# First-order differenticial equation

###### tags:`工程數學`

## 分離變數型微分方程式(Separable Equation)

### 直接分離變數型

#### Type 1: $y' = g(x) \times h(y)$

$\Rightarrow \dfrac{dy}{dx} = g(x) \times h(y)$

則可分離成 $\dfrac{dy}{h(y)} = g(x)\ dx$

接著對兩邊做積分

$\displaystyle\int\dfrac{dy}{h(y)} = \displaystyle\int g(x)\ dx + c$

> Example
>
> $y' = \dfrac{y}{1+x}$
>
> $\dfrac{dy}{dx} = \dfrac{y}{1+x}$
>
> 原式可分離成 $\dfrac{dy}{y}=\dfrac{1}{1+x}\ dx$
>
> 兩邊同時積分 $\displaystyle\dfrac{dy}{y}=\int\dfrac{dx}{1+x}$
>
> $\Rightarrow \ln{|y|} = \ln |(1+x)|+c,\ c=constant$
>
> $\Rightarrow y = e^c(1+x)$[^1][^2][^3]

[^1]: `general solution` (通解)、`explicit solution` (詳盡解)，$c$ 可以是任意的常數
[^2]: $c$ 在某個條件下的解稱為particular solution (特解)
[^3]: $y = 0$ 時的解稱為 `trivial solution` (平凡無價值之解、零解)，it is not coming from the general solution and also called `singular solution`(發散、奇異)

#### Type 2: $M_1(x)\ M_2(y)\ dx+ N_1(x)\ N_2(y)\ dy=0$

則可分離成 $\dfrac{M_1(x)}{N_1(x)}\ dx+\dfrac{N_2(y)}{M_2(y)}\ dy = 0$

即 $\displaystyle\int\dfrac{M_1(x)}{N_1(x)}\ dx+\displaystyle\int\dfrac{N_2(y)}{M_2(y)}\ dy = c$

> Example
>
> $(xy+3x-y-3)\ dx=(xy-2x+4y-8)\ dy$
>
> $(x-1)(y+3)\ dx=(x+4)(y-2)\ dy$
>
> $\dfrac{(x-1)}{(x+4)}\ dx=\dfrac{y-2}{(y+3)}\ dy$
>
> **多項式分解**
>
> $1 - \dfrac{5}{(x+4)}\ dx=1 - \dfrac{5}{(y+3)}\ dy$
>
> $\displaystyle\int\Big(1 - \dfrac{5}{(x+4)}\Big)\ dx-\displaystyle\int\Big(1 - \dfrac{5}{(y+3)}\Big)\ dy=c$
>
> 兩邊積分$\Rightarrow x-5\ln |x+4| = y-5\ln |y+3|+c$

### 可簡化成分離變數型

* 變數變換

#### 齊次方程式

目的：若一階 ODE 為 $M(x, y)\ dx+N(x, y)\ dy = 0$，其中 $M(x, y)$ 與 $N(x, y)$ 同為 $m$ 次齊次函數，則稱 ODE 為一階齊次 ODE

> Example: $(x+y)\ dx-x\ dy=0$ 為齊1次ODE

將原式代為 $M(\dfrac{y}{x})\ dx+N(\dfrac{y}{x})\ dy=0$

令 $\dfrac{y}{x}=u$，則$dy=u\ dx+x\ du$ （以u取代y）

代入原ODE得 $M(u)\ dx+N(u)(u\ dx+x\ du)=0$

$\Rightarrow \displaystyle\int\dfrac{dx}{x}+\displaystyle\int\dfrac{N(u)}{uN(u)+M(u)}\ du=c$

