
\end{equation*}%
pri\v{c}om st\'{l}pec prav\'{y}ch str\'{a}n sa nach\'{a}dza v k-tom st\'{l}%
pci. Funkcie $\varphi _{k},\,\,k=1,\dots ,n$ s\'{u} origin\'{a}ly, preto\v{z}%
e maj\'{u} tvar $\varphi \left( t\right) =t^{\alpha }e^{\beta t},$ kde $%
\alpha ,\,\beta $ s\'{u} vhodn\'{e} kon\v{s}tanty. Rie\v{s}enie $x$ spolu s
deriv\'{a}ciami do r\'{a}du $\left( n-1\right) $ dostaneme z funkci\'{\i} $%
\varphi _{1},\dots ,\varphi _{n},f$ pou\v{z}it\'{\i}m oper\'{a}ci\'{\i}, ktor%
\'{e} zachov\'{a}vaj\'{u} exponenci\'{a}lny rast, to znamen\'{a} \v{z}e
funkcie $x,x^{\prime },\dots ,x^{\left( n-1\right) }$ s\'{u} origin\'{a}ly.
Funkcia $x^{\left( n\right) }$ je tie\v{z} origin\'{a}l, preto\v{z}e sa d%
\'{a} vyjadri\v{t} z rovnice (1) v tvare
\begin{equation*}
x^{\left( n\right) }\left( t\right) =a_{n}^{-1}\left[ f\left( t\right)
-\sum_{k=0}^{n-1}a_{k}x^{\left( k\right) }\left( t\right) \right] ,\,t\geq 0.
\end{equation*}%
Potom aplik\'{a}ciou Laplaceovej transform\'{a}cie na \'{u}lohu (1) s vyu%
\v{z}it\'{\i}m podmienok (2) dostaneme
\begin{equation*}
a_{n}\left[ p^{n}X\left( p\right) -p^{n-1}b_{0}-p^{n-2}b_{1}-\dots
-pb_{n-2}-b_{n-1}\right] +
\end{equation*}%
\begin{equation*}
+a_{n-1}\left[ p^{n-1}X\left( p\right) -p^{n-2}b_{0}-p^{n-3}b_{1}-\dots
-pb_{n-3}-b_{n-2}\right] +
\end{equation*}%
\begin{equation*}
+\dots +
\end{equation*}%
\begin{equation*}
+a_{1}\left[ pX\left( p\right) -b_{0}\right] +a_{0}X\left( p\right) =F\left(
p\right)
\end{equation*}%
kde $F\left( p\right) =\mathbf{L}\left[ f\left( t\right) \right] $ a $%
X\left( p\right) =\mathbf{L}\left[ x\left( t\right) \right] .$ Rie\v{s}en%
\'{\i}m tejto algebrickej rovnice dostaneme
\begin{equation*}
X\left( p\right) =\frac{F\left( p\right) }{Q_{n}\left( p\right) }+\frac{%
P_{n-1}\left( p\right) }{Q_{n}\left( p\right) }.
\end{equation*}%
V poslednom v\'{y}raze je $Q_{n}\left( p\right) $ charakteristick\'{y} polyn%
\'{o}m rovnice (1) n-t\'{e}ho stup\v{n}a a $P_{n-1}\left( p\right) $ je polyn%
\'{o}m najviac $\left( n-1\right) $ stup\v{n}a. Pou\v{z}it\'{\i}m vety o
konvol\'{u}cii dostaneme origin\'{a}l v tvare
\begin{equation*}
x\left( t\right) =\int_{0}^{t}f\left( s\right) g\left( t-s\right) ds+h\left(
t\right) ,\,t\geq 0,
\end{equation*}%
kde
\begin{equation*}
\mathbf{L}\left[ g\right] =\frac{1}{Q_{n}},\,\mathbf{L}\left[ h\right] =%
\frac{P_{n-1}}{Q_{n}}.
\end{equation*}
\begin{example}
Rie\v{s}te za\v{c}iato\v{c}n\'{u} \'{u}lohu:
\begin{equation*}
x^{\prime \prime }-2x^{\prime }+5x=e^{t}\cos 2t,\,x\left( 0+\right)
=C_{1},\,x^{\prime }\left( 0+\right) =C_{2}.
