[Analysis] Catch up to present

semester3/analysis-ii/parts/diffeq/linear-ode/00_intro.tex

@@ -5,19 +5,21 @@ An ODE is considered linear if and only if the $y$s are only scaled and not part
 $y^{(k)} + a_{k - 1}y^{(k - 1)} + \ldots + a_1 y' + a_0 y = b$, with $a_i$ and $b$ functions in $x$.
 If $b(x) = 0 \smallhspace \forall x$, \bi{homogeneous}, else \bi{inhomogeneous}\\
 %
-\shorttheorem For open $I \subseteq \R$ and $k \geq 1$, for lin. ODE over $I$ with cont. $a_i$ we have:
-\begin{enumerate}
+\shorttheorem For open $I \subseteq \R$ and $k \geq 1$, for lin. ODE over $I$ with continuous $a_i$ we have:
+\rmvspace
+\begin{enumerate}[noitemsep]
     \item Set $\mathcal{S}$ of $k \times$ diff. sol. $f: I \rightarrow \C (\R)$ of the eq. is a complex (real) subspace of complex (real)-valued func. over $I$
     \item $\dim(\mathcal{S}) = k \smallhspace\forall x_0 \in I$ and any $(y_0, \ldots, y_{k - 1}) \in \C^k$, exists unique
           $f \in \mathcal{S}$ s.t. $f(x_0) = y_0, f'(x_0) = y_1, \ldots, f^{(k - 1)}(x_0) = y_{k - 1}$.
           If $a_i$ real-valued, same applies, but $\C$ replaced by $\R$.
-    \item Let $b$ cont. on $I$. Exists solution $f_0$ to inhom. lin. ODE and $\mathcal{S}_b$ is set of funct. $f + f_0$ where $f \in \mathcal{S}$\\
+    \item Let $b$ continuous on $I$. Exists solution $f_0$ to inhom. lin. ODE and $\mathcal{S}_b$ is set of funct. $f + f_0$ where $f \in \mathcal{S}$
 \end{enumerate}
 The solution space $\mathcal{S}$ is spanned by $k$ functions, which thus form a basis of $\mathcal{S}$. If inhomogeneous, $\mathcal{S}$ not vector space.
 
 \shade{gray}{Finding solutions (in general)}
-\begin{enumerate}[label=\bi{(\arabic*)}]
-    \item Find basis $\{ f_1, \ldots, f_k \}$ for $\mathcal{S}_0$ for homogeneous equation (set $b(x) = 0$).
-    \item If inhom. find $f_p$ that solves the equation. The set of solutions $\mathcal{S}_b = \{ f_h + f_p \divides f_h \in \mathcal{S_0} \}$.
-    \item If initial conditions, find equations $\in \mathcal{S}_b$ which fulfill conditions using SLE (as always)
+\rmvspace
+\begin{enumerate}[label=\bi{(\arabic*)}, noitemsep]
+    \item Find basis $\{ f_1, \ldots, f_k \}$ for $\mathcal{S}_0$ for homogeneous equation (set $b(x) = 0$) (i.e. find homogeneous part, solve it)
+    \item If inhomogeneous, find $f_p$ that solves the equation. The set of solutions is then $\mathcal{S}_b = \{ f_h + f_p \divides f_h \in \mathcal{S}_0 \}$.
+    \item If there are initial conditions, find equations $\in \mathcal{S}_b$ which fulfill conditions using SLE (as always)
 \end{enumerate}

semester3/analysis-ii/parts/diffeq/linear-ode/01_order-one.tex

@@ -1,7 +1,12 @@
 \newsectionNoPB
 \subsection{Linear differential equations of first order}
-\shade{gray}{Homogeneous equation} Move all $y$ to one side and all other vars to other. Integrate both
-
+\rmvspace
 \shortproposition Solution of $y' + ay = 0$ is of form $f(x) = z e^{-A(x)}$ with $A$ anti-derivative of $a$
 
