## Differential equation – Part IV ODE system

ODE system $\left\{ \begin{array}{l} {x^{\prime}} = f\left( {x,y,t} \right)\\ {y^{\prime}} = g\left( {x,y,t} \right) \end{array} \right.,x\left( {{t_0}} \right) = {x_0},y\left( {{t_0}} \right) = {y_0}$ Linear system: $\left\{ \begin{array}{l} {x^{\prime}} = ax + by + {r_1}\left( t \right)\\ {y^{\prime}} = cx + dy + {r_2}\left( t \right) \end{array} \right.$ $a$, $b$, $c$, $d$ can be functions of $t$. If they are constant, then the system is called constant coefficient system. When $r_1\left( t \right) = {r_2}\left( t \right) = 0$, the system is linearly homogeneous. Example:

## Differential equation – Part III Fourier series

Fourier Series To solve the ODE ${y^{\prime\prime}} + a{y^{\prime}} + by = f\left( t \right)$, in which $f\left( t \right)$ is often the combination of ${e^t}$, $\sin t$ and $\cos t$, then any reasonable $f\left( t \right)$ which is periodic, can be discontinuous but not terribly discontinuous, can be written as Fourier Series. $f\left( t \right) = {c_0} + \sum\limits_{n = 1}^\infty {{a_n}\cos nt + {b_n}\sin nt}$ Theorem: $u\left( t \right),v\left( t \right)$ are functions with period $2\pi$ on the real number,

## Differential equation – Part II Second-Order ODE

Linear 2nd order ODE with constant coefficients Homogeneous: ${y^{\prime\prime}} + A{y^{\prime}} + By = 0$ Solution: $y = {c_1}{y_1} + {c_2}{y_2}$ in which, ${y_1}$ and ${y_2}$ are two independent solutions of the homogeneous ODE. The basic method to solve this ODE is to try $y = {e^{rt}}$. Plug in and we can get ${r^2}{e^{rt}} + Ar{e^{rt}} + B{e^{rt}} = 0 \to {r^2} + Ar + B = 0$ Case 1: two different real roots $y = {c_1}{e^{{r_1}t}} + {c_2}{e^{{r_2}t}}$ Case

## Differential equation – Part I First-Order ODE

Euler equation: \left\{ \begin{align} & {{x}_{n+1}}={{x}_{n}}+h \\ & {{y}_{n+1}}={{y}_{n}}+{{A}_{n}}h \\ \end{align} \right. \left\{ \begin{align} & h=\text{step}\text{size} \\ & {{A}_{n}}=f\left( {{x}_{n}},{{y}_{n}} \right) \\ \end{align} \right. Error of solution error = exact solution – approximate solution Method 1: smaller step Euler first-order: $e\sim {{c}_{1}}h$, proportional to step size Method 2: find a better slop ${{A}_{n}}$ Heun’s method = improved Euler method = modified Euler method = RK2 \left\{ \begin{align} & {{x}_{n+1}}={{x}_{n}}+h \\ & {{y}_{n+1}}={{y}_{n}}+h\left( \frac{{{A}_{n}}+{{B}_{n}}}{2} \right) \\ \end{align} \right. in which, \({{B}_{n}}=f\left(

### Site Footer

Recording Life, Sharing Knowledge, Be Happy~