\end{equation*}
\end{example}
\begin{solution}
Pou\v{z}it\'{\i}m vz\v{t}ahu $X\left( p\right) =\mathbf{L}\left[ x\left(
t\right) \right] $ dostaneme $\mathbf{L}\left[ x\,^{\prime }\left( t\right) %
\right] =pX\left( p\right) -C_{1},\,\mathbf{L}\left[ x\,^{\prime \prime
}\left( t\right) \right] =p^{2}X\left( p\right) -pC_{1}-C_{2}.$\ $\mathbf{L}%
\left[ e^{t}\cos 2t\right] =\frac{p-1}{\left( p-1\right) ^{2}+4}.$\ Aplik%
\'{a}ciou Laplaceovej transform\'{a}cie na rovnicu dostaneme:%
\begin{equation*}
p^{2}X\left( p\right) -pC_{1}-C_{2}-2[pX\left( p\right) -C_{1}]+5X\left(
p\right) =\frac{p-1}{\left( p-1\right) ^{2}+4},
\end{equation*}%
odkia\v{l}
\begin{equation*}
X\left( p\right) \left[ p^{2}-2p+5\right] =\frac{p-1}{\left( p-1\right)
^{2}+4}+pC_{1}+C_{2}-2C_{1}.
\end{equation*}%
\v{c}o d\'{a}va
\begin{equation*}
X\left( p\right) =\frac{p-1}{\left[ \left( p-1\right) ^{2}+4\right] ^{2}}%
+C_{1}\frac{p}{\left( p-1\right) ^{2}+4}+(C_{2}-2C_{1})\frac{1}{\left(
p-1\right) ^{2}+4}.
\end{equation*}%
Aplik\'{a}ciou vety o inverznej transform\'{a}cii alebo vety o konvol\'{u}%
cii dostaneme:
\begin{equation*}
\mathbf{L}\left[ \left( \frac{1}{4}te^{t}\sin 2t\right) \right] =\frac{p-1}{%
\left[ \left( p-1\right) ^{2}+4\right] ^{2}}
\end{equation*}%
a
\begin{equation*}
\mathbf{L}\left[ \left( e^{t}\cos 2t\right) \right] =\frac{p-1}{\left(
p-1\right) ^{2}+4},\,\mathbf{L}\left[ \left( \frac{1}{2}e^{t}\sin 2t\right) %
\right] =\frac{1}{\left( p-1\right) ^{2}+4},
\end{equation*}%
\v{c}o implikuje, \v{z}e
\begin{equation*}
x\left( t\right) =\frac{1}{4}te^{t}\sin 2t+C_{1}e^{t}\cos 2t+(C_{2}-2C_{1})%
\frac{1}{2}e^{t}\sin 2t.\,\square
\end{equation*}
\end{solution}
Laplaceovu transform\'{a}ciu m\^{o}\v{z}me pou\v{z}i\v{t} pri rie\v{s}en%
\'{\i} oby\v{c}ajn\'{y}ch diferenci\'{a}lnych rovn\'{\i}c s kon\v{s}tantn%
\'{y}mi koeficientami, ktor\'{y}ch prav\'{a} strana m\'{a} tvar posunutej
funkcie alebo tvar kone\v{c}n\'{e}ho impulzu. Uva\v{z}ujme teda za\v{c}iato%
\v{c}n\'{u} \'{u}lohu:
\begin{equation*}
a_{n}x^{\left( n\right) }\left( t\right) +a_{n-1}x^{\left( n-1\right)
}\left( t\right) +\dots +a_{1}x^{\prime }\left( t\right) +a_{0}x\left(
t\right) =f\left( t\right)
\end{equation*}%
pre $t>0,$ ktor\'{e} bude sp\'{l}\v{n}a\v{t} podmienky
\begin{equation*}
x\left( 0+\right) =b_{0},x^{\prime }\left( 0+\right) =b_{1},\dots ,x^{\left(
n-1\right) }\left( 0+\right) =b_{n-1},
\end{equation*}%
t.j. Z\'{U} (1), (2) s pravou stranou tvaru:
\begin{equation}
f\left( t\right) =\eta \left( t-\tau _{1}\right) f_{1}\left( t-\tau
_{1}\right) +\dots +\eta \left( t-\tau _{k}\right) f_{k}\left( t-\tau
_{k}\right) \tag{(3)}
\end{equation}%
kde funkcie $f_{1},\dots ,f_{k}$ s\'{u} origin\'{a}ly a $0\leq \tau
_{1}<\dots <\tau _{k}.$ Pou\v{z}it\'{\i}m Laplaceovej transform\'{a}cie na t%
\'{u}to Z\'{U} dostaneme:
\begin{equation*}
Q_{n}\left( p\right) X\left( p\right) -P_{n-1}\left( p\right) =e^{-\tau
_{1}p}F_{1}\left( p\right) +\dots +e^{-\tau _{k}p}F_{k}\left( p\right) .
\end{equation*}%
odkia\v{l}
\begin{equation*}
X\left( p\right) =\frac{P_{n-1}\left( p\right) }{Q_{n}\left( p\right) }%
+e^{-\tau _{1}p}G_{1}\left( p\right) +\dots +e^{-\tau _{k}p}G_{k}\left(
p\right) ,
\end{equation*}%
kde
\begin{equation*}
G_{j}=\frac{F_{j}}{Qn},\,j=1,2,\dots ,k.
