線性微分方程

# 線性微分方程 Linear differential equation ![Creative Commons License](https://i.creativecommons.org/l/by/4.0/88x31.png) This work by Jephian Lin is licensed under a [Creative Commons Attribution 4.0 International License](http://creativecommons.org/licenses/by/4.0/). $\newcommand{\trans}{^\top} \newcommand{\adj}{^{\rm adj}} \newcommand{\cof}{^{\rm cof}} \newcommand{\inp}[2]{\left\langle#1,#2\right\rangle} \newcommand{\dunion}{\mathbin{\dot\cup}} \newcommand{\bzero}{\mathbf{0}} \newcommand{\bone}{\mathbf{1}} \newcommand{\ba}{\mathbf{a}} \newcommand{\bb}{\mathbf{b}} \newcommand{\bc}{\mathbf{c}} \newcommand{\bd}{\mathbf{d}} \newcommand{\be}{\mathbf{e}} \newcommand{\bh}{\mathbf{h}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bq}{\mathbf{q}} \newcommand{\br}{\mathbf{r}} \newcommand{\bx}{\mathbf{x}} \newcommand{\by}{\mathbf{y}} \newcommand{\bz}{\mathbf{z}} \newcommand{\bu}{\mathbf{u}} \newcommand{\bv}{\mathbf{v}} \newcommand{\bw}{\mathbf{w}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\nul}{\operatorname{null}} \newcommand{\rank}{\operatorname{rank}} %\newcommand{\ker}{\operatorname{ker}} \newcommand{\range}{\operatorname{range}} \newcommand{\Col}{\operatorname{Col}} \newcommand{\Row}{\operatorname{Row}} \newcommand{\spec}{\operatorname{spec}} \newcommand{\vspan}{\operatorname{span}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\idmap}{\operatorname{id}} \newcommand{\am}{\operatorname{am}} \newcommand{\gm}{\operatorname{gm}} \newcommand{\mult}{\operatorname{mult}} \newcommand{\iner}{\operatorname{iner}}$ ```python from lingeo import random_int_list, random_good_matrix ``` ## Main idea In this section, every function, such as $x$ and $y$, are from $\mathbb{R}$ to $\mathbb{R}$ in variable $t$. We will use the notation $\dot{x}$ and $x'$ interchangeably for $\frac{dx}{dt}$. Note that the derivative is taken with respect to $t$ but not to $x$. Some differential equations are easy to be solved. For example, the solution to the equation $$ \dot{x} = kx $$ is $x = C_1e^{kt}$ for any constant $C_1\in\mathbb{R}$. And one may try that the solution to the equation $$ \dot{x} = kx + C_1e^{kt} $$ is $x = (C_2 + C_1t)e^{kt}$ for any constant $C_2\in\mathbb{R}$. (To see these are _all_ the solutions relies on the Picard–Lindelöf theorem.) Let $x_1$ and $x_2$ be functions in variable $t$. The system $$ \begin{aligned} \dot{x}_1 &= ax_1 + bx_2 \\ \dot{x}_2 &= cx_1 + dx_2 \end{aligned} $$ is called a **homogeneous system of first-order linear differential equation**. One may write $$ \bx = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}, \quad \dot{\bx} = \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix}, \text{ and}\quad A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}. $$ Then the system becomes $$ \dot{\bx} = A\bx. $$ If $A$ is diagonalizable as $D = Q^{-1}AQ$, we may set $Q\by = \bx$ for some $\by = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}$. Since $Q$ is a constant matrix, $\dot{\bx} = \dot{(Q\by)} = Q\dot{\by}$. Thus, the original system becomes $$ \dot{\by} = Q^{-1}AQ \by = D\by, $$ which is much easier to be solved. Once $\by$ is solved, $\bx$ can be found by $\bx = Q\by$. ## Side stories - higher order differential equation ## Experiments ##### Exercise 1 執行以下程式碼。  Run the code below.  ```python ### code set_random_seed(0) print_ans = False lam1,lam2 = random_int_list(2, 3) Q = random_good_matrix(2,2,2) A = Q * diagonal_matrix([lam1, lam2]) * Q.inverse() pretty_print(LatexExpr(r"\dot{x}_1 = (%s)x_1 + (%s)x_2"%(A[0,0],A[0,1]))) pretty_print(LatexExpr(r"\dot{x}_2 = (%s)x_1 + (%s)x_2"%(A[1,0],A[1,1]))) t = var("t") y = vector([exp(lam1 * t), exp(lam2 * t)]) x = Q * y pretty_print(LatexExpr("x_1 ="), x[0]) pretty_print(LatexExpr("x_2 ="), x[1]) if print_ans: pretty_print(LatexExpr(r"\dot{x}_1 ="), x[0].derivative(t)) pretty_print(LatexExpr(r"\dot{x}_2 ="), x[1].derivative(t)) ``` :::warning Record the output equations here. ::: By executing the given code above, we get $\dot{x}_1 = (−3)x_1+(−30)x_2$ $\dot{x}_2 = (0)x_1+(3)x_2$ $x_1 = -5e^{3t}+e^{-3t}$ $x_2= e^{3t}.$ ##### Exercise 1(a) 計算 $\dot{x}_1$ 及 $\dot{x}_2$。  Find $\dot{x}_1$ and $\dot{x}_2$.  Ans: By computation, $$ \begin{aligned} \dot{x}_1 &= \dot{(−5e^{3t}+e^{−3t})} = (-15)e^{3t}-3e^{-3t} \\ \dot{x}_2 &= \dot{(e^{3t})} =3e^{3t}. \end{aligned} $$ --- ##### Exercise 1(b) 驗證題目給的 $x_1$ 和 $x_2$ 是否滿足給定的微分方程組。  Verify that $x_1$ and $x_2$ are solutions to the system of differential equations.  Ans: $\dot{x}_1 = (-15)e^{3t}-3e^{-3t} = (−3)x_1+(−30)x_2 = \dot{x}_1$ $\dot{x}_2 = 3e^{3t} = (0)x_1+(3)x_2 = \dot{x}_2$ By the equation above, we can know that the given $x_1$ and $x_2$ satisfy the differential equations. :::info What do the experiments try to tell you? (open answer) We can know that $x_1$ and $x_2$ satisfy the differential equations. ::: :::warning This part is not yet done. ::: ## Exercises ##### Exercise 2 解以下的微分方程組。（請記得加上該有的常數。）  Solve the following differential equation. (Note that constant terms are important.)  ##### Exercise 2(a) 已知 $$ \begin{bmatrix} 4 & 0 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} 5 & -1 \\ -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ 解微分方程組： $$ \begin{aligned} \dot{x}_1 &= 5x_1 - x_2 \\ \dot{x}_2 &= -x_1 + 5x_2 \end{aligned} $$  It is known that $$ \begin{bmatrix} 4 & 0 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} 5 & -1 \\ -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ Solve the system of differential equations: $$ \begin{aligned} \dot{x}_1 &= 5x_1 - x_2 \\ \dot{x}_2 &= -x_1 + 5x_2 \end{aligned} $$  Ans: We begin by writing the system of differential equations in matrix form: $$ \dot{\mathbf{x}} = \begin{bmatrix} 5 & -1 \\ - 1 & 5 \end{bmatrix} \mathbf{x}. $$ While $$ \begin{bmatrix} 4 & 0 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} 5 & -1 \\ -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}, $$ we let $$ \by = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \bx $$ such that $$ \begin{bmatrix} \dot{y}_1 \\ \dot{y}_2 \\ \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 6 \\ \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ \end{bmatrix}, $$ and we can have $$ \begin{bmatrix} y_1 \\ y_2 \\ \end{bmatrix} = \begin{bmatrix} c_1e^{4t} \\ c_2e^{6t} \\ \end{bmatrix}. $$ Since $$ \mathbf{x} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \mathbf{y}, $$ We can know that $$ \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} = \begin{bmatrix} c_1e^{4t}+c_2e^{6t} \\ c_1e^{4t}-c_2e^{6t} \\ \end{bmatrix}. $$ :::warning - [x] we can assume that there exists $$ \mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} $$ --> we let $$ \by = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \bx $$ ::: --- ##### Exercise 2(b) 已知 $$ \begin{bmatrix} 4 & 0 \\ 0 & -6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} -1 & 5 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ 解微分方程組： $$ \begin{aligned} \dot{x}_1 &= -x_1 + 5x_2 \\ \dot{x}_2 &= 5x_1 - x_2 \end{aligned} $$  It is known that $$ \begin{bmatrix} 4 & 0 \\ 0 & -6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} -1 & 