409 lines
13 KiB
TeX
409 lines
13 KiB
TeX
% !TeX root = ../../script.tex
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\documentclass[../../script.tex]{subfiles}
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\begin{document}
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\section{Solution Methods}
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\begin{defi}
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An ordinary differential equation (ODE) is an equation of the form
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\[
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F(x, y, y', \cdots, y^{(n)}) = 0
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\]
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with $F: \realn^{n+2} \rightarrow \realn$. $n$ is the order of the ODE.
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Let $I$ be an open interval. A function $y: I \rightarrow \realn$ is a solution of the ODE if $y \in C^n(\realn)$ and
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\[
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F(x, y(x), y'(x), \cdots, y^{(n)}(x)) = 0 ~~\forall x \in I
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\]
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\end{defi}
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\begin{eg}
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\begin{align*}
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y'' = -\frac{1}{y^2} && \text{Gravitational field} \\
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y'' = -\sin y && \text{Pendulum}
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\end{align*}
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\end{eg}
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\begin{rem}
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\begin{enumerate}[(i)]
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\item Often times $F$ is only defined on subsets of $\realn^{n+2}$
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\item ODEs are not simple to solve
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\item Even if we can't calculate explicit solutions, we can inspect the following properties
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\begin{itemize}
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\item Existence of solutions
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\item Uniqueness of solutions
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\item Dependency of solutions from initial conditions
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\item Sability
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\end{itemize}
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\end{enumerate}
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\end{rem}
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\begin{eg}
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\begin{enumerate}[(i)]
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\item Let $I$ be an open interval and $f: I \rightarrow \realn$ continuous. Then the solution of
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\[
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y' = f(x)
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\]
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is the antiderivative of $f$. Let $x_0 \in I$, then
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\[
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y(x) = \int_{x_0}^x f(t) \dd{t} + c ~~c \in \realn
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\]
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\item Consider the ODE
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\[
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y' = y
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\]
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The functions $x \mapsto c e^x$ are solutions $\forall c \in \realn$. Are those all the solutions that exist?
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Let $y: I \rightarrow \realn$ be any solution, and consider
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\[
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u(x) = y(x)e^{-x}
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\]
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Then
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\begin{align*}
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u'(x) &= y'(x) e^{-x} - y(x)e^{-x} \\
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&= \left(y'(x) - y(x)\right) e^{-x} = 0 ~~\forall x \in I
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\end{align*}
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So $u(x) = c$.
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\end{enumerate}
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\end{eg}
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\begin{defi}[Initial Value Problem]
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Let $y_0, \cdots, y_{n-1} \in \realn$ and also $F: \realn^{n+2} \rightarrow \realn$. The system of equations
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\begin{align*}
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F(x, y, y', \cdots, y^{(n)}) = 0 && \begin{cases}
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y(0) = y_0 \\
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y'(0) = y_1\\
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\cdots \\
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y^{(n-1)}(0) = y_{n-1}
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\end{cases}
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\end{align*}
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is said to be an initial value problem (IVP).
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\end{defi}
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\begin{eg}
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Consider the problem
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\begin{align*}
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y'' = -\rec{y^2} && \begin{cases}
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y(0) = y_0 \\
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y'(0) = y_1
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\end{cases}
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\end{align*}
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This describes the movement of a point mass in the gravitational field of the earth along a straight line
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through the center of the earth with the initial position $y_0$ and the initial velocity $y_1$.