\end{equation*}%
Pou\v{z}it\'{\i}m
%TCIMACRO{\hyperref{vety o posune v origin\'{a}le}{}{}{K44.tex#10} }%
%BeginExpansion
\msihyperref{vety o posune v origin\'{a}le}{}{}{K44.tex#10}
%EndExpansion
dostaneme origin\'{a}l v tvare
\begin{equation*}
x\left( t\right) =h\left( t\right) +\eta \left( t-\tau _{1}\right)
g_{1}\left( t-\tau _{1}\right) +\dots +\eta \left( t-\tau _{k}\right)
g_{k}\left( t-\tau _{k}\right) ,
\end{equation*}%
pri\v{c}om
\begin{equation*}
\mathbf{L}\left[ h\right] =\frac{P_{n-1}}{Q_{n}},\,\mathbf{L}\left[ g_{j}%
\right] =G_{j},\,j=1,2,\dots ,k.
\end{equation*}
\begin{example}
N\'{a}jdite rie\v{s}enie Z\'{U}:
\begin{equation*}
x^{\prime \prime }+4x=f\left( t\right) ,\,x\left( 0+\right) =1,\,x^{\prime
}\left( 0+\right) =0,\,\text{\ kde \ }\,f\left( t\right) =\left\{
\begin{tabular}{ccc}
$0$ & pre & $t<0\,$ \\
$\cos t$ & pre & $0\leq t<\frac{\pi }{2}$ \\
$0$ & pre & $t\geq \frac{\pi }{2}$%
\end{tabular}%
,\right.
\end{equation*}
\end{example}
\begin{solution}
Pou\v{z}it\'{\i}m vz\v{t}ahu $X\left( p\right) =\mathbf{L}\left[ x\left(
t\right) \right] $ dostaneme $\mathbf{L}\left[ x\,^{\prime }\left( t\right) %
\right] =pX\left( p\right) -1,\,\mathbf{L}\left[ x\,^{\prime \prime }\left(
t\right) \right] =p^{2}X\left( p\right) -p.$\ Prav\'{u} stranu rovnice mo%
\v{z}no zap\'{\i}sa\v{t} v tvare
\begin{equation*}
f\left( t\right) =\cos t\left[ \eta \left( t\right) -\eta \left( t-\frac{\pi
}{2}\right) \right] =\eta \left( t\right) \cos t+\eta \left( t-\frac{\pi }{2}%
\right) \sin \left( t-\frac{\pi }{2}\right) ,
\end{equation*}%
\begin{equation*}
\,\mathbf{L}\left[ f\left( t\right) \right] =\frac{p}{p^{2}+1}+e^{-p\frac{%
\pi }{2}}\frac{1}{p^{2}+1}.
\end{equation*}%
Aplik\'{a}ciou Laplaceovej transform\'{a}cie na rovnicu dostaneme:
\begin{equation*}
p^{2}X\left( p\right) -p+4X\left( p\right) =\frac{p}{p^{2}+1}+e^{-p\frac{\pi
}{2}}\frac{1}{p^{2}+1},
\end{equation*}%
t.j.
\begin{equation*}
X\left( p\right) =\frac{p}{p^{2}+4}+\frac{p}{\left( p^{2}+1\right) \left(
p^{2}+4\right) }+e^{-p\frac{\pi }{2}}\frac{1}{\left( p^{2}+1\right) \left(
p^{2}+4\right) },
\end{equation*}%
teda
\begin{equation*}
x\left( t\right) =x_{1}\left( t\right) +x_{2}\left( t\right) +\eta \left( t-%
\frac{\pi }{2}\right) x_{3}\left( t-\frac{\pi }{2}\right) ,
\end{equation*}%
pri\v{c}om
\begin{equation*}
\mathbf{L}\left[ x_{1}\left( t\right) \right] =\frac{p}{p^{2}+4},\,\mathbf{L}%
\left[ x_{2}\left( t\right) \right] =\frac{p}{\left( p^{2}+1\right) \left(
p^{2}+4\right) },\,\mathbf{L}\left[ x_{3}\left( t\right) \right] =\frac{1}{%
\left( p^{2}+1\right) \left( p^{2}+4\right) }.