5 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ Solve the system of differential equations: $$ \begin{aligned} \dot{x}_1 &= -x_1 + 5x_2 \\ \dot{x}_2 &= 5x_1 - x_2 \end{aligned} $$  ##### Exercise 2(b)Answer here:) Write the system of differential equations in matrix form: $$ \dot{\bx} = A\bx. $$ $$ \dot{\mathbf{x}} = \begin{bmatrix} -1 & 5 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. $$ And it is known that $$ \begin{bmatrix} 4 & 0 \\ 0 & -6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} -1 & 5 \\ 5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ Because $A$ is diagonalizable as $D = Q^{-1}AQ$, we can set $Q\by = \bx$ for some $\by = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}$. $$ \begin{bmatrix} \dot{y}_1 \\ \dot{y}_2 \\ \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & -6 \\ \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ \end{bmatrix}, $$ $$ \begin{bmatrix} y_1 \\ y_2 \\ \end{bmatrix} = \begin{bmatrix} c_1e^{4t} \\ c_2e^{-6t} \\ \end{bmatrix}. $$ Since $Q$ is a constant matrix, $\dot{\bx} = \dot{(Q\by)} = Q\dot{\by}$. $\dot{\by} = Q^{-1}AQ \by = D\by,$then $\bx = Q\by$. $$ \bx=\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\by. $$ $$ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} c_1e^{4t} + c_2e^{-6t} \\ c_1e^{4t} - c_2e^{-6t} \\ \end{bmatrix}. $$ ##### Exercise 2(c) 已知 $$ \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ 解微分方程組： $$ \begin{aligned} \dot{x}_1 &= x_1 + x_2 \\ \dot{x}_2 &= x_1 + x_2 \end{aligned} $$  It is known that $$ \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. $$ Solve the system of differential equations: $$ \begin{aligned} \dot{x}_1 &= x_1 + x_2 \\ \dot{x}_2 &= x_1 + x_2 \end{aligned} $$  Ans: We begin by writing the system of differential equations in matrix form: $$ \dot{\mathbf{x}} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \mathbf{x}. $$ While $$ \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}, $$ we can assume that there exists $$ \mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} $$ we let $$ \by = \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} \bx $$ such that $$ \begin{bmatrix} \dot{y}_1 \\ \dot{y}_2 \\ \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 0 \\ \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ \end{bmatrix}, $$ and we can have $$ \begin{bmatrix} y_1 \\ y_2 \\ \end{bmatrix} = \begin{bmatrix} c_1e^{2t} \\ c_2e^{0t} \\ \end{bmatrix}. $$ Since $$ \mathbf{x} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \mathbf{y}, $$ We can know that $$ \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} = \begin{bmatrix} c_1e^{2t}+c_2e^{0} \\ c_1e^{2t}-c_2e^{0} \\ \end{bmatrix} = \begin{bmatrix} c_1e^{2t}+c_2 \\ c_1e^{2t}-c_2 \\ \end{bmatrix}. $$ :::warning See 2(a). ::: ##### Exercise 3 解以下的微分方程組。（請記得加上該有的常數。）  Solve the following differential equation. (Note that constant terms are important.)  ##### Exercise 3(a) 已知 $$ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1} \begin{bmatrix} 4 & 0 & -1 \\ 0 & 4 & -1 \\ -1 & -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}. $$ 解微分方程組： $$ \begin{aligned} \dot{x}_1 &= 4x_1 + 0x_2 - x_3 \\ \dot{x}_2 &= 0x_1 + 4x_2 - x_3 \\ \dot{x}_3 &= -x_1 - x_2 + 5x_3 \end{aligned} $$  It is known that $$ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1} \begin{bmatrix} 4 & 0 & -1 \\ 0 & 4 & -1 \\ -1 & -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}. $$ Solve the system of differential equations: $$ \begin{aligned} \dot{x}_1 &= 4x_1 + 0x_2 - x_3 \\ \dot{x}_2 &= 0x_1 + 4x_2 - x_3 \\ \dot{x}_3 &= -x_1 - x_2 + 5x_3 \end{aligned} $$  **[由孫心提供]** ##### Exercise 3(a)-- answer here: We rewrite the questions as $$ \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} 4 & 0 & -1 \\ 0 & 4 & -1 \\ -1 & -1 & 5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}.