\part{Fundamentals}


\chapter{Introduction to ODES}

At the beginning of one's mathematical journey, they learn the 4 arithmetic operations and how to do calculations with them. Later on, the 4 operations (and possibly more functions) connect variables in algebraic equations, where one needs to find a number (or set of numbers) that satisfies the equation, possibly restricted to some set. When first solving algebraic equations, students are unfamiliar with complex numbers and hence implicitly restrict themselves to real number solutions.


\[ x^3 -6x^2 +11x -6 = 0, x \in \mathbb{R}\]
\[ x \in \{1,2,3\}\]

Similarly when one learns mathematical analysis, they learn about the derivative and integral of functions and the techniques to calculate them. Similarly, we can consider \emph{differential equations} which relate a variable function with its derivatives.

Our goal is to find such a function that satisfies the equation.
In this part, we will only consider differential equations with solutions that are real, complex, and $\mathbb{R}^n$ functions.

\[ y' = y, y \in C^{1}(\mathbb{C})\]
\[ y \in \{ f(x) = e^{x} +c, c \in \mathbb{C} \}\]

\begin{definition}[Ordinary differential equation (ODE)]
\[ F(y,y',\hdots ,y^{(n)}) = 0)\]
	\[\]
\end{definition}


\begin{definition}[Initial condition]
\end{definition}

\section{Types of differential equations}

\subsection{Order and degree}
\begin{definition}[Order of a differential equation]
Highest order derivative
\end{definition}

\begin{definition}[Degree of a differential equation]
Highest power applied to a derivative
\end{definition}

\subsection{Forms of equations}
\begin{definition}[Degree of a differential equation]
\end{definition}

There exist some more exotic types of differential equatiosn such as fractional differential equations 




\section{First order separable ODE}
\[y' = f(y)g(x)\]
\[\frac{y'}{f(y)} = g(x)\]
\[ \int \frac{y'}{f(y)}dx = \int g(x) dx\]
\[ \int \frac{1}{f(y)}dy = \int g(x) dx\]

\section{First order linear ODE}
\subsection{Integrating factor}
\[y' + q_1 (x)y = q_2 (x)\]
\[I(x) = e^{\int q_1 (x) dx}\]
\[I(x)y' + q_1 (x)I(x)y = I(x) q_2 (x)\]
\[ [ I(x)y]' = I(x) q_2 (x)\]
\[  I(x) y =\int  I(x) q_2 (x) dx\]
\[  y(x) =\frac{ \int  I(x) q_2 (x) dx}{I(x)}\]




\section{Second order linear, homogeneous, constant coefficient ODE}

\[y'' + my' +ny = 0 \]

\begin{proposition}
\end{proposition}

\[ y(x) = Ae^{\lambda_1 x} + Be^{\lambda_2 x} \]
\[ y(x) = (Ax+b)e^{\lambda_1 x} \]
\[ y(x) = e^{\alpha x}[A\cos (\omega x) + B\sin(\omega x)] \]





\section{Exact equations}
\[I(x,y)dx + J(x,y)dy = 0\]

			<h3 class=blue>Exact equation</h3>
			<p>\( P(x,y) dx + Q(x,y)dy = 0 \)</p>
			<p>\( y' = \frac{P(x,y)}{Q(x,y)} \)</p>
			<h4 class=blue>With conservative function</h4>
			<p>\( P(x,y) dx + Q(x,y)dy = 0 \land \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} \implies \exists F(x,y) ( Q = \frac{\partial F}{\partial y} \land P = \frac{\partial F}{ \partial y} ) \land y(x) : F(x,y)=C \text{ satisfies the DE}\)</p>
			<h4 class=blue>With homogeneous functions of same degree</h4>
			<p>\(P,Q \text{ are homogenous of same degree } \implies y=xv(x) \text{ makes ODE separable} \)</p>
			<p>Notice how this substitution abuses homogeneity to cancell out the dependent variable \(x\)</p>
			<ul>
				<li>Make the substitution \(y(x)=xv(x)\]
				<li>Solve the first order separable ODE</li>
			</ul>




\section{Bernoulli equations}
\[y' + q_1 (x) y = q_2 (x) y^n\]

A common technique in differential equations (and even algebraic equations!) is to transform a difficult equation into an equation that can be solved.
The Bernoulli equation can be transformed into a first order linear equation in the following way.
\begin{itemize}
	\item Make the substitution $u = y^{1-n}$
	\item Make the substitution $u' = (1-n) y'y^{-n}$
	\item Solve the first order linear equation
\end{itemize}

Fortunately, we are able to calculate analytic solutions for first order linear differential equations, so we can therefore analytically solve Bernoulli equations! However, sometimes we cannot transform our DE into something so nice.

\section{Riccati equations}
\[q_1 (x)y' + q_2 (x) y + q_3 (x) y^2 = q_4 (x)\]



\begin{itemize}
	\item Make the substitution $y(x) = \frac{A(x)u'(x)}{u(x)}$
	\item Determine the $A(x)$ requireD TO cancel nonlinear terms
	\item Solve the second order linear ODE
\end{itemize}

Second order linear ODEs are generally much harder to solve; we have studied the case of constant coefficients but a general solution is not yet available to us. We will soon study various techniques to solve such problems.


An interesting sidenote is the following formula for generating a new solution for the Ricatti equation using 3 knowl solutions.

\begin{proposition}[Nonlinear superposition of Riccati solutions]
\[ y_4 = \frac{y_1 (y_3 - y_2) + a y_2 (y_1 -y_3)}{y_3 -y_2 + a(y_1 - y_3)} \]
\end{proposition}

Proving such a proposition requires the machinery of Lie theory; this will be covered in the advanced part of this book.

\section{Basic tools for solving ODEs}
Method of undetermined coefficients
Annihilator method
Differential operators
Picard's theorem



			<h3 class=blue>Method of undetermined coefficients</h3>
			<p>Method of forming an ansatz to reduce differential equations into algebraic equations.</p>
			<table>
				<tr><th>\(g(x)\)</th><th>\(y(x)\)</th></tr>
				<tr><td>\(ke^{ax}\)</td><td>\(Ce^{ax}\)</td></tr>
				<tr><td>\(k \sin (ax)\)</td><td>\(K \cos (ax) + M \sin (ax)\)</td></tr>
				<tr><td>\(k x^n\)</td><td>\( \sum^{n}_{i=0} K_i x^i \)</td></tr>
			</table>


<h3 class=blue>Differential operator</h3>
<p>Unary operator (function) defined in terms of functions and differentials</p>
<ul>
	<li><b>Laplacian</b> $\nabla^2 f = \sum_{i=1}^{n} \frac{\partial_{i}^2 f}{\partial x_{i}^{2}}$
	<li><b>Differentiation</b> $Df = \frac{df}{dx}$
	<li><b>n differentiation</b> $D^{n}f = \frac{d^n f}{dx^n}$
</ul>


















\chapter{Linear ODEs}


These constitute a class of ODEs that are particularly well behaved, and there exist a myriad of tools that make use of the linearity of their solutions.
Perhaps the most fundamental property of such ODEs is the following.


\begin{proposition}
The general solution of a linear ODE of degree $n$ is the linear combination of $n$ distinct functions.
\end{proposition}

\begin{corollary}
The set $F = \{f \in C^{n}(I) : \forall x \in I [ \sum^{n}_{k=1} q_k (x) f^{(k)}(x) = 0 ] \}$ forms a linear space over $\mathbb{R}$ (where addition and scalar multiplication are as usual)
\end{corollary}

Knowing that solutions of Linear ODEs form a linear space suggests the philosophy that solving our ODE amounts to finding a basis for this space.


\section{Wronskian}



\begin{definition}[Wronskian]
Given a vector of real or complex functions $\mathbf{y}$, their \emph{Wronskian} is the function \(W\) which is the determinant of the matrix where each column is a vector with a solution to a linear DE and its \(n-1\) derivatives.
\[ W(\mathbf{y})(x) = \begin{vmatrix} y_1 & y_2 & \cdots & y_n \\ y'_1 & y'_2 & \cdots & y'_n \\ \vdots & \vdots & \ddots & \vdots \\ y^{(n-1)}_{1} & y^{(n-1)}_{2} & \cdots & y^{(n-1)}_{n} \end{vmatrix} \]

Where  $\mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix}$ is a vector of solutions to a linear DE
\end{definition}


\subsection{Properties of the Wronskian}
\[\mathbf{y} \text{ are linearly dependent on } I \implies W(I) =\{0\} \]
Where $ \mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix}$ is a vector of solutions to a linear DE


\[\mathbf{y} \text{ are analytic on } I \implies [ \mathbf{y} \text{ are linearly dependent on } I \iff W(I) =\{0\} ] \]
\[\mathbf{y} \text{ satisfy } y''+q_1 (x)y'+q_2(x)=0 \text{ on } I \implies [ \mathbf{y} \text{ are linearly dependent on } I \iff W(I) =\{0\} ] \]

\subsection{Abel's formula}
\begin{theorem}[Abel's formula]
<p>\(y_1,y_2 \text{ satisfy } y'' + q_1 (x) y' + q_2 (x) y = 0  \implies W(y_1 , y_2) = K_{12}e^{-\int q_1 (x) dx}\)</p>
\end{theorem}


\begin{corollary}
\[ y_1 \text{ satisfies } y'' + q_1 (x) y' + q_2 (x) y = 0  \land \neg ( \forall x [y_1(x)=0] )  \implies y_2 = y_1 \int \frac{e^{-\int q_1 (x) dx}}{y_1 (x)^2} dx \text{ satisfies } y'' + q_1 (x) y' + q_2 (x) y = 0 \land y_1 , y_2 \text{ are linearly independent} \]
\end{corollary}

As wonderful as it is that we can abuse the linear space formed by the linear ODE to generate novel solutions using known solutions, this still requires that we know some solutions for the ODE.

To get an initial foothold on the problem, the Annihilator method or Cauchy-Euler method can be used to obtain a first set of solutions before expanding our solutions using Abel's formula. Unfortunately such methods are not always possible to apply due to assumptions not being met, but there exists some more potent methods for evaluating solutions to linear ODEs.




\section{Variation of Parameters (VOP)}
VOP is essentially just assuming the ansatz $y(x) = u(x)y_1(x) + v(x) y_2 (x)$ and $u' y_1 + v' y_2 = 0$



\section{Series method}


One powerful idea is to assume that the solution to the linear ODE is an analytic function and simply apply the ansatz $y(x) = \sum^{\infty}_{n=0} a_n x^n$. It's quite simple, yet incredibly effetive. Often the solutions to an ODE are nonelementary, so this method comes in clutch where others fail.







For differential equations such that $y^{(n)} + \sum q_k (x) y^{(k)} = 0 \land q_k \text{ are analytic on }I \land 0 \in I$, solutions are always analytic and hence has a Taylor series representation and hence a power series representation. This is due to considering the DE on a complex neighborhood and by noting that holomorphic functions are analytic, see Complex Analysis.
\begin{itemize}
	\item Assume the ansatz $y(x) = \sum^{\infty}_{n=0} a_n x^n$
	\item Substitute the ansatz into the DE and manipulate into one single sum
			$ [ \sum^{\infty}_{n=0} a_n x^n ]^{(k)} = \sum^{\infty}_{n=k} a_n \frac{n!}{(n-k)!} x^{n-k}$</ul><li>
			<li>Since the RHS equals zero, equate the coefficient of the new sum to 0<
			<li>Solve recursive formula for $a_n$
			</ul>

\begin{definition}[Ordinary point]
	\[ x_0 \text{ is an ordinary point of } \sum q_n (x) y^{n} (x) = 0 \iff \forall k \in \{0,1,..,n\}a_k \text{ is analytic at }x_0) \]
\end{definition}



			<p>\( \prod^{n}_{k=1} 2k = 2^n n!\)</p>
			<p>\( \prod^{n}_{k=1} (2k-1) = \frac{(2n-1)!}{2^(n-1) (n-1)!}\)</p>

			<h4 class=red>Singular point</h4>
			<p>\(x_0 \text{ is a singular point of } \sum q_n (x) y^{n} (x) = 0 \iff \neg [ x_0 \text{ is a singular point of } \sum q_n (x) y^{n} (x) = 0]\)</p>
			<p>All coefficient function analytic at a point means all DEs solutions taylor series equal function itself</p>



Series method
Method of Frobenius
\section{Bessel functions}

Bessel functions
Modified Bessel functions
Airy functions

THe following proposition essentially serves as a shortcut for the series method; the series method is indeed the reasoning for such a proposition.
\begin{propositition}[Bessel function substitution method]
\[x^2 y'' + (1-2s)xy' + [(s^2 -r^2 \alpha^2 )  + a^2 r^2 x^{2r}]y = 0\]
\[ y(x) = c_1 x^s J_{\alpha}(ax^r) + c_2 x^s Y_{\alpha}(ax^r)\]
\end{proposition}













\chapter{Laplace transform}

\begin{definition}[Heaviside step function]
The characteristic function of real numbers greater than or equal to 0, \(\chi_{\mathbb{R}_{+}}\), it serves useful for the Laplace inversion of function with a factor $e^{-sa}$
\[ H(t) = \begin{cases} 1 & t \geq 0 \\ 0 & t \lt 0 \end{cases} \]
\end{definition}


\begin{definition}[Laplace transform]
Integral transform on \((0,\infty)\) with kernel $e^{-st}$
\[ \mathcal{L} \{ f \} (s) = \int^{\infty}_{0} f(t) e^{-st} dt \]
\begin{itemize}
\item $f : [0,\infty) \to \mathbb{R}$ is a real valued function dominated by an exponential function
\item $\mathcal{L}\{f\} : S \to \mathbb{C} $ is the Laplace transform of $f$
\end{itemize}
\end{definition}

				\[\mathcal{L}\{f\} \in L^\infty([0,\infty])\]
				\[ \mathcal{L} \{ af+bg \} = a \mathcal{L}\{f\} + b \mathcal{L}\{g\} \]
				\[ \mathcal{L} \{ f \} \text{ is analytic on } \text{dom}( \mathcal{L}\{ f \})  \]
				\[ \mathcal{L} \{ t^n f \}(s) = (-1)^n \frac{d^n}{ds^n}\mathcal{L}\{f\}(s) \]
				\[ \mathcal{L} \{ e^{-at} f \}(s) = \mathcal{L}\{f\}(s+a) \]
				\[ \sum^{\infty}_{n=0} \frac{a_n}{n!}t^n \text{ is absolutely convergent } \implies \mathcal{L} \{ \sum^{\infty}_{n=0} \frac{a_n}{n!}t^n \}(s) = \sum^{\infty}_{n=0} \frac{a_n}{s^{n+1}} \]
				\[ \mathcal{L} \{ f(at) \}(s) = \frac{1}{a}\mathcal{L}\{f\}(\frac{s}{a}) \]
				\[\exists \mathcal{L}\{f'\}, \mathcal{L}\{f\} \implies \mathcal{L} \{ f' \}(s) = -f(0) + s \mathcal{L}\{f\}(s) \]
				\[\mathcal{L} \{ f^{(n)} \}(s) = s^{n} \mathcal{L}\{f\}(s) - \sum^{n-1}_{i=0} s^{n-i-1}f^{(i)}(0) \]
				\[ \mathcal{L} \{ f \} = 0 \implies f = 0 \text{ almost everywhere} \]
				\[ \mathcal{L} \{ f \} = \mathcal{L}\{g\} \implies f = g \text{ almost everywhere} \]
				\[  \mathcal{L}\{H(t-a) f(t-a) \} (s) = e^{-sa} \mathcal{L}\{f\}(s)\]


\begin{itemize}
			\item $\mathcal{L}\{t^n\}(s) = \frac{\Gamma (n +1 )}{s^{n+1}} $
			\item $\mathcal{L}\{e^{-at}\}(s) =  \frac{1}{s+a} $
			\item $\mathcal{L}\{\sin (at)\}(s) = \frac{a}{s^2 + a^2} $
			\item $\mathcal{L}\{\cos (at)\}(s) = \frac{s}{s^2 + a^2} $
			\item $\mathcal{L}\{J_0 (t)\}(s) = \frac{1}{\sqrt{1 +s^2}} $
\end{itemize}




\begin{definition}[Inverse Laplace transform]
Since a property of Laplace transforms is that two functions with the same transform must be equal almost everywhere, Laplace transforms are unique and hence invertible.
\[  \mathcal{L}^{-1} \{ \mathcal{L}\{f\} \} = f \]
\end{definition}
			<h4 class=green>Properties</h4>
			<p>\( \mathcal{L}^{-1} \{ af+bg \} = a \mathcal{L}^{-1}\{f\} + b \mathcal{L}^{-1}\{g\} \)</p>
			<h4 class=green>Inversion by series method</h4>
			<p>Convert the Laplace transform into a series, use linearity of inverse Laplace operator to resolve each term directly</p>

\begin{theorem}[Convolution theorem (Laplace transform)]
\[ \mathcal{L}\{f * g\}(s) = \mathcal{L}\{f\}(s) \mathcal{L}\{g\}(s) \]
\end{theorem}


The Laplace transform can be used to transform ODEs into algebraic equations! Here is the outline of the method.
\begin{itemize}
	\item Compute the Laplace transform for both sides of the equation and equate them
	\item Solve algebraically for $y_1$
	\item Compute $\mathcal{L}^{-1}\{y_1\}(x)$
\end{itemize}



\begin{definition}[Confluent hypergeometric functions]
Linearly independent solutions to the ODE \(zw''+(b-z)w'-aw=0\) (Kummer's equation)
First kind
\[ M(a,b,z)= \sum^{\infty}_{n=0} \frac{a^{(n)} z^n }{ b^{(n)} n! } \]
\begin{itemize}
	\item $x^{(n)}=\prod^{n-1}_{k=0}(x+k)$ is the rising factorial
	\item $ (x)_{n}=\prod^{n-1}_{k=0}(x-k)$ is the falling factorial
\end{itemize}
Second kind
\[ U(a,b,z) \]
\end{definition}

\chapter{PDEs}
\section{Separation of variables}
Separation of variables. This works when the partial derivatives are known to reduce to constants rather than more complex functions. Here is the general method.

\begin{itemize}
	\item Assume the ansatz $f(x,y) = X(x)Y(y)$ and substitute into equation
	\item Separate variables to both sides and equate both sides to a constant $\pm \lambda$
	\item Solve each problem using SL theory or otherwise and then substitute the solutions into the original ansatz; the general solution is a linear combination of these eigenfunctions
	\item Equate the general solution to the final boundary condition and find the projection.
\end{itemize}
\end{definition}

Maximum principle
Harnack's theorem

\part{Advanced}

\chapter{Existence and uniqueness of differential equations}
\begin{theorem}[Peano existence theorem]
\end{theorem}

\begin{theorem}[Picard–Lindelöf theorem]
\end{theorem}

\begin{theorem}[Cauchy-Kovalevskaya theorem]
\end{theorem}

\chapter{Sturm-Liouville theory}
Operator are an object from functional analysis; familiar with differential operators is required for this chapter.
- Sturm-liouville operator


\chapter{Fuchsian theory}
- Further study of metho of frobenius


\chapter{Applications of Lie theory for differential equations}
-Lie point symmetry


- Holonomic function
- D-module



















			<h3 class=blue>Basic ODEs</h3>
			<p>See M1</p>
			<ul>
				<li>Differential equations</li>
				<li>Integrating factor</li>
				<li>First order linear equation</li>
				<li>First order separable equation</li>
				<li>Second order linear, homogeneous, constant coefficients</li>
				<li>Second order linear, constant coefficients</li>
				<li>IVP</li>
				<li>BVP</li>
			</ul>




			<h2 class=yellow>Second order DE</h2>


			<h3 class=yellow>Picard's theorem</h3>



			<h3 class=yellow>Second order linear, homogeneous equations</h3>
			<p>\(y'' + q_1 (x) y' + q_2 (x) y = 0\)</p>
			<ul>
				<li>Obtain a solution \(y_1\)
					<ul><li>Series method</li><li>Cauchy-Euler equation</li><li>Method of Frobenius</li><li>Laplace transfrom method</li><li>Bessel function substitution method</li></ul>
				</li>
				<li>Construct a linearly independent solution \(y_2 = y_1 \int \frac{e^{-\int q_1 (x) dx}}{y_1 (x)^2} dx\]
				\[y(x) = c_1 y_1 (x) + c_2 y_2 (x) \]
			</ul>
			<h4 class=yellow>Second order Cauchy-Euler equation</h4>

			<h3 class=yellow>Second order linear equations</h3>
			<p>\(y'' + q_1 (x) y' + q_2 (x) y = R (x) \)</p>
			<ul>
				<li>Compute the linearly independent \(y_1 ,y_2\) that solve \(y'' + q_1 (x) y' + q_2 (x) y = 0 \) and let \(y_c (x)  = c_1 y_1 (x) +c_2 y_2 (x) \]
				<li><b>Variation of parameters</b> implies that we have \(y_p (x) = u(x)y_1 (x) + v(x)y_2 (x)\)
					<ul>
						\[u(x) = - \int \frac{y_2 R (x)}{W(y_1 , y_2)} dx\]
						\[v(x) = \int \frac{y_1 R(x)}{W(y_1 , y_2)} dx\]
					</ul>
				</li>
				\[y(x) = y_c (x) + y_p (x)\]
			</ul>


			<h2 class=red>Series methods</h2>





			<h3 class=red>Airy function</h3>
			<p>Linearly independent solutions to the ODE \(y'' -xy =0\) (Airy equation)</p>
			<h4 class=red>First kind</h4>

			<p>\(\text{Ai}(x) = \frac{3^{-\frac{2}{3}}}{\Gamma (\frac{2}{3})} [ 1 + \sum^{\infty}_{n=1} \frac{\prod^{n}_{k=1}(3k-2)}{(3n)!}x^{3n} ] - \frac{3^{-\frac{1}{3}}}{\Gamma (\frac{1}{3})} [ x + \sum^{\infty}_{n=1} \frac{\prod^{n}_{k=1}(3k-1)}{(3n+1)!}x^{3n+1} ] \)</p>

			<p>\(\text{Ai}(x) = \frac{1}{\pi} \int^{\infty}_{0} \cos ( \frac{t^3}{3} +xt)dt\)</p>

			<h4 class=red>Second kind</h4>
			<p>\(\text{Bi}(x) = \sqrt{3} [ \frac{3^{-\frac{2}{3}}}{\Gamma (\frac{2}{3})} [ 1 + \sum^{\infty}_{n=1} \frac{\prod^{n}_{k=1}(3k-2)}{(3n)!}x^{3n} ] + \frac{3^{-\frac{1}{3}}}{\Gamma (\frac{1}{3})} [ x + \sum^{\infty}_{n=1} \frac{\prod^{n}_{k=1}(3k-1)}{(3n+1)!}x^{3n+1} ] ] \)</p>


			<p>\(\text{Bi}(x) = \frac{1}{\pi} \int^{\infty}_{0} [ \exp ( -\frac{t^3}{3} +xt) + \sin ( \frac{t^3}{3} +xt ) ] dt\)</p>
			<h3 class=red>Regular singular point</h3>

			<h3 class=red>Method of Frobenius</h3>
			<p>Series method variant when singularity at 0</p>
			<ul>
				<li>Assume the ansatz \(y(x) = \sum^{\infty}_{n=0} a_n x^{n+s}\) where \(s \in \mathbb{C}\]
				<li>Substitute the ansatz into the DE and manipulate into one single sum<ul>
				\[ [ \sum^{\infty}_{n=0} a_n x^{n+s} ]^{(k)} = \sum^{\infty}_{n=0} a_n [ \prod^{k-1}_{i=0} (n+s-i) ] x^{n+s-k}\]</ul><li>
				<li>Since the RHS equals zero, equate the coefficient of the new sum to 0</li>
				<li>Noting that letting \(a_0 = 0\) leads to the trivial solution, hence the precise value(s) of \(s\) must be determined to make the coefficient of \(x^s\) equal to 0</li>
				<li>Solve recursive formula for \(a_n\]
			</ul>
			<h4 class=red>Single solution for \(s\)</h4>
			<p>\(y'' + q_1 (x) y' + q_2 (x) y = 0\)</p>
			<p>\( s_1 \text{ is the only value of} s \text{ and produces solution } y_1(x) \implies \)</p>
			<p>\(y_2(x) = y_1 (x) \ln (x) + x^{s_1} \sum^{\infty}_{n=1}a'_n (s_1)x^n\)</p>
			<h4 class=red>\(s_1 -s_2 \in \mathbb{Z}\)</h4>
			<p>\(y'' + q_1 (x) y' + q_2 (x) y = 0\)</p>
			<p>\( s_1 -s_2 \in \mathbb{Z} \land s_1 \text{ produces solution } y_1(x) \implies \)</p>
			<p>\(y_2(x) = \frac{b_N}{a_0}y_1 (x) \ln (x) + x^{s_2} \sum^{\infty}_{n=0}b'_n (s_2)x^n\)</p>
			<p>\( b_n = (s-s_2) a_n( s) \)</p>