\end{equation*}%
Preto\v{z}e plat\'{\i}
\begin{equation*}
\frac{p}{\left( p^{2}+1\right) \left( p^{2}+4\right) }=\frac{1}{3}\frac{p}{%
p^{2}+1}-\frac{1}{3}\frac{p}{p^{2}+4}
\end{equation*}%
a
\begin{equation*}
\frac{1}{\left( p^{2}+1\right) \left( p^{2}+4\right) }=\frac{1}{3}\frac{1}{%
p^{2}+1}-\frac{1}{3}\frac{1}{p^{2}+4},
\end{equation*}%
tak
\begin{equation*}
x_{1}\left( t\right) =\cos 2t,\,x_{2}\left( t\right) =\frac{1}{3}\cos t-%
\frac{1}{3}\cos 2t,\,x_{3}\left( t\right) =\frac{1}{3}\cos t-\frac{1}{6}\cos
2t
\end{equation*}%
a
\begin{equation*}
x\left( t\right) =\frac{1}{3}\cos t+\frac{2}{3}\cos 2t+\eta \left( t-\frac{%
\pi }{2}\right) \left[ \frac{1}{3}\sin \left( t-\frac{\pi }{2}\right) -\frac{%
1}{6}\sin 2\left( t-\frac{\pi }{2}\right) \right] .\,\square
\end{equation*}
\end{solution}
V elektrotechnike sa \v{c}asto stret\'{a}vame s \'{u}lohou rie\v{s}i\v{t} oby%
\v{c}ajn\'{u} diferenci\'{a}lnu rovnicu s periodickou funkciou na pravej
strane.
\begin{equation}
a_{n}x^{\left( n\right) }\left( t\right) +a_{n-1}x^{\left( n-1\right)
}\left( t\right) +\dots +a_{1}x^{\prime }\left( t\right) +a_{0}x\left(
t\right) =\eta \left( t\right) f\left( t\right) \tag{(4)}
\end{equation}%
\begin{equation}
x\left( 0+\right) =b_{0},x^{\prime }\left( 0+\right) =b_{1},\dots
,x^{(n-1)}\left( 0+\right) =b_{n-1} \tag{(5)}
\end{equation}%
s periodickou funkciou $f,$ $f\left( t\right) =f\left( t+T\right) .$ Rie\v{s}%
enie takejto Z\'{U} h\v{l}ad\'{a}me v tvare
\begin{equation*}
x\left( t\right) =y\left( t\right) +z\left( t\right) ,\,t>0;
\end{equation*}%
kde $y$ je periodick\'{e} rie\v{s}enie rovnice (4) a $z$ je rie\v{s}en\'{\i}%
m Z\'{U}
\begin{equation*}
a_{n}z^{\left( n\right) }\left( t\right) +a_{n-1}z^{\left( n-1\right)
}\left( t\right) +\dots +a_{1}z^{\prime }\left( t\right) +a_{0}z\left(
t\right) =0,
\end{equation*}%
s podmienkami
\begin{equation*}
z\left( 0+\right) =c_{0},z^{\prime }\left( 0+\right) =c_{1},\dots ,z^{\left(
n-1\right) }\left( 0+\right) =c_{n-1},
\end{equation*}%
kde
\begin{equation*}
c_{j}=b_{j}-y^{\left( j\right) }\left( 0+\right) ,\,j=0,1,\dots ,n-1.
\end{equation*}%
Sk\'{u}majme teda najsk\^{o}r \'{u}lohu
\begin{equation}
a_{n}y^{\left( n\right) }\left( t\right) +a_{n-1}y^{\left( n-1\right)
}\left( t\right) +\dots +a_{1}y^{\prime }\left( t\right) +a_{0}y\left(
t\right) =\eta \left( t\right) f\left( t\right) , \tag{(6)}
\end{equation}%
kde $f:\mathbf{R\longrightarrow C}$ je funkcia s periodou $T,$ t.j. $f\left(
t+T\right) =f\left( t\right) $ a my h\v{l}ad\'{a}me rie\v{s}enie $y\left(
t\right) ,$ tak aby platilo $y\left( t\right) =y\left( t+T\right) $ pre ka%
\v{z}d\'{e} $t\geq 0.$ Aplik\'{a}ciou Laplaceovej transform\'{a}cie na
rovnicu (4) dostaneme
\begin{equation*}
Y\left( p\right) =\frac{P_{n-1}\left( p\right) }{Q_{n}\left( p\right) }+%
\frac{F_{T}\left( p\right) }{Q_{n}\left( p\right) \left( 1-e^{-pT}\right) },
\end{equation*}%
kde $Y\left( p\right) =\mathbf{L}\left[ y\left( t\right) \right] $ a $%
F_{T}\left( p\right) $ je Laplaceova transform\'{a}cia funkcie $%
f_{\left\langle 0,T\right\rangle }\left( t\right) ,$ ktor\'{a} je definovan%
\'{a} vz\v{t}ahom
\begin{equation*}
f_{\left\langle 0,T\right\rangle }\left( t\right) =\left\{
\begin{tabular}{ccc}
$0$ & pre & $t<0,\,T\leq t$ \\
$f\left( t\right) $ & pre & $0\leq t