> Example:
>
> Find $y'= \dfrac{x-y}{x+y}$
>
> $\Rightarrow (x-y)\ dx=(x+y)\ dy$
>
> 同除$x$，$(1-\dfrac{y}{x})\ dx=(1+\dfrac{y}{x})\ dy$
>
> 令 $\dfrac{y}{x}=u$，則$dy=u\ dx+x\ du$
>
> $\Rightarrow (1-u)\ dx=(1+u)\ (u\ dx+x\ du)$
>
> $\Rightarrow \dfrac{(1-u)}{(1+u)}=\dfrac{u\ dx+x\ du}{dx}$
>
> $\Rightarrow \dfrac{(1-2u-u^2)}{(1+u)}\ dx=x\ du$
>
> $\Rightarrow -\dfrac{dx}{x}=\dfrac{(u+1)}{(u^2+2u-1)}\ du$
>
> $\Rightarrow \displaystyle\int-\dfrac{dx}{x}=\displaystyle\int\dfrac{(u+1)}{(u^2+2u-1)}\ du$
>
> $\Rightarrow -\ln|x|+c_1=\dfrac{1}{2}ln|u^2+2u-1|$
>
> $\Rightarrow -2\ln|x|+2c_1=ln|u^2+2u-1|$
>
> $x^2(u^2+2u-1)=c,\ \text{where}\ c=e^{2c_1}$
>
> 將 $u=\dfrac{y}{x}$ 代入可得 $x^2(\dfrac{y^2}{x^2}+2\dfrac{y}{x}-1)=c$
>
> $y^2+2xy-x^2=c$

#### $(a_1x+b_1y+c_1)\ dx+(a_2x+b_2y+c_2)\ dy=0$ 型

屬於$M(x, y)\ dx+N(x, y)\ dy=0$

$\Rightarrow$ 非齊次方程式 $\rightarrow$ 令 $u=\dfrac{y}{x}$ 無法作變數分離

##### Type 1: 若 $\dfrac{a_1}{a_2}\neq \dfrac{b_1}{b_2}$

找到兩條線 $\begin{cases}
a_1x+b_1y+c_1=0 \\
a_2x+b_2y+c_2=0
\end{cases}$ 的交點 $(\alpha,\ \beta)$

令 $\begin{cases}
u=x-\alpha \\
v=y-\beta
\end{cases}$ 以$u$取代$x$，以$v$取代$y$ $\Rightarrow \begin{cases}
dx=du \\
dy=dv
\end{cases}$

將上式替換代入 ODE，可得一階齊次 ODE

$(a_1u+b_1v)\ du+(a_2u+b_2v)\ dv=0$

利用一階齊次 ODE 之求解方式來求解

> Example
>
> Find $(2x-5y+3)\ dx-(2x+4y-6)\ dy=0$
>
> 由 $\begin{cases}
2x-5y+3=0 \\
2x+4y-6=0
\end{cases} \Rightarrow$ 交點為 $(1, 1)$
>
> 令 $\begin{cases}
u=x-1 \\
v=y-1
\end{cases}\Rightarrow \begin{cases}
dx=du \\
dy=dv
\end{cases}$
>
> 代回原本的公式$(2u-5v)\ du-(2u+4v)\ dv=0$
>
> 令$\dfrac{v}{u}=s\Rightarrow dv=s\ du+u\ ds$
>
> 兩邊同除 $u\Rightarrow (2-5\dfrac{v}{u})\ du-(2+4\dfrac{v}{u})\ dv=0$
>
> $(2-5s)\ du=(2+4s)\ (s\ du+u\ ds)$
>
> $\dfrac{2-5s}{2+4s}=s+u\dfrac{ds}{du}$
>
> $\dfrac{-4s^2-7s+2}{4s+2}=\dfrac{u\ ds}{du}$
>
> $\dfrac{du}{u}=\dfrac{4s+2}{-4s^2-7s+2}\ ds$
>
> $\displaystyle\int\dfrac{du}{u}=\displaystyle\int\dfrac{4s+2}{-4s^2-7s+2}\ ds$
>
> 因式分解[^4]：$\dfrac{4s+2}{-4s^2-7s+2}=\dfrac{4s+2}{-(4s-1)(s+2)}$=$-\Big(\dfrac{\tfrac{4}{3}}{4s-1}+\dfrac{\tfrac{2}{3}}{s+2}\Big)$
>
> 代入原式可得 $\dfrac{1}{u}\ du=-\Big(\dfrac{\frac{4}{3}}{4s-1}+\dfrac{\frac{2}{3}}{s+2}\Big)\ ds$
>
> $\Rightarrow \dfrac{4}{3}\dfrac{ds}{4s-1}+\dfrac{2}{3}\dfrac{ds}{s+2}+\dfrac{1}{u}\ du=0$
>
> $\Rightarrow\displaystyle\int \dfrac{4}{3}\dfrac{ds}{4s-1}+\displaystyle\int\dfrac{2}{3}\dfrac{ds}{s+2}+\displaystyle\int\dfrac{1}{u}\ du=0$
>
> $\Rightarrow\dfrac{1}{3}\ln|4s-1|+\dfrac{2}{3}\ln|s+2|+\ln|u| = c_1$
>
> $\Rightarrow\ln|4\dfrac{y-1}{x-1}-1|+2\ln|\dfrac{y-1}{x-1}+2|+3\ln|x-1| = 3c_1$
>
> $\Rightarrow\ln|4y-x-3|+2\ln|y+2x-3| = 3c_1$
>
> $\Rightarrow(4y-x-3)(y+2x-3)^2=c,\text{ where } c=e^{3c_1}$

