%我的 tex 模版 \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 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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 \begin{tikzpicture} [ grow=right, r/.style={rectangle,draw,fill=red!20,align=center,rounded corners=.8ex}, g/.style={rectangle,draw,fill=green!20,align=center,rounded corners=.8ex}, b/.style={rectangle,draw,fill=blue!20,align=center,rounded corners=.8ex}, grow via three points={one child at (4,-4) and two children at (4,-4) and (4,-8)}, edge from parent path={(\tikzparentnode.south)|-(\tikzchildnode.west)}, ]%属性定义 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \node(-1)at(0,12)[r]{解的存在唯一性\\$/x-x_0/\leq h$} child{node(g)[g]{解的延拓}} child{node(h)[g]{解对初值的连续可微性}} child[missing]{} child[missing]{} child[missing]{} child[missing]{}; \node(-2)at(10,8)[b]{解的存在空间(局部)\\$|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$}}; \draw[->](-2)--(-3); \draw[->](-3)--(-6); \draw[->](-4)--(-7); \draw[->](-3)to node[above]{初值$(x_0,y_0)$}(-4); \draw[->](-4)--(-5); \draw[->](-2)--(10,9)to node[above]{延拓}(25,9)--(-5); \draw[->](-6)--(-8); \node(0)at(0,0)[r]{一阶常微分方程\\$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); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{tikzpicture} \end{center} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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} [ grow=right, r/.style={rectangle,draw,fill=red!20,align=center,rounded corners=.8ex}, g/.style={rectangle,draw,fill=green!20,align=center,rounded corners=.8ex}, b/.style={rectangle,draw,fill=blue!20,align=center,rounded corners=.8ex}, grow via three points={one child at (4,-2) and two children at (4,-2) and (4,-4)}, edge from parent path={(\tikzparentnode.south)|-(\tikzchildnode.west)}, ]%属性定义 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \node(0)at(0,0)[r]{高阶常微分方程} child[missing]{} child[missing]{} 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}$}}} child[missing]{} child[missing]{} child{node[b]{重根} child{node[b]{实$k$重根\\$\underbrace{e^{\lambda_1t},te^{\lambda_1t},\dots,t^{k-1}e^{\lambda_1t}}_k$}} child[missing]{} child{node[b]{$e^{\lambda t}=e^{\alpha t}(cos(\beta t)+sin(\beta t))$\\复$k$重根\\(复$=2$,单$=k\Rightarrow 2\times k$的矩阵)\\$\underbrace{e^{\lambda_1t},te^{\lambda_1t},\dots,t^{k-1}e^{\lambda_1t}}_k$\\$\underbrace{e^{\overline{\lambda_1}t},te^{\overline{\lambda_1}t},\dots,t^{k-1}e^{\overline{\lambda_1}t}}_k$}}}}}} child[missing]{} child[missing]{} child[missing]{} child[missing]{} child[missing]{} child[missing]{} child[missing]{} child[missing]{} child[missing]{} child[missing]{} child{node(b)[g]{非齐次微分方程} child{node[b]{常系数非线性齐次微分方程} child{node[b]{待定系数法求特解} child{node[b]{$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})$}}}}}}; \node(1)at(4,0)[g]{降阶法\\$x=x(t)$} child{node[b]{不显含$x$,降$k$阶\\$F(t,x^{(k)},x^{(k+1)},\dots,x^{(n)})=0$}} child{node[b]{不显含$t$,降$1$阶\\$F(x,x‘,x‘‘,\dots,x^{(n)})=0$}}; \draw[->](0)--(1); \draw[->,green](a)to node[right]{常数变易法$c\Rightarrow c(x)$}(b); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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 \begin{tikzpicture} [ grow=right, r/.style={rectangle,draw,fill=red!20,align=center,rounded corners=.8ex}, g/.style={rectangle,draw,fill=green!20,align=center,rounded corners=.8ex}, b/.style={rectangle,draw,fill=blue!20,align=center,rounded corners=.8ex}, grow via three points={one child at (4,-2) and two children at (4,-2) and (4,-4)}, edge from parent path={(\tikzparentnode.south)|-(\tikzchildnode.west)}, ]%属性定义 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 123 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{tikzpicture} \end{center} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{document}
时间: 2024-11-09 17:12:04