[Analysis] Add example

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RobinB27
2026-01-11 11:40:21 +01:00
parent 58b70a2ec6
commit f0c472bb87
2 changed files with 46 additions and 35 deletions

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@@ -31,6 +31,8 @@
\method \textbf{Educated Guess}\\
Usually, $y$ has a similar form to $b$:
\begin{footnotesize}
\begin{center}
\begin{tabular}{ll}
\hline
$b(x)$ & \text{Guess} \\
@@ -45,19 +47,17 @@ Usually, $y$ has a similar form to $b$:
$P_n(x) \cdot e^{\alpha x}\cos(\beta x)$ & $e^{\alpha x}\left( R_n(x) \sin(\beta x) + S_n(x) \cos(\beta x) \right)$\\
\hline
\end{tabular}
\end{center}
\end{footnotesize}
\remark If $\alpha, \beta$ are roots of $P(X)$ with multiplicity $j$, multiply guess with a $P_j(x)$.
\subsection{Linear Solutions: Constant Coefficients}
\textbf{Form:}
$$
y^{(k)} + a_{k-1}y^{(k-1)} + \ldots + a_1y' + a_0y = b
$$
\textbf{Form:} $ y^{(k)} + a_{k-1}y^{(k-1)} + \ldots + a_1y' + a_0y = b$\\
\subtext{Where $a_0, \ldots, a_{k-1} \in \C$ are constants, $b(x)$ is continuous.}
\subsubsection{Homogeneous Equations}
The idea is to find a Basis of $\S$:
\smalltext{The idea is to find a Basis of $\S$:}
\definition \textbf{Characteristic Polynomial} $P(X) = \prod_{i=1}^{k} (X-\alpha_i)$
@@ -82,7 +82,6 @@ $$
\end{subbox}
\subtext{Solutions exist $\forall Z = (z_1,\ldots,z_k)$ since that system's $\det(M_Z) \neq 0$.}
\newpage
\subsubsection{Inhomogeneous Equations}
\method \textbf{Undetermined Coefficients}: An educated guess.
@@ -98,13 +97,24 @@ $$
\remark \textbf{Applying Linearity}\\
If $b(x) = \sum_{i=1}^{n} b_i(x)$, A solution for $b(x)$ is $f(x) = \sum_{i=1}^{n} f_i(x)$\\
\subtext{Sometimes called \textit{Superposition Principle} in this context}
\subtext{Sometimes called \textit{Superposition Principle} in this context.}
\subsection{Other Methods}
\method \textbf{Change of Variable}\\
If $f(x)$ is replaced by $h(y) = f(g(y))$, then $h$ is a sol. too.\\
\subtext{Changes like $h(t) = f(e^t)$ may lead to useful properties.}
\subtext{Changes like $h(t) = f(e^t)$ may lead to, i.e. ODEs in constant coeffs}
\begin{footnotesize}
\textbf{Example:} $2xy' - y = 0$ \\
Using substitution: $x = e^t,\quad h(t) = y(e^t),\quad h'(t) = e^t \cdot y'(e^t)$
\begin{enumerate}
\item $2x \cdot y'(x) = 2 \cdot h'(t)$
\item $-y(x)$ = $-h(t)$
\end{enumerate}
So: $2xy' - y \overset{\text{sub}}{=} 2h'(t) - h(t) = 0$ \\
Yields: $h(t) = \alpha \cdot e^\frac{t}{2} \overset{\text{resub}}{\implies} y(x) = \alpha \cdot e^\frac{\ln(x)}{2} = \alpha \cdot \sqrt{x}$
\end{footnotesize}
\begin{subbox}{Separation of Variables}
Form:
@@ -119,7 +129,7 @@ If $f(x)$ is replaced by $h(y) = f(g(y))$, then $h$ is a sol. too.\\
\subtext{Usually $\int 1/a(y)\ \text{d}y$ can be solved directly for $\ln|a(y)|+c$.}
\subsection{Method Overview}
\begin{footnotesize}
\begin{center}
\begin{tabular}{l|l}
\textbf{Method} & \textbf{Use case} \\
@@ -131,3 +141,4 @@ If $f(x)$ is replaced by $h(y) = f(g(y))$, then $h$ is a sol. too.\\
Change of Variables & e.g. $y' = f(ax + by + c)$ \\
\end{tabular}
\end{center}
\end{footnotesize}