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Differential Equation Basics.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large Differential Equation Basics} \begin{align*} \text{\bf General:}\quad& \text{Equations that contain an unknown function and one or more of its derivatives. The highest derivative}\\ &\text{in the equation is called its \it order. \rm The function that satisfies the differential equation is called a \it solution.}\\ \\ \text{\bf Direction }&\text{\bf Field:}\quad \text{If the differential equation can be expressed as $y'=F(x,y)$, it is called a \it direction field. \rm It means,}\\ &\text{for every point on the $x$-$y$ plane, a direction (slope -- vector without magnitude) can be defined.}\\ \\ \text{\bf Separable }&\text{\bf Equations:}\quad \text{A first-order differential equation which can be expressed as $\frac{dy}{dx}=g(x)f(y)$.}\\ &\text{If $f(y)\neq 0$, this can be expressed as $\int\frac{1}{f(y)}~dy=\int g(y)~dx$}\quad\text{and therefore solved.}\\ &\text{e.g.}\quad\frac{dy}{dx}=x^2 y~,\quad\int\frac{dy}{y}=\int x^2 dx~~(y\neq 0),\quad\ln|y|=\frac{x^3}{3}+C~,\\ &\qquad |y|=e^{\frac{x^3}{3}+C}=e^C e^{\frac{x^3}{3}}~,\quad y=A e^{\frac{x^3}{3}}~~(A=\pm e^C,~A\neq 0,~y\neq 0).\\ \\ \text{\bf First order }&\text{\bf linear ODEs (Ordinary Differential Equations):}\quad\frac{dy}{dx}+f(x)y=g(x)~.\\ &\text{Let $h(x)=e^{\int f(x)dx}$, called the \it integrating factor, \rm then}\quad\frac{d}{dx}h(x)=h(x)f(x)~.\\ &\text{Examine:}\quad\frac{d}{dx}\Big(h(x)y\Big)=h(x)\frac{dy}{dx}+y\frac{d}{dx}h(x)=h(x)\frac{dy}{dx}+h(x)f(x)y=h(x)g(x)~.\\ &\text{Solve:}\quad\frac{d}{dx}\Big(h(x)y\Big)=h(x)g(x)~.\\ &\text{e.g.}\quad\frac{dy}{dx}+3y=e^{-x}.\quad\text{Let }h(x)=e^{3x}.\quad\frac{d}{dx}(e^{3x}y)=e^{3x}e^{-x}=e^{2x}.\\ &\qquad e^{3x}y=\frac{1}{2}e^{2x}+C.\quad y=\frac{1}{2}e^{-x}+Ce^{-3x}.\\ \\ \text{\bf Exact }&\text{\bf ODEs:}\quad H(x,y)=C~.\\ &\text{Consider:}\quad dH(x,y)=\frac{\partial H}{\partial x}dx+\frac{\partial H}{\partial y}dy=0.\quad \frac{\partial H}{\partial x}+\frac{\partial H}{\partial y}\frac{dy}{dx}=0.\\ &\text{This is in the form of}\quad F(x,y)+G(x,y)\frac{dy}{dx}=0,\quad\text{where $F=\frac{\partial H}{\partial x}$ and $G=\frac{\partial H}{\partial y}$.}\\ &\text{Given that}\quad\frac{\partial^2H}{\partial y\partial x}=\frac{\partial^2H}{\partial x\partial y}\quad\text{needs to hold true,}\quad\frac{\partial F}{\partial y}=\frac{\partial G}{\partial x}\quad\text{is required.}\\ &\boxed{\text{An ODE of the form}\quad F(x,y)+G(x,y)\frac{dy}{dx}=0\quad\text{is \it exact \rm if}\quad\frac{\partial F}{\partial y}=\frac{\partial G}{\partial x}.}\\ &\text{An exact ODE can be solved by solving $F=\frac{\partial H}{\partial x}$ and $G=\frac{\partial H}{\partial y}$ for $H$, and the solution is }H(x,y)=C.\\ &\text{e.g.}\quad 2xy+(x^2+y^2)\frac{dy}{dx}=0.\qquad\text{Solution:}\quad\text{Let}\quad F(x,y)=2xy\quad\text{and}\quad G(x,y)=x^2+y^2.\\ &\qquad\text{The ODE is exact as}\quad\frac{\partial F}{\partial y}=2x\quad\text{and}\quad\frac{\partial G}{\partial x}=2x~.\\ &\qquad\text{Solve }\frac{\partial H}{\partial x}=2xy,\quad H=x^2y+C_1(y)\ldots(1)\qquad \text{Solve }\frac{\partial H}{\partial y}=x^2+y^2,\quad H=x^2y+\frac{1}{3}y^3+C_2(x)\ldots(2)\\ &\qquad\text{To find a common $H$:}\quad C_1(y)=\frac{1}{3}y^3,\quad\text{and}\quad C_2(x)=0.\qquad\therefore H(x,y)=x^2y+\frac{1}{3}y^3.\\ &\qquad\text{The solution is}\quad 3x^2y+y^3=A.\\ \end{align*} % % % \begin{align*} \\ \text{\bf Second }&\text{\bf order linear ODEs:}\quad\frac{d^2y}{dx^2}+a\frac{dy}{dx}+by=f.\\ \\ &\text{The ODE is \it homogeneous \rm if $f=0$. i.e.}\quad\frac{d^2y}{dx^2}+a\frac{dy}{dx}+by=0.\\ &\text{If $y1$ and $y2$ are two solutions for a homogeneous ODE, any linear combination $Ay_1+By_2~~(A,B\in\mathbf{R})$ is also a solution.}\\ &\frac{d^2}{dx^2}(Ay_1+By_2)+a\frac{d}{dx}(Ay_1+By_2)+b(Ay_1+By_2)=A\left(\frac{d^2y_1}{dx^2}+a\frac{dy_1}{dx}+by_1\right)+B\left(\frac{d^2y_2}{dx^2}+a\frac{dy_2}{dx}+by_2\right)=0.\\ &\text{If $y=e^{\lambda x}$, then $y'=\lambda y$ and $y''=\lambda^2 y$, and the ODE becomes }\lambda^2 y+a\lambda y+by=0.\quad \text{As $y\ne 0$,}\quad\lambda^2+a\lambda+b=0.\\ % &\text{Let $\lambda_1$ and $\lambda_2$ are the two solutions of $\lambda^2 y+a\lambda y+by=0$, there are three cases:}\\ \text{Case 1: }&\text{If $\lambda_1$ and $\lambda_2$ are two distinct real roots, the general solution to the ODE is}\quad y=Ae^{\lambda_1 x}+Be^{\lambda_2 x}.\\ \text{Case 2: }&\text{If $\lambda_1$ and $\lambda_2$ are equal (must be both real), the general solution to the ODE is}\quad y=Ae^{\lambda x}+Bxe^{\lambda x}.\\ \text{Case 3: }&\text{If $\lambda_1$ and $\lambda_2$ are two distinct complex roots $\alpha+\beta i$ and $\alpha-\beta i$ ($\alpha,\beta\in\mathbf{R}$ and $\beta\ne 0$),}\quad y=Ce^{(\alpha+\beta i)x}+De^{(\alpha-\beta i)x}.\\ &\text{By choosing complex numbers $C$ and $D$, we can find the general solution that is real.}\\ &y=Ce^{\alpha x}e^{i\beta x}+De^{\alpha x}e^{-i\beta x}=e^{\alpha x}\big(C(\cos\beta x+i\sin\beta x)+D(\cos\beta x-i\sin\beta x))\big)=e^{\alpha x}\big((C+D)\cos\beta x+i(C-D)\sin\beta x\big)\\ &\text{The general solution is therefore}\quad y=e^{\alpha x}\big(A\cos\beta x+B\sin\beta x\big).\\ &\text{(Note: This is by chooing $C=\tfrac{A}{2}-i\tfrac{B}{2}$ and $D=\tfrac{A}{2}+i\tfrac{B}{2}$~.)}\\ \\ &\text{One type of non-homogeneous ODE is that $f$ is a polynomial of degree $P$. i.e.}\quad\frac{d^2y}{dx^2}+a\frac{dy}{dx}+by=f_P(x).\\ &\text{If $y_P$ is a solution, then the general solution is $y_H+y_P$, where $y_H$ is the general solution of }\frac{d^2y}{dx^2}+a\frac{dy}{dx}+by=0.\\ &\text{Given that $y_P$ must be of degree $P$, we can deduce that}\quad\frac{dy}{dx}=\sum_{r=1}^{P}c_rrx^{r-1}\quad\text{and}\quad\frac{d^2y}{dx^2}=\sum_{r=2}^{P}c_rr(r-1)x^{r-2}.\\ &\text{We can solve for $y_P$ by finding all zeros of polynomial}\quad\sum_{r=2}^{P}c_rr(r-1)x^{r-2}+a\sum_{r=1}^{P}c_rrx^{r-1}+b\sum_{r=0}^{P}c_rx^r=f_P(x).\\ \\ &\text{e.g.}\quad y''-5y'+6y=12x-4.\\ &\text{First solve}\quad y''-5y'+6y=0\text{ (for $y_H$).}\quad\lambda^2-5\lambda+6=0\text{ has roots $\lambda=2$ and $\lambda=3$}.\quad y_H=Ae^{2x}+Be^{3x}.\\ &\text{Let $y_P=ax+b$, $y'=a$ and $y''=0$. So}\quad 0-5(a)+6(ax+b)=12x-4.\quad y_P=2x+1.\\ &\text{Solution:}\quad y=y_H+y_P=Ae^{2x}+Be^{3x}+2x+1.\\ \\ &\text{If $f$ is not a polynomial, you may need to ``guess'' what $y_P$ looks like.}\\ \end{align*} \end{document}