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\end{eg}
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\begin{eg}
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Consider the problem
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\begin{align*}
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y' = -y^2 && y(0) = 1
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\end{align*}
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Assume $y: I \rightarrow \realn$ is a solution and $y(x) > 0 ~~\forall x \in I$. Then
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\[
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1 = -\frac{1}{y(t)^2} ~y'(t) ~~\forall t \in I
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\]
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By integrating we get
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\begin{align*}
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x = -\int_0^x \frac{1}{y(t)^2} y'(t) \dd{t} &\equalexpl{Substitution} -\int_1^{y(x)} \rec{y^2} \dd{y} \\
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&= \left. \rec{y} \right\vert_1^{y(x)} = \rec{y(x)} - 1 ~~\forall x \in I
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\end{align*}
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So a solution is
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\[
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y(x) = \frac{1}{1+x}
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\]
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The biggest domain that makes sense is $(-1, \infty)$. Analogously one can approach equations with "separated variables", so of the form
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\begin{align*}
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y' = f(y)g(x) && y(x_0) = y_0
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\end{align*}
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\end{eg}
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\begin{thm}[Separation of Variables]
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Let $I, J$ be open intervals, and let
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\begin{align*}
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f: I \longrightarrow \realn && g: J \longrightarrow \realn
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\end{align*}
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be continuous with $0 \ne f(I)$. Let $x_0 \in J, ~y_0 \in I$.
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Then there exists an open interval $I_2 \subset J$ and $x_0 \in I_2$ such that the IVP
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\begin{align*}
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y' = f(y)g(x) && y(x_0) = y_0
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\end{align*}
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has exactly one solution on $I_2$. Set
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\[
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F(y) = \int_{y_0}^y \rec{f(t)} \dd{t}
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\]
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Then $y: I_2 \rightarrow I$ is uniquely defined by
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\[
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F(y(x)) = \int_{x_0}^x g(t) \dd{t}
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\]
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\end{thm}
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\begin{proof}
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$f$ does not have any roots, thus w.l.o.g. $f > 0$.
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\begin{equation}
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F'(y) = \rec{f(y)} > 0 \implies F \text{ strictly monotonically increasing}
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\end{equation}
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Therefore there exists an inverse function $H: F(I) \rightarrow I$. According to the theorem about inverse functions, $H$ is $C^1$ and
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\begin{equation}
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H'(z) = \rec{F'(H(z))} ~~\forall z \in F(I)
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\end{equation}
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$F(I)$ is an open interval containing the $0$. Then we have
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\begin{equation}
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y(x) = H(G(x)) ~~x \in I_2
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\end{equation}
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where
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\begin{equation}
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G(x) = \int_{x_0}^x g(t) \dd{t}
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\end{equation}
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Now choose $I_2$ such that $x_0 \in I_2$ and $G(I_2) \subset F(I)$. Then
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\begin{equation}
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\begin{split}
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y'(x) &= H'(G(x)) \cdot G'(x) \\
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&= \rec{F'(H(G(x)))} \cdot G'(x) \\
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&= \rec{F'(y(x))} \cdot G'(x) \\
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&= f(y(x)) g(x)
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\end{split}
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\end{equation}
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So $y$ solves the ODE. However, if $\tilde{y}: I \rightarrow \realn$ some solution of the IVP, then $\forall x \in I_2$
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\begin{equation}
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\begin{split}
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G(x) = \int_{x_0}^x g(x) \dd{x} = \int_{x_0}^x \frac{\tilde{y}(x)}{f(\tilde{y}(x))} \dd{x} = \int_{\tilde{y}(x_0)}^{\tilde{y}(x)} \rec{f(y)} \dd{y} = F(\tilde{y}(x))
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\end{split}
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\end{equation}
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So $\tilde{y}(x) = H(G(x))$
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\end{proof}
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\begin{rem}
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$I_2$ is obviously not unique. We can find the biggest possible domain with
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\[
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\bigcup_{\substack{x \in I_2 \\ I_2 \text{ open} \\ G(I_2) \subset F(I)}} I_2 = I_{2, \max}
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\]
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\end{rem}
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\begin{thm}
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Let $f: \realn \rightarrow \realn$ be a continuous function, $a, b, c \in \realn$ and $I$ an open interval.
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Then $y: I \rightarrow \realn$ is a solution of the ODE
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\[
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y' = f(ax + by + c)
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\]
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if and only if $ u(x) := ax + by + c$ is a solution of
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\[
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u' = a + bf(u)
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\]
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\end{thm}
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\begin{hproof}
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Consider
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\[
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u'(x) = a + by'(x)
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\]
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\end{hproof}
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\begin{eg}[Euler Homogeneous ODE]
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Let $f: \realn \rightarrow \realn$ be a function and $I$ an open interval not containing the $0$.