[^4]: 為了可形成$\ln$之積分項

##### Type 2: 若 $\dfrac{a_1}{a_2}= \dfrac{b_1}{b_2}=m\neq \dfrac{c_1}{c_2}時$ [^5]

[^5]: 原ODE為$(a_1x+b_1y+c_1)\ dx+(a_2x+b_2y+c_2)\ dy=0$

令 $a_2x+b_2y=t$，則 $a_1x+b_1y=mt$ ($t$ 取代 $y$) [^6]，且 $dy=\dfrac{dt-a_2\ dx}{b_2}$

[^6]: 取代 $x$ 或 $y$ 都可以

將以 $t$ 之變換代入原ODE，則 $(mt+c_1)\ dx+(t+c_2)\dfrac{dt-a_2\ dx}{b_2}=0$

即$\displaystyle\int\dfrac{t+c_2}{a_2t+a_2c_2-b_2mt-b_2c_1}\ dt=\displaystyle\int dx+c$

> Example
>
> Find $(2x-4y-3)dy+(x-2y+3)dx=0$
>
> $t = x-2y,\ dx = dt+2dy$ (以 $t$ 取代 $x$)
>
> 代入原式 $(2t-3)dy+(t+3)(dt+2dy)=0$
>
> $\Rightarrow (\dfrac{2t-3}{t+3})dy=-dt-2dy$
>
> $(\dfrac{4t+3}{t+3})dy=-dt$
>
> $\Rightarrow \dfrac{t+3}{4t+3}dt+dy=0$
>
> 分式分解 $\dfrac{t+3}{4t+3}=\dfrac{1}{4}+\dfrac{\frac{9}{4}}{4t+3}$ 代回上式
>
> $\displaystyle\int(\dfrac{1}{4}t+\dfrac{9}{4}\times\dfrac{1}{4t+3})dt +\displaystyle\int dy=c_1$
>
> $\Rightarrow\dfrac{1}{4}t+\dfrac{9}{16}\ln|4t+3|+y=c_1$
>
> $\Rightarrow 4t+9\ln|4t+3|+16y=16c_1$
>
> 將 $t=x-2y$ 代入得：
>
> $4x-8y+9\ln|4x-8y+3|+16y=c, \text{ where }c=16c_1$[^7]

[^7]: $\ln$ 內的內容可能會因為過程而有所不同，但都可以套進常數 $c$

#### $y'=f(ax+by+c)$ 型

令 $t=ax+by+c$，則 $dy=\dfrac{dt-adx}{b}$

代入原 ODE 得 $$\dfrac{dy}{dx}=\dfrac{\tfrac{dt-adx}{b}}{dx}=f(t)$$

即 $\displaystyle\int\dfrac{dx}{bf(t)+a}=\displaystyle\int dx+c$

> Example
>
> Find $y'=\tan^2(x+y)$
>
> $\therefore$ 令 $x+y=t$
>
> $\dfrac{dt-dx}{dx}=\tan^2t \Rightarrow \dfrac{dt}{dx}=1+\tan^2t=\text{sec}^2t$
>
> 移項 $\Rightarrow \dfrac{1}{\text{sec}^2t}dt=dx\Rightarrow\cos^2t\ dt=dx$
>
> $\because \cos^2t=\dfrac{1+\cos\ 2t}{2} \therefore$ 原式變為 $\displaystyle\int\dfrac{1+cos\ 2t}{2}=\displaystyle\int dx+c$

## 正合型（Exact Equation[^8]）

[^8]: 恰好、剛好

* exact equation
* 直接另解 $(x,\ y)$
* modified equation
* 積分因子(integrating factor) $I(x,\ y)$

### Exact Equation

#### 定義

設 1st-order ODE 為 $M(x,\ y)dx+N(x,\ y)dy=0$
若存在一函數 $\phi(x,\ y)$ 滿足 $M(x,\ y)dx+N(x,\ y)dy=d\phi=0$
則稱 $M(x,\ y)dx+N(x,\ y)dy=0$ 為一階正合 ODE

> Example
>
> $6xy\ dx+3x^2\ dy=0$
>
> $d\ (3x^2y)=0$
>
> $3x^2y=c$ 為解

此時 $d\phi=\Big(\dfrac{\partial\phi}{\partial x}\Big)_ydx+\Big(\dfrac{\partial\phi}{\partial y}\Big)_xdy=M(x,\ y)dx+N(x, y)dy$

#### 如何知道此ODE是否為正合

假設 ODE $M(x,\ y)dx+N(x, y)dy=0$ 為正合 equation

則由定義知，其必存在一函數中 $(x,\ y)$

使得 $\Big(\dfrac{\partial\phi}{\partial x}\Big)_y=M(x,\ y)$，且 $\Big(\dfrac{\partial\phi}{\partial y}\Big)_x=N(x, y)$

假設 $\phi=\phi(x,\ y)$ 為具有連續二偏導數之函數則$$\begin{cases}
\dfrac{\partial}{\partial y}\bigg[\Big(\dfrac{\partial \phi}{\partial x}\Big)_y\bigg]_x =\dfrac{\partial^2\phi}{\partial x\partial y}=\Big(\dfrac{\partial M}{\partial y}\Big)_x\\\\
\dfrac{\partial}{\partial x}\bigg[\Big(\dfrac{\partial \phi}{\partial y}\Big)_x\bigg]_y =\dfrac{\partial^2\phi}{\partial x\partial y}=\Big(\dfrac{\partial N}{\partial x}\Big)_y
\end{cases}$$

因此 ODE $\phi(x,\ y)$ 滿足 $M(x,\ y)dx+N(x,\ y)dy=0$ 為正合 quation 時
其判別式為 $$\Big(\dfrac{\partial M(x,\ y)}{\partial y}\Big)_x=\Big(\dfrac{\partial N(x,\ y)}{\partial x}\Big)_y$$

#### 解題方法

若 ODE $M(x,\ y)dx+N(x, y)dy=0$ 為正合
則由定義知 ODE 可改為 $M(x,\ y)dx+N(x, y)dy=d\phi=0$
因此 ODE 之解為 $\phi(x,\ y)=c$
此時 $d\phi=\Big(\dfrac{d\phi}{\partial x}\Big)_ydx+\Big(\dfrac{d\phi}{\partial y}\Big)_xdy=M(x,\ y)dx+N(x, y)dy$

