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\documentclass[UTF8,a1paper,landscape]{ctexart}%UTF8 中文支持,a1paper 纸张大小,landscape 横向版面,ctexart 中文文章
\usepackage{tikz}%图包
\usetikzlibrary{trees}%树包
\usepackage{amsmath}
\usepackage{geometry}%页边距设置
\geometry{top=5cm,bottom=5cm,left=5cm,right=5cm}
\usepackage{fancyhdr}%页头页尾页码设置
\pagestyle{fancy}
\begin{document}
\title{常微分方程图解}
\author{dengchaohai}
\maketitle
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\newpage%另起一页
\part{一阶常微分方程}
\section*{一阶常微分方程逻辑关系图解}
\begin{center}%居中
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\node(-1)at(0,12)[r]{解的存在唯一性\\$/x-x_0/\leq h$}
child{node(g)[g]{解的延拓}}
child{node(h)[g]{解对初值的连续可微性}}
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child[missing]{}
child[missing]{}
child[missing]{};
\node(-2)at(10,8)[b]{解的存在空间（局部）\\$|x-x_0|\leq h$};
\node(-3)at(15,8)[b]{通解\\$y=y(c,x)$};
\node(-4)at(20,8)[b]{定解\\$y=y(x_0,y_0,x)$};
\node(-5)at(25,8)[b]{解的存在空间（饱和）\\$(c,d)$};
\node(-6)at(15,0)[b]{包络};
\node(-7)at(20,4)[b]{解对初值的连续可微性\\$y=y(x_0,y_0,x,\lambda)$};
\node(-8)at(40,0)[b]{奇解}
child{node[b]{$c-$判别曲线\\$\Phi(c,x,y)=0,\Phi'_c=0$}}
child{node[b]{$p-$判别曲线\\$F(x,y,p)=0,F'_p=0$}};
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\node(0)at(0,0)[r]{一阶常微分方程\\$F(x,y,y')=0$}
child{node[g]{显式\\$y'=f(x,y)$}
child{node(a)[b]{3分式微分方程\\$\frac{dy}{dx}=\frac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2}$}}
child{node(b)[b]{7伯努利微分方程\\$\frac{dy}{dx}=P(x)y+Q(x)y^n$}}}
child[missing]{}
child[missing]{}
child{node[g]{隐式\\$F(x,y,y')=0$}
child{node(c)[b]{8显解$x$\\$x=f(y,y')$}}
child{node(d)[b]{9显解$y$\\$y=f(x,y')$}}
child{node(e)[b]{10不含$x$\\$F(y,y')=0$}}
child{node(f)[b]{11不含$y$\\$F(x,y')=0$}}};
\node(1)at(15,-8)[b]{2齐次微分方程\\$\frac{dy}{dx}=f(\frac{y}{x})$};
\node(2)at(20,-8)[b]{1变量分离方程\\$\frac{dy}{dx}=\psi(x)\varphi(y)$};
\node(3)at(25,-8)[b]{4恰当微分方程\\$M(x,y)dx+N(x,y)dy=0$};
\node(4)at(15,-12)[b]{6非齐次线性微分方程\\$\frac{dy}{dx}=P(x)y+Q(x)$};
\node(5)at(20,-12)[b]{5齐次线性微分方程\\$\frac{dy}{dx}=P(x)y$};
\draw[->](a)--(1);
\draw[->](1)--(2);
\draw[->](2)--(3);
\draw[->](b)--(4);
\draw[->](4)--(5);
\draw[->](5)--(2);
\draw(-1)--(0);
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\end{tikzpicture}
\end{center}
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\newpage
\section*{一阶常微分方程图解对应解法}
\begin{itemize}
\item 1.
\[
\begin{split}
given:\quad\frac{dy}{dx}=\psi(x)\varphi(y).\\
if\quad\varphi(y)=0\\
&\Rightarrow \varphi(y)=0\\
&\Rightarrow y=y_0\\
if\quad\varphi(y)\ne0\\
&\Rightarrow \frac{1}{\varphi(y)}dy=\psi(x)dx\\
&\Rightarrow \int\frac{1}{\varphi(y)}dy=\int\psi(x)dx+c\\
&\Rightarrow \Phi(c,x,y)=0\\
&\Rightarrow y=y(c,x)\\
\end{split}
\]
\item 2.
