The Product: HERE

Please note that the math are very non-rigorous… I don’t even know what I was doing

\documentclass{article}
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\title{2nd order LODE}

\author{
    \textbf{Henry Yip} and
    \textbf{Lorian Richmond}\\
}

\date{March 2022}

\begin{document}

\maketitle
\begin{abstract}
    This appendix is related to solving ordinary diffential equations with a mathematically rigorous style.
\end{abstract}
\section{Introduction}
Consider the homogeneous equation:
\begin{equation}
\ddot{y}+ \alpha\dot{y}+ \beta y = 0
\end{equation}
By basic differentiation techniques, we have 
 \begin{align}
 \frac{d}{dx}(e^x)=e^x\\
 \implies \frac{d}{dx}(ae^x)=ae^x
 \end{align}
 
 Suppose that f: $\mathbb{R} \rightarrow \mathbb{R}$ is differentiable everywhere and $f'(x)=af(x)$.
 There are no other functions in which the derivative is equal to itself. 
 \subsection{Proof 1}

 If we let $g(x)=\frac{af(x)}{e^{ax}}$


$ f(x)$ and $e^{ax}$ are differentiable $\rightarrow g'(x)$ is defined.
\begin{align}
     g'(x)&=\frac{f'(x)e^{ax}-af(x)e^{ax}}{(e^{ax})^2}\\
     g'(x)&=\frac{af(x)e^{ax}-af(x)e^{ax}}{(e^{ax})^2}\\
     g'(x)&=0
\end{align}
 By identity theorem, $g'(x)=0\rightarrow g(x)= c$, where $c\in \mathbb{R}$\\
 $\implies$ $f(x)=Ae^{ax}$, where $c=Aa$. We take $a=1$ and hence $A=c$.
 The above statement can be proved by Picard–Lindelöf theorem, however we won't cover it here.
 \subsection{Proof 2}
 Suppose that f: $\mathbb{R} \rightarrow \mathbb{R}$ is differentiable everywhere and $f(x)=f'(x)$
 \begin{align}
     1&=\frac{f(x)}{f'(x)}\\
     x+c&=\int{\frac{f(x)}{f'(x)}}, c\in \mathbb{R}\\
     x+c&=ln(f(x))\\
     f(x)&=e^{x+c}\\
     f(x)&=\gamma e^x, \gamma =e^c
 \end{align}
 \subsection{Ansatz}
 Therefore, we take the Ansatz: $y=e^{\lambda x}$ and obtain:
 \begin{align}
 \ddot{y}+ \alpha\dot{y}+ \beta y &= 0\\
     ({\lambda}^2 + \alpha\lambda  + \beta)(e^{\lambda x}) &=0
\end{align}
     Obviously $e^{\lambda x}=0$ is undefined, so 
     \begin{equation}
     {\lambda}^2 + \alpha\lambda  + \beta=0
     \end{equation}
     Note that the equation above is called the auxillary (characteristic) equation.
     \begin{equation}
     \lambda_1=\frac{a}{2}+\sqrt{\frac{a^2}{4}-b} 
     \end{equation}
     \begin{equation}
     \lambda_2=\frac{a}{2}-\sqrt{\frac{a^2}{4}-b}
     \end{equation}
     
     We can imagine both $\lambda_1$ and $\lambda_2$ satisfies the equation, so $y$ is a linear combination of $e^{\lambda_1 x}$ and $e^{\lambda_2 x}$, we can rewrite this as:
     \begin{equation}
         y=Ae^{\lambda_1 x}+Be^{\lambda_1 x}
     \end{equation}
     Where A \& B $\in \mathbb{R}$
     
     A Rigorous Statement:
     Let S be the solution space of 
     $$\ddot{y}+ \alpha\dot{y}+ \beta y = 0$$
     If $\lambda_1 \ne \lambda_2$, then $\{e^{\lambda_1 x} , e^{\lambda_2 x}\}$ is a basis for S.
     
     Note that the above is only true for $\lambda_1 \ne \lambda_2$. For $\lambda_1 = \lambda_2$,$\{e^{\lambda_1 x} , te^{\lambda_1 x}\}$ is a basis for S. The proof will be written some other time.
\section{Substitution}
\subsection{Motivation}
If we want to solve a non-homogeneous 2nd order differential equation, say
\begin{eqnarray*}
y"+ky+c=0 & & \textrm{where}\ c\in \mathbb{R}
\end{eqnarray*}
Getting rid of the constant would allow us to solve using the equation (17).
\subsection{Construction}
Let 
\begin{equation}
u=y+\frac{c}{k}
\end{equation}
By basic diffentiation techniques, we can obtain the first and second derivative:
\begin{align}
u'&=y'\\
u"&=y"
\end{align}
So the equation becomes
\begin{align}
u"+k(u-\frac{c}{k})+c&=0\\
u"+ku=0
\end{align}
From equation(14), we can see that the auxillary equation becomes:
\begin{equation}
\lambda^2+k=0
\end{equation}
\begin{equation}
(\lambda+\sqrt{k}i)(\lambda-\sqrt{k}i)=0
\end{equation}

So:
\begin{align}
\lambda_1&=\sqrt{k}i\\
\lambda_2&=-\sqrt{k}i
\end{align}
Therefore, the General Solution:\footnote{This equation reminds me of quantum mechanics, in particular, the Schrödinger equation. I may write a paper on this later.}
\begin{equation}
u=Ae^{i\sqrt{k} x}+Be^{-i\sqrt{k} x}
\end{equation}
However, this is not our final answer. For $y$:
\begin{align}
y&=u-\frac{c}{k}\\
y&=Ae^{i\sqrt{k} x}+Be^{-i\sqrt{k} x}-\frac{c}{k}
\end{align}
 \end{document}

Updated: