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\documentclass[a4paper,12pt,twoside]{book}\usepackage{amsmath,amssymb,amsfonts,geometry} % Typical maths resource packages\usepackage{xcolor} % For creating coloured text and background\usepackage{hyperref} % For creating hyperlinks in cross references\usepackage{graphicx}\newtheorem{theorem}{Theorem}[section]\newtheorem{proposition}[theorem]{Proposition}\newtheorem{corollary}[theorem]{Corollary}\newtheorem{lemma}[theorem]{Lemma}\newtheorem{remark}[theorem]{Remark}\newtheorem{definition}[theorem]{Definition}\newtheorem{teo}{Theorem}[section]\usepackage{fancyhdr}\pagestyle{fancy}\renewcommand{\chaptermark}[1]{\markboth{\MakeUppercase{\chaptername \ \thechapter \ \ #1}}{}}\fancyhf{}%clear all header and footer fields\fancyhead[LO]{\nouppercase{\rightmark}}\fancyhead[RE]{\nouppercase{\leftmark}}\fancyhead[RO,LE]{\thepage}\def\chem#1#2{ {{\scriptscriptstyle#1\atop\longrightarrow}\atop{\longleftarrow\atop \scriptscriptstyle#2}} }\begin{document}...\subsection{Linearization stability parallel flow (Rayleigh equation)}...Now, seek the solution with form $ \overline{\psi}=\phi(y)e^{i\alpha(x-ct)} $ where $ \alpha $ and c are constants (c is complex in general) so continue derivation\begin{align*}\\-\dfrac{\partial}{\partial t}{[-\alpha^{2}\phi(y)+\phi_{yy}(y)]e^{{\textrm{i}}\alpha(x-ct)}}+U(y)\dfrac{\partial}{\partial x}{[-\alpha^{2}\phi(y)+\phi_{yy}(y)e^{{\textrm{i}}\alpha(x-ct)}]}+U_{yy}(y)\dfrac{\partial}{\partial x}[\phi(y)e^{{\textrm{i}}\alpha(x-ct)}]&=0,\\\end{align*}\begin{align*}\\ \alpha c{\textrm{i}}[\phi_{yy}(y)-\alpha^{2}\phi(y)]e^{{\textrm{i}}\alpha(x-ct)}-\alpha {\textrm{i}}U(y)[\phi_{yy}(y)-\alpha^{2}\phi(y)]e^{{\textrm{i}}\alpha(x-ct)}+\alpha {\textrm{i}}U_{yy} (y)\phi(y)e^{{\textrm{i}}\alpha(x-ct)}&=0,\\\end{align*}\begin{align*}