$$ According to the known conditions, we can get the eigendecomposition $QDQ^{-1}$ of the matrix can be obtained as $$ \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix} \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1}. $$ Let $$ \dot{\bx} = \dot{(Q\by)} = Q\dot{\by},\ \dot{\by} = Q^{-1}AQ \by = D\by. $$ Then $$ \begin{bmatrix} \dot{y}_1 \\ \dot{y}_2 \\ \dot{y}_3 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix},$$ $$ \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} c_1e^{3t} \\ c_2e^{4t} \\ c_3e^{6t} \end{bmatrix}$$ $$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} c_1e^{3t}+c_2e^{4t}+ c_3e^{6t} \\ c_1e^{3t}-c_2e^{4t}+ c_3e^{6t} \\ c_1e^{3t}-2c_3e^{6t} \end{bmatrix}. $$ --- ##### Exercise 3(b) 已知 $$ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & -6 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1} \begin{bmatrix} 2 & -2 & 3 \\ -2 & 2 & 3 \\ 3 & 3 & -3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}. $$ 解微分方程組： $$ \begin{aligned} \dot{x}_1 &= 2x_1 - 2x_2 + 3x_3 \\ \dot{x}_2 &= -2x_1 + 2x_2 + 3x_3 \\ \dot{x}_3 &= 3x_1 + 3x_2 - 3x_3 \end{aligned} $$  It is known that $$ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & -6 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1} \begin{bmatrix} 2 & -2 & 3 \\ -2 & 2 & 3 \\ 3 & 3 & -3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}. $$ Solve the system of differential equations: $$ \begin{aligned} \dot{x}_1 &= 2x_1 - 2x_2 + 3x_3 \\ \dot{x}_2 &= -2x_1 + 2x_2 + 3x_3 \\ \dot{x}_3 &= 3x_1 + 3x_2 - 3x_3 \end{aligned} $$  **[由孫心提供]** ##### Exercise 3(b)-- answer here: Equivalent to the previous question, we rewrite the question as $$\dot{\bx}== \begin{bmatrix} 2 & -2 & 3 \\ -2 & 2 & 3 \\ 3 & 3 & -3 \end{bmatrix}\bx.$$ We know that $$ \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix} \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & -6 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 0 & -2 \end{bmatrix}^{-1} $$ is the eigen decomposition of $QDQ^{-1}$$$\begin{bmatrix} 2 & -2 & 3 \\ -2 & 2 & 3 \\ 3 & 3 & -3 \end{bmatrix}.$$ Let $$\dot{\bx} = \dot{(Q\by)} = Q\dot{\by},\dot{\by} = Q^{-1}AQ \by = D\by. $$ THen $$ \begin{bmatrix} \dot{y}_1 \\ \dot{y}_2 \\ \dot{y}_3 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & -6 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix},$$ $$ \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} c_1e^{3t} \\ c_2e^{4t} \\ c_3e^{-6t} \end{bmatrix}$$ $$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} c_1e^{3t}+c_2e^{4t}+ c_3e^{-6t} \\ c_1e^{3t}-c_2e^{4t}+ c_3e^{-6t} \\ c_1e^{3t}-2c_3e^{-6t} \end{bmatrix}. $$ ##### Exercise 4 對於高階的線性微分方程 $y'' + c_1 y' + c_2 y = 0$，我們也可以寫成矩陣形式。令 $$ \by = \begin{bmatrix} y \\ y' \end{bmatrix}, \quad \dot{\by} = \begin{bmatrix} y' \\ y'' \end{bmatrix},\text{ and}\quad A = \begin{bmatrix} 0 & 1 \\ -c_2 & -c_1 \end{bmatrix}. $$ 則原方程式可以改寫為 $$ \dot{\by} = A\by. $$ 解以下的微分方程。  For the higher-order differential equation $y'' + c_1 y' + c_2 y = 0$, we may let $$ \by = \begin{bmatrix} y \\ y' \end{bmatrix}, \quad \dot{\by} = \begin{bmatrix} y' \\ y'' \end{bmatrix},\text{ and}\quad A = \begin{bmatrix} 0 & 1 \\ -c_2 & -c_1 \end{bmatrix}. $$ Then the differential equation can be written as $$ \dot{\by} = A\by. $$ Solve the following differential equations.  ##### Exercise 4(a) 已知 $$ \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -6 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}. $$ 考慮微分方程 $y'' - 5y' + 6y = 0$。  It is known that $$ \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -6 & 5 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 3 & 2 \end{bmatrix}. $$ Consider the differential equation $y'' - 5y' + 6y = 0$.  **[由孫心提供]** ##### Exercise 4(a)-- answer here: Let $$ A = \begin{bmatrix} 0 & 1 \\ -6 & 5 \end{bmatrix},\ Q = \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix}. $$ And it can be seen from the question $$ Q^{-1}AQ = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}. $$ Set $\bx$ let $$ \bz = Q\bx, \\ \bz = \begin{bmatrix} z \\ z' \end{bmatrix},\ \bx = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. $$ And $\bz' = (Q\bx)' = Q\bx'$ . By the differential equations $y'' - 5y' + 6y = 0$, we can know that $\bz' = A\bz$, and we can get that $$ Q\bx' = AQ\bx\ ,\ \bx' = Q^{-1}AQ\bx. $$ It is $$ \begin{bmatrix} {x_1}' \\ {x_2}' \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. $$ So we would know $x_1 = c_1e^{2t}, \ x_2 = c_2e^{3t}$ . Since $\bz = Q\bx$ and $y$ is the first entry of $\bz$, we can get $$ y = c_1e^{2t} + c_2e^{3t}\ . $$ ##### Exercise 4(b) 已知 $$ \begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -4 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix}. $$ 考慮微分方程 $y'' + 0y' + 4y = 0$。  It is known that $$ \begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -4 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix}. $$ Consider the differential equation $y'' + 0y' + 4y = 0$.  **[由孫心提供]** ##### Exercise 4(b)-- answer here: Let $$ A = \begin{bmatrix} 0 & 1 \\ 4 & 0 \end{bmatrix},\ Q = \begin{bmatrix} 1 & 1 \\ 2 & -2 \end{bmatrix}. $$ And it can be seen from the question $$ Q^{-1}AQ = \begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix}. $$ Set $\bx$ let $$ \bz = Q\bx, \\ \bz = \begin{bmatrix} z \\ z' \end{bmatrix},\ \bx = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. $$ And $\bz' = (Q\bx)' = Q\bx'$ 。 By the differential equations $y'' + 0y' + 4y = 0$, we can know that $\bz' = A\bz$, and we can get that $$ Q\bx' = AQ\bx\ ,\ \bx' = Q^{-1}AQ\bx. $$ It is $$ \begin{bmatrix} {x_1}' \\ {x_2}' \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. $$ So we would know $x_1 = c_1e^{2t},\ x_2 = c_2e^{-2t}$. Since $\bz = Q\bx$ and $y$ is the first entry of $\bz$, we can get $$ y = c_1e^{2t} + c_2e^{-2t}\ . $$ ##### Exercise 4(c) 已知 $$ \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -4 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}. $$ 考慮微分方程 $y'' - 4y' + 4y = 0$。  It is known that $$ \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -4 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}. $$ Consider the differential equation $y'' - 4y' + 4y = 0$.  **[由孫心提供]** ##### Exercise 4(c)-- answer here: Let $$ A = \begin{bmatrix} 0 & 1 \\ -4 & 4 \end{bmatrix},\ Q = \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}. $$ And it can be seen from the question $$ Q^{-1}AQ = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}. $$ Set $\bx$ let $$ \bz = Q\bx, \\ \bz = \begin{bmatrix} z \\ z' \end{bmatrix},\ \bx = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. $$ And $\bz' = (Q\bx)' = Q\bx'$. By the differential equations $y'' - 4y' + 4y = 0$, we can know that $\bz' = A\bz$, and we can get that $$ Q\bx' = AQ\bx\ ,\ \bx' = Q^{-1}AQ\bx. $$ It is $$ \begin{bmatrix} {x_1}' \\ {x_2}' \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. $$ So we would know $x_2 = c_2e^{2t}$. Since ${x_1}' = 2x_1 + c_2e^{2t}$, we would know that$x_1 = (c_2 + tc_1)e^{2t}$. Since $\bz = Q\bx$ and $y$ is the first entry of $\bz$, we can get $$ y = 2(c_2 + tc_1)e^{2t} - c_2e^{2t}\ . $$ ##### Exercise 4(d) 已知 $$ \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 9 & 0 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -9 & 6 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 9 & 0 \end{bmatrix}. $$ 考慮微分方程 $y'' - 6y' - 9y = 0$。  