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Then $y: I \rightarrow \realn$ is a solution of the ODE
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\[
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y' = f(\frac{y}{x})
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\]
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if and only if
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\[
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u(x) = \frac{y(x)}{x}
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\]
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solves the ODE
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\[
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u' = \frac{f(u) - u}{x}
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\]
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\end{eg}
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\begin{eg}
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Let $f: \realn \rightarrow \realn$ be continuous and $a_1, a_2, b_1, b_2, c_1, c_2 \in \realn$ such that
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\[
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\begin{vmatrix}
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a_1 && b_1 \\
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a_2 && b_2
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\end{vmatrix} \ne 0
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\]
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Now let $\tilde{x}, \tilde{y}$ be the solutions of the equation system
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\begin{align*}
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a_1\tilde{x} + b_1\tilde{y} + c_1 &= 0 \\
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a_2\tilde{x} + b_2\tilde{y} + c_2 &= 0
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\end{align*}
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Let $I$ be an open interval not containing the $0$. Then $y: I \rightarrow \realn$ is a solution to
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\[
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y' = f\left(\frac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}\right)
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\]
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if and only if
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\begin{align*}
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u: I - \tilde{x} &\longrightarrow \realn \\
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x &\longmapsto y(x + \tilde{x}) - \tilde{y}
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\end{align*}
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is a solution to
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\[
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u' = f\left(\frac{a_1 + b_y \frac{u}{x}}{a_2 + b_2 \frac{u}{x}}\right)
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\]
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\end{eg}
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\begin{proof}
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Let $y: I \rightarrow \realn$ be a solution to the initial equation. Then
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\begin{equation}
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\begin{split}
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u'(x) &= y'(x + \tilde{x}) = f\left( \frac{a_1(x + \tilde{x}) + b_1 y(x + \tilde{x}) + c_1}{a_2(x + \tilde{x}) + b_2y(x + \tilde{x}) + c_2} \right) \\
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&= f\left( \frac{a_1x + b_1u(x) + a_1\tilde{x} + b_1\tilde{y} + c_1}{a_2x + b_2u(x) + a_2\tilde{x} + b_2\tilde{y} + c_2} \right) \\
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&= f\left(\frac{a_1 + b_1 \frac{u(x)}{x}}{a_2 + b_2 \frac{u(x)}{x}}\right)
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\end{split}
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\end{equation}
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The other direction is left to the reader.
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\end{proof}
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\begin{defi}[Exact ODE]
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Let $D \subset \realn^2$ be open, and $p, q: D \rightarrow \realn$ continuous.
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The ODE
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\[
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p(x, y) + q(x, y) y' = 0
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\]
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is said to be exact if there exists a $C^1$-function $H: D \rightarrow \realn$, such that
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\begin{align*}
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\partial_1 H = p && \partial_2 H = q
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\end{align*}
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Such a function is called a potential function.
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\end{defi}
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\begin{thm}
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Let $D \subset \realn^2$ be open, and $p, q: D \rightarrow \realn$ continuous. Let
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\[
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p(x, y) + q(x, y)y' = 0
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\]
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be exact and $H$ a potential function. Furthermore let $I$ be an open interval and $y: I \rightarrow \realn$ a $C^1$-function such that
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\[
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\set[x \in I]{(x, y(x))} \subset D
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\]
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Then $y$ solves the ODE if and only if $\exists c \in \realn$ such that
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\[
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H(x, y(x)) = c
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\]
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\end{thm}
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\begin{proof}
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\begin{equation}
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\begin{split}
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\dv{x} H(x, y(x)) &= \partial_1 H(x, y(x)) + \partial_2 H(x, y(x)) y'(x) \\
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&= p(x, y) + q(x, y)y'(x)
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\end{split}
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\end{equation}
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\end{proof}
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\begin{thm}
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Let $D \subset \realn^2$ be open, and $p, q: D \rightarrow \realn$ continuously differentiable. If
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\[
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p(x, y) + q(x, y)y' = 0
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\]
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is exact, then
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\[
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\partial_2 p = \partial_1 q
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\]
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\end{thm}
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\begin{proof}
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Let $H$ be a potential $C^2$-function. Then
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\begin{equation}
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\partial_2 p = \partial_2 \partial_1 H = \partial_1 \partial_2 H = \partial_1 q
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\end{equation}
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\end{proof}
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\begin{rem}
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The above condition is merely necessary! However, for "nice" $D$ it can be considered sufficient.