			<h2 class=orange>Bessel functions</h2>




<!--<h3 class=orange>Bessel's equation</h3>
<p>Differential equation that is encountered in solving the Helmholtz equation in polar coordinates</p>
<p>Solution to \(  x^2 \frac{d^2 y}{dx^2} + x \frac{d y}{dx} - (x^2 + \alpha^2) y = 0 \)</p>
<p>Solutions to Bessel's equations found by the method of Frobenius</p>
<p>\(  x^2 \frac{d^2 y}{dx^2} + x \frac{d y}{dx} + (x^2 - \alpha^2) y = 0 \)</p>-->

<h3 class=orange>Bessel function</h3>
<p>Linearly independent solutions to the ODE \(  x^2 y'' + x y' + (x^2 - \alpha^2) y = 0 \) (Bessel equation)</p>
<h4 class=orange>First kind</h4>
<p>\( J_{\alpha} (x) = \sum^{\infty}_{k=0} \frac{(-1)^k  }{k! \Gamma (k + \alpha + 1)}(\frac{x}{2} )^{2k+\alpha} \)</p>
<p>\( J_{n} (x) = \frac{1}{\pi} \int^{\pi}_{0} \cos(nt - x \sin t)dt\)</p>
<p>\(\alpha \notin \mathbb{Z} \implies J_{\alpha} , J_{-\alpha} \text{ are linearly independent solutions to Bessel equation}\)</p>
<p>\( n \in \mathbb{Z} \implies J_{-n}(x) = (-1)^n J_{n} (x) \)</p>
<h4 class=orange>Second kind</h4>
<p>\( Y_{n} (x) = \frac{2}{\pi} [ J_{n}(x)(\gamma + \ln( \frac{x}{2}) ) - \frac{1}{2}\sum^{n-1}_{k=0}\frac{(n-k-1)!(\frac{x}{2})^{2k-n}}{k!} - \frac{1}{2} \sum^{\infty}_{k=0} \frac{(-1)^k [H_k +H_{k+n}](\frac{x}{2})^{2k+n}}{k! (k+n)!} ]\)</p>
<p>\( Y_{\alpha} (x) = \frac{J_{\alpha}(x) \cos ( \alpha x) - J_{-\alpha}(x)}{\sin ( \alpha x)} \)</p>
<p>\(\alpha \notin \mathbb{Z} \implies J_{\alpha} , Y_{\alpha} \text{ are linearly independent solutions to Bessel equation}\)</p>

<h3 class=orange>Modified Bessel function</h3>
<p>Linearly independent solutions to the ODE \(  x^2 y'' + x y' + (x^2 + \alpha^2) y = 0 \) (Modified Bessel equation)</p>
<h4 class=orange>First kind</h4>
<p>\(I_{\alpha}(x) = \sum^{\infty}_{k=0} \frac{1}{k! \Gamma (k+\alpha+1)} (\frac{x}{2})^{2k+\alpha}\)</p>
<p>\(  x^2 \frac{d^2 y}{dx^2} + x \frac{d y}{dx} - (x^2 + \alpha^2) y = 0 \) has general solution \(y(x) = c_1 I_{\alpha}(x) + c_2 I_{-\alpha}(x)\)</p>
<p>\(J_{\alpha} (ix) = i^{\alpha}I_{\alpha}(x)\)</p>
<p>\(\alpha \notin \mathbb{Z} \implies I_{\alpha} , I_{-\alpha} \text{ are linearly independent solutions to modified Bessel equation}\)</p>
<h4 class=orange>Second kind</h4>
<p>\(K_{n}(x)\)</p>
<p>\(K_{\alpha}(x) = \frac{\pi}{2} \frac{I_{-\alpha}(x)- I_{\alpha}(x)}{\sin \alpha \pi}\)</p>
<p>\( n \in \mathbb{Z} \implies I_{n} , K_{n} \text{ are linearly independent solutions to modified Bessel equation}\)</p>

<h3 class=orange>Digamma function</h3>
<p>\(\psi (z) = \frac{d}{dz} \ln \Gamma (z) = \frac{\Gamma' (z)}{\Gamma (z)} \)</p>