It is known that $$ \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 9 & 0 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 \\ -9 & 6 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 9 & 0 \end{bmatrix}. $$ Consider the differential equation $y'' - 6y' - 9y = 0$.  **[由孫心提供]** ##### Exercise 4(d)-- answer here: Let $$ A = \begin{bmatrix} 0 & 1 \\ -9 & 6 \end{bmatrix}, \ Q = \begin{bmatrix} 3 & -1 \\ 9 & 0 \end{bmatrix}. $$ And it can be seen from the question $$ Q^{-1}AQ = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix}. $$ Set $\bx$ let $$ \bz = Q\bx, \\ \bz = \begin{bmatrix} z \\ z' \end{bmatrix},\ \bx = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. $$ And $\bz' = (Q\bx)' = Q\bx'$. By the differential equations $y'' - 6y' - 9y = 0$, we can know that $\bz' = A\bz$, and we can get that $$ Q\bx' = AQ\bx\ ,\ \bx' = Q^{-1}AQ\bx. $$ It is $$ \begin{bmatrix} {x_1}' \\ {x_2}' \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. $$ So we would know $x_2 = c_2e^{3t}$ 。 Since ${x_1}' = 3x_1 + c_2e^{3t}$, we would know that $x_1 = (c_2 + tc_1)e^{3t}$. Since $\bz = Q\bx$ and $y$ is the first entry of $\bz$, we can get $$ y = 3(c_2 + tc_1)e^{3t} - c_2e^{3t}\ . $$ --- ##### Exercise 5 已知 $$ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 9 & 4 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -11 & 6 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 9 & 4 & 1 \end{bmatrix}. $$ 解微分方程 $y''' - 6y'' + 11y' - 6y = 0$。  It is known that $$ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 9 & 4 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -11 & 6 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 9 & 4 & 1 \end{bmatrix}. $$ Consider the differential equation $y''' - 6y'' + 11y' - 6y = 0$.  **[由孫心提供]** ##### Exercise 5-- answer here: Set $$\bz =\begin{bmatrix} z \\ z' \\ z'' \end{bmatrix}, \dot{\bz}=\begin{bmatrix} z' \\ z'' \\ z''' \end{bmatrix} ,\bz=Q\bx. $$ $$ \bx=\begin{bmatrix} x \\ x' \\ x'' \end{bmatrix},\dot{\bx}=\begin{bmatrix} x' \\ x'' \\ x''' \end{bmatrix}. $$ So, $\dot{\bz}=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -11 & 6 \end{bmatrix}\bz=\dot{(Q\bx)} =Q\dot\bx= AQ\bx.$ So, $\dot\bx=Q^{-1}AQ\bx=\begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}\bx=\begin{bmatrix} x' \\ x'' \\ x''' \end{bmatrix}=\begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x \\ x' \\ x'' \end{bmatrix}=\begin{bmatrix} 3x \\ 2x' \\ x'' \end{bmatrix}$. We can solve the equation $x=c_1e^{3t},x'=c_2e^{2t},x''=c_3e^{t}$， take into $\bz=Q\bx=\begin{bmatrix} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 9 & 4 & 1 \end{bmatrix}\begin{bmatrix} c_1e^{3t} \\ c_2e^{2t} \\ c_3e^{t} \end{bmatrix}=\begin{bmatrix} c_1e^{3t}+c_2e^{2t}+ c_3e^{t} \\ 3c_1e^{3t}+2c_2e^{2t}+c_3e^{t} \\ 9c_1e^{3t}+4c_2e^{2t}+c_3e^{t} \end{bmatrix}$ Since $\bz = Q\bx$ and $y$ is the first entry of $\bz$, we can get $$ y=c_1e^{3t}+c_2e^{2t}+ c_3e^{t}. $$ :::info collaboration: 2 3 problems: 3 - 2abc moderator: 1 qc: 1 :::

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