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88 lines
1.7 KiB
88 lines
1.7 KiB
3 years ago
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\documentclass{article}
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\usepackage[margin=2cm]{geometry}
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\usepackage{titlesec}
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\usepackage{titling}
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\usepackage[hidelinks]{hyperref}
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\usepackage{multicol}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{braket}
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\titleformat{\section}
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{\Large\bfseries}
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{}
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{0em}
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{}[\titlerule]
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\titleformat{\subsection}
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{\large\bfseries}
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{}
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{0em}
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{}
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\titlespacing{\subsection}
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{0em}{2em}{.4em}
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\titleformat{\subsubsection}[runin]
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{\bfseries}
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{}
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{0em}
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{}
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\titlespacing{\subsubsection}
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{0em}{2em}{1em}
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\renewcommand{\maketitle}{
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\begin{center}
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{\huge\bfseries\thetitle}\\
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\vspace{1em}
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{\Large\theauthor} \\
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\vspace{1em}
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elalmqvist@gmail.com --- \url{https://wych.dev}
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\end{center}
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}
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\begin{document}
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\title{Anteckningar 2022-04-05}
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\author{Elias Almqvist}
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\maketitle
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\newpage
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\section{Inhomogena diff-ekvationer av första-ordningen bevis}
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$$
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y' + ay = f(x)
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$$
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$$
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\text{Låt } g = y_{p1} - y_{p2}
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$$
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Där $y_{p1}$ och $y_{p2}$ (antagande) är partikulärlösningar till ekvationen $y' + ay = f(x)$. Alltså är $y_{pn}' + ay_{pn} = f(x)$ Detta ger den homogena ekvationen för $g$:
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$$
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g' + ag = y_{p1}' - y_{p2}' + a\left(y_{p1} - y_{p2}\right) = \left( y_{p1}' + ay_{p1} \right) - \left( y_{p2}' + ay_{p2} \right)
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$$
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Eftersom $\left( y_{p1}' + ay_{p1} \right) - \left( y_{p2}' + ay_{p2} \right) = f(x) - f(x)$ får vi att:
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$$
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g' + ag = \left( y_{p1}' + ay_{p1} \right) - \left( y_{p2}' + ay_{p2} \right) = f(x) - f(x) = 0
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$$
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Dvs $g = y_{p1} - y_{p2}$ är en lösning till $y' + ay = 0$
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$$
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\implies g + y_{p2} = y_{p1}
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$$
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$$
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\therefore y = g + y_{p2} \because g = y_{p1} - y_{p2}
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$$
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Då $g = y_h$ och $y_{p2} = y_{p}$
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$$
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y = y_h + y_p \quad \text{V.S.V}
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$$
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\end{document}
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