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\end{rem}
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\begin{eg}\label{eg:817}
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Consider
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\begin{align*}
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\underbrace{(2x + y^2)}_p + \underbrace{2xyy'}_q = 0 && y(1) = 1
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\end{align*}
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Then
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\begin{align*}
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\partial_2 p = 2y && \partial_1 q = 2y
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\end{align*}
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So $\partial_2 p = \partial_1 q$. If $H$ is a potential function, then
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\begin{align*}
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\partial_1 H(x, y) &= p(x, y) = 2x + y^2 \\
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\implies H(x, y) &= \int p(x, y) \dd{x} = x^2 + y^2x + G(y)
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\end{align*}
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and
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\begin{align*}
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\partial_2 H(x, y) &= q(x, y) = 2xy = 2xy + G'(y) \\
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\implies G(y) &= c
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\end{align*}
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So the potential function is
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\[
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H(x, y) = x^2 + y^2 x
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\]
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We can insert the initial condition
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\[
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H(1, 1) = 2
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\]
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So the solution has to fulfil
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\[
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x^2 + y(x)^2 x = 2 ~~\forall x \in I
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\]
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and thus
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\[
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y(x) = \pm \sqrt{\frac{2}{x} - x}
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\]
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Only the positive sign fulfils the initial conditions, so the solution is
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\[
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y(x) = \sqrt{\frac{2}{x} - x}
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\]
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This function is defined on $(-\infty, -\sqrt{2}] \cup (0, \sqrt{2}]$, however due to the initial conditions $(0, \sqrt{2}]$ is the only useful domain.
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\end{eg}
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\begin{rem}
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If
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\[
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p(x, y) + q(x, y)y' = 0
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\]
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is not exact one can try and find an "integrating factor", i.e. $h: D \rightarrow \realn$ such that
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\[
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h(x, y)p(x, y) + h(x, y)q(x, y)y' = 0
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\]
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is exact. A necessary condition is
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\[
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\left(\partial_2 h(x, y)\right)p(x, y) + h(x, y) \partial_2 p(x, y) = \left(\partial_1 h(x, y)\right) q(x, y) + h(x, y) \partial_1 q(x, y)
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\]
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This is a partial differential equation and won't be discussed further in this chapter.
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\end{rem}
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\begin{defi}[Ordinary Differential Equation System]
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An ordinary differential equation system (ODES) is an equation of the form
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\[
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F(x, y, y', \cdots, y^{(n)}) = 0
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\]
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with
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\[
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F: \realn \times \realn^L \times \realn^L \times \cdots \times \realn^L \longrightarrow \realn^m
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\]
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\end{defi}
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\begin{eg}
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\begin{enumerate}[(i)]
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\item Let $z = (z_1, z_2, z_3)$, then
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\[
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z'' = -\frac{z}{\norm{z}^3} = -\rec{\norm{z}^2} \frac{z}{\norm{z}}
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\]
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is the Kepler problem.
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\item The equation
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\begin{align*}
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b' &= \alpha_1 b - \gamma_1 br \\
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r' &= -\alpha_2 r + \gamma_2 br
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\end{align*}
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is called the "Lotka-Volterra-Equation" and it models the population of prey and predators.
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\end{enumerate}
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\end{eg}
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\begin{rem}
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The ODES
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\[
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F(x, y, y', y'', \cdots, y^{((n)}) = 0
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\]
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is equivalent to the ODES of first order
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\begin{align*}
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F(x, y, y_1, y_2, \cdots, y_{n-1}) = 0 && \begin{cases}
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y_1 = y' \\
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y_2 = y_1' \\
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\vdots
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\end{cases}
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\end{align*}
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\end{rem}
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\end{document} |