\[
\begin{split}
given:\quad\frac{dy}{dx}=f(\frac{y}{x}).\\
let\quad u=\frac{y}{x}
&\Rightarrow y=ux\\
&\Rightarrow \frac{dy}{dx}=\frac{du}{dx}x+u\\
&\Rightarrow \frac{du}{dx}x+u=f(u)\\
&\Rightarrow \frac{du}{dx}=\frac{f(u)-u}{x}\\
&\Rightarrow \frac{du}{dx}=\psi(u)\varphi(x)\\
\end{split}
\]
\item 3.
\[
\begin{split}
given:\quad\frac{dy}{dx}=\frac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2}.\\
if\quad c_1=c_2=0\\
&\Rightarrow \frac{dy}{dx}=\frac{a_1+b_1\frac{y}{x}}{a_2+b_2\frac{y}{x}}\\
&\Rightarrow \frac{dy}{dx}=f(\frac{y}{x})\\
let\quad u=\frac{y}{x}\\
&\Rightarrow \frac{dy}{dx}=\psi(u)\\
if\quad \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k\\
&\Rightarrow \frac{dy}{dx}=k\\
if\quad \frac{a_1}{a_2}=\frac{b_1}{b_2}=k\ne\frac{c_1}{c_2}\\
let\quad u=a_1x+b_1y\\
&\Rightarrow \frac{du}{dx}=a_1+b_1\frac{dy}{dx}=a_1+b_1\frac{ku+c_1}{u+c_2}\\
&\Rightarrow \frac{du}{dx}=\psi(u)\\
if\quad \frac{a_1}{a_2}\ne\frac{b_1}{b_2}\\
&\Rightarrow a_1x+b_1y+c_1=0,a_2x+b_2y+c_2=0\\
&\Rightarrow x=x_0,y=y_0\\
let\quad x=X+x_0,y=Y+y_0\\
&\Rightarrow \frac{dy}{dx}=\frac{dY}{dX}=\frac{a_1X+b_1Y}{a_2X+b_2X}=\frac{a_1+b_1\frac{Y}{X}}{a_2+b_2\frac{Y}{X}}\\
let\quad u=\frac{Y}{X}\\
&\Rightarrow \frac{du}{dX}=\psi(u)\varphi(X)\\
\end{split}
\]
\item 4.
\[
\begin{split}
given:\quad M(x,y)dx+N(x,y)dy=0.\\
if\quad \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{-M}=\varphi(y)\\
&\Rightarrow \mu=e^{\int\varphi(y)dy}\\
if\quad \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=\psi(x)\\
&\Rightarrow \mu=e^{\int\varphi(x)dx}\\
let\quad \mu Mdx+\mu Ndy=0\\
&\Rightarrow u=\int\mu Mdx+\varphi(y)\\
&\Rightarrow \frac{du}{dy}=\mu N\\
&\Rightarrow \varphi(y)\\
&\Rightarrow u=\Phi(x,y)\\
&\Rightarrow \Phi(x,y)=c\\
\end{split}
\]
\item 5.
\[
\begin{split}
given:\quad\frac{dy}{dx}=P(x)y.\\
&\Rightarrow y=ce^{\int P(x)dx}\\
\end{split}
\]
\item 6.
\[
\begin{split}
given:\quad\frac{dy}{dx}=P(x)y+Q(x).\\
&\Rightarrow y=e^{\int P(x)dx}(\int Q(x)e^{-\int P(x)dx}dx+c)\\
\end{split}
\]
\item 7.
\[
\begin{split}
given:\quad\frac{dy}{dx}=P(x)y+Q(x)y^n.\\
y^{-n}\\
&\Rightarrow y^{-n}\frac{dy}{dx}=y^{-n}P(x)y+y^{-n}Q(x)y^n\\
&\Rightarrow y^{-n}\frac{dy}{dx}=P(x)y^{1-n}+Q(x)\\
let\quad z=y^{1-n}\\
&\Rightarrow \frac{dz}{dx}=(1-n)y^{-n}\frac{dy}{dx}\\
&\Rightarrow \frac{dz}{dx}=(1-n)P(x)z+(1-n)Q(x)\\
&\Rightarrow \frac{dz}{dx}=\psi (x)z+\varphi(x)\\
\end{split}
\]
\item 8.