-\TODO Improve procedure with notes from session \& SPAM
+\rmvspace
+\shade{gray}{Imhomogeneous equation} 
+\rmvspace
+\begin{enumerate}[noitemsep]
+    \item Plug all values into $y_p = \int b(x) e^{A(x)}$ ($A(x)$ in the exponent instead of $-A(x)$ as in the homogeneous solution)
+    \item Solve and the final $y(x) = y_h + y_p$. For initial value problem, determine coefficient $z$
+\end{enumerate}

semester3/analysis-ii/parts/diffeq/linear-ode/02_constant-coefficient.tex

@@ -2,6 +2,29 @@
 \subsection{Linear differential equations with constant coefficients}
 The coefficients $a_i$ are constant functions of form $a_i(x) = k$ with $k$ constant, where $b(x)$ can be any function.\\
 %
-\shade{gray}{Homogeneous Equation} Find \textit{characteristic polynomial} (of form $\lambda^k + a_{k - 1} \lambda^{k - 1} + \ldots + a_1 \lambda + a_0$ for order $k$ lin. ODE with coefficients $a_i$).
-Find the roots of polynomial. The solution space is given by $\{ x^{v_j - 1} e^{\gamma_i x} \divides v_j \in \N, \gamma_i \in \R \}$ where $v_j$ is the multiplicity of the root $\gamma_i$.
-For $\gamma_i = \alpha + \beta i \in \C$, we have $e^{\alpha x}\cos(\beta x)$, $e^{\alpha x}\sin(\beta x)$.
+\shade{gray}{Homogeneous Equation}\rmvspace
+\begin{enumerate}[noitemsep]
+    \item Find \textit{characteristic polynomial} (of form $\lambda^k + a_{k - 1} \lambda^{k - 1} + \ldots + a_1 \lambda + a_0$ for order $k$ lin. ODE with coefficients $a_i \in \R$).
+    \item Find the roots of polynomial. The solution space is given by $\{ z_j \cdot x^{v_j - 1} e^{\gamma_i x} \divides v_j \in \N, \gamma_i \in \R \}$ where $v_j$ is the multiplicity of the root $\gamma_i$.
+          For $\gamma_i = \alpha + \beta i \in \C$, we have $z_1 \cdot e^{\alpha x}\cos(\beta x)$, $z_2 \cdot e^{\alpha x}\sin(\beta x)$, representing the two complex conjugated solutions.
+\end{enumerate}
+
+\rmvspace
+\shade{gray}{Inhomogeneous Equation}\rmvspace
+\begin{enumerate}[noitemsep]
+    \item \bi{(Case 1)} $b(x) = c x^d e^{\alpha x}$, with special cases $x^d$ and $e^{\alpha x}$:
+          $f_p = Q(x) e^{\alpha x}$ with $Q$ a polynomial with $\deg(Q) \leq j + d$, where $j$ is multiplicity of root $\alpha$ (if $P(\alpha) \neq 0$, then $j = 0$) of characteristic polynomial
+    \item \bi{(Case 2)} $b(x) = c x^d \cos(\alpha x)$, or $b(x) = c x^d \sin(\alpha x)$:
+          $f_p = Q_1(x) \cdot \cos(\alpha x) + Q_2(x9 \cdot \sin(\alpha x))$,
+          where $Q_i(x)$ a polynomial with $\deg(Q_i) \leq d + j$, where $j$ is the multiplicity of root $\alpha i$ (if $P(\alpha i) \neq 0$, then $j = 0$) of characteristic polynomial
+\end{enumerate}
+
+\rmvspace\shade{gray}{Other methods}\rmvspace
+\begin{itemize}[noitemsep]
+    \item \bi{Change of variable} Apply substitution method here, substituting for example for $y' = f(ax + by + c)$ $u = ax + by$ to make the integral simpler.
+          Mostly intuition-based (as is the case with integration by substitution)
+    \item \bi{Separation of variables} For equations of form $y' = a(y) \cdot b(x)$ (NOTE: Not linear),
+          we transform into $\frac{y'}{a(y)} = b(x)$ and then integrate by substituting $y'(x) dx = dy$, changing the variable of integration.
+          Solution: $A(y) = B(x) + c$, with $A = \int \frac{1}{a}$ and $B(x) = \int b(x)$.
+          To get final solution, solve for the above equation for $y$.
+\end{itemize}