\item 9.
\[
\begin{split}
given:\quad y=f(x,y').\\
let\quad p=y'\\
&\Rightarrow y=f(x,p)\\
&\Rightarrow \frac{dy}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial p}\frac{dp}{dx}\\
&\Rightarrow p=\varphi(c,x)\\
&\Rightarrow x=\psi(c,p)\\
&\Rightarrow y=f(\psi(c,p),p)\\
\end{split}
\]
\item 10.
\item 11.
\[
\begin{split}
given:\quad F(x,y')=0.\\
let\quad p=y'=p(t,x)\\
&\Rightarrow x=\psi(t)\\
&\Rightarrow p=\varphi(t)\\
&\Rightarrow dy=pdx=\varphi(t)\psi'(t)dt\\
&\Rightarrow y=\int\varphi(t)\psi'(t)dt+c\\
\end{split}
\]
\end{itemize}
\newpage
\part{高阶常微分方程}
\section*{高阶常微分方程逻辑关系图解}
\begin{center}%居中
\begin{tikzpicture}
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child{node(a)[g]{齐次微分方程}
child{node[b]{常系数线性齐次微分方程}
child{node[b]{特征根法求通解}
child{node[b]{单根}
child{node[b]{实单根\\$e^{\lambda_1t}$}}
child{node[b]{$e^{\lambda t}=e^{\alpha t}(cos(\beta t)+sin(\beta t))$\\复单根\\(复$=2$,单$=1\Rightarrow 2\times1$的矩阵)\\$e^{\lambda_1t}$\\$e^{\overline{\lambda_1}t}$}}}
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child{node(b)[g]{非齐次微分方程}
child{node[b]{常系数非线性齐次微分方程}
child{node[b]{待定系数法求特解}
child{node[b]{$f(t)=(b_0t^m+b_1t^{m-1}+\dots+b_m)e^{\lambda t}$}
child{node[b]{$k$是特征值$\lambda$的重数,若$\lambda$不是特征值,那么$k=0$\\\~{x}$=t^k((b_0t^m+b_1t^{m-1}+\dots+b_m)e^{\lambda t})$}}}
child[missing]{}
child[missing]{}
child{node[b]{$f(t)=((P(t)cos(\beta t)+Q(t)sin(\beta t))e^{\alpha t}$}
child{node[b]{$e^{\lambda t}=e^{\alpha t}(cos(\beta t)+sin(\beta t))$\\$k$是特征值$\lambda$的重数,若$\lambda$不是特征值,那么$k=0$\\\~{x}$=t^k(((P(t)cos(\beta t)+Q(t)sin(\beta t))e^{\alpha t})$}}}}}};
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child{node[b]{不显含$t$,降$1$阶\\$F(x,x',x'',\dots,x^{(n)})=0$}};
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{tikzpicture}
\end{center}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newpage
\section*{高阶常微分方程图解对应解法}
\begin{itemize}
\item 1.降$1$阶
\[
\begin{split}
given\quad F(x,x',x'',\dots,x^{(n)})=0.\\
let\quad y=x'\\
&\Rightarrow x''=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=y'\frac{dy}{dx}\\
&\Rightarrow F(x,y,y'\frac{dy}{dx},\dots,\cdots\frac{dy}{dx})=0\\
&\Rightarrow y=y(t,c_1,c_2,\dots,c_{n-1})\\
&\Rightarrow \int ydx=\int x'dx\\
&\Rightarrow x\\
\end{split}
\]
\item 2.降$k$阶
\[
\begin{split}
given\quad F(t,x^{(k)},x^{(k+1)},\dots,x^{(n)})=0.\\
let\quad y=x^{(k)}\\
&\Rightarrow F(t,y,y',\dots,y^{(n-k)})=0\\
&\Rightarrow y=y(t,c_1,c_2,\dots,c_{n-k})\\
&\Rightarrow \int ydx=\int x^{(k)}dx\\
&\Rightarrow x=\underbrace{\idotsint}_k x^{(k)}dx\\
\end{split}
\]
\end{itemize}
\newpage
\part{常微分方程组}
\section*{常微分方程组逻辑关系图解}
\begin{center}%居中
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123
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\end{tikzpicture}
\end{center}
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\end{document}