滿足 $$\begin{cases}
\Big(\dfrac{\partial\phi}{\partial x}\Big)_y=M(x,\ y)\\\\
\Big(\dfrac{\partial\phi}{\partial y}\Big)_x=N(x, y)
\end{cases}\Rightarrow\begin{cases}
\phi(x,\ y)=\displaystyle\int^xM(x,\ y)\ dx+g(y)\\\\
\phi(x,\ y)=\displaystyle\int^yN(x,\ y)\ dy+h(x)
\end{cases}$$

比較由積分所求出之兩式取**聯集**後，即可求出 $g(y)$ 跟 $h(x)$ 即可得 $\phi(x,\ y)$

> Example
>
> Find $(\sin y-y\sin x)\ dx+(\cos x+x\cos y-y)\ dy=0$
>
> 令 $\begin{cases}
M=\sin y-y\sin x\\
N=\cos x+x\cos y-y
\end{cases}$
>
> 判斷是否正合 $\Rightarrow\begin{cases}
\Big(\dfrac{\partial M}{\partial y}\Big)_x=\cos y-\sin x\\\\
\Big(\dfrac{\partial N}{\partial x}\Big)_y=-\sin x+\cos y
\end{cases}\Rightarrow\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$
>
> $\begin{cases}
\Big(\dfrac{\partial\phi}{\partial x}\Big)_y=M(x,\ y)=\sin y-y\sin x\\\\
\Big(\dfrac{\partial\phi}{\partial y}\Big)_x=N(x,\ y)=\cos x+x\cos y-y
\end{cases}$
>
> 將上式偏積分後可得
>
> $\begin{cases}
\phi(x,\ y)=\displaystyle\int^x(sin y-y\sin x)\ dx+g(y)=x\sin y+y\cos x+g(y)\\\\
\phi(x,\ y)=\displaystyle\int^y(\cos x+x\cos y-y)\ dy+h(x)=y\cos x+x\sin y-\dfrac{y^2}{2}+h(x)
\end{cases}$$
>
> 比較上式，取兩式之聯集，可得 $\phi(x,\ y)=x\sin y+y\cos x-\dfrac{y^2}{2}$
(即 $g(y)=-\dfrac{y^2}{2},\ h(x)=0$)
>
> 代入 $\phi(x,\ y)=c$ 中，可得 ODE 之通解為 $x\sin y+y\cos x-\dfrac{y^2}{2}=$
>
>> 另解略過

### Modified Exact Equations 修正型正合 (find $I(x,\ y)積分因子$)

#### 定義

設一階 ODE $M(x,\ y)dx+N(x, y)dy=0$ 為非正合
若存在一函數 $I(x,\ y)$ 乘回上式後，能使原 ODE 變為正合
即 $I(x,\ y)M(x,\ y)dx+I(x,\ y)N(x, y)dy=d\phi(x,\ y)=0$
則 $I(x,\ y)$ 稱為原 ODE 之積分因子 (integrating factor)

#### 如何找 $I(x,\ y)$

由定義知 $I(x,\ y)M(x,\ y)dx+I(x,\ y)N(x, y)dy=0$ 為正合

故滿足 $\dfrac{\partial}{\partial y}\Big[I(x,\ y)M(x,\ y)\Big]=\dfrac{\partial}{\partial x}\Big[I(x,\ y)N(x, y)\Big]$

將上式展開可得： $I\dfrac{\partial M}{\partial y}+M\dfrac{\partial I}{\partial y}=I\dfrac{\partial N}{\partial x}+N\dfrac{\partial I}{\partial x}$

$\Rightarrow N(x,\ y)\dfrac{\partial I}{\partial x}-M(x,\ y)\dfrac{\partial I}{\partial y}=I\Big(\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\Big)$ (以 $I$ 為函數之一階線性PDE)

想辦法做移項（分離變數法）
if $I=I(x,\ y)$，上式無法做移項，則無解（trivial）
, but if $I=I(x)$ only, $I=I(y)$ only，則上式可簡化成 $$\begin{cases}
I=I(x):\ N\dfrac{d I}{d x}=I\Big(\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\Big) -\text{\textcircled 1}\\
or\\
I=I(y):\ -M\dfrac{d I}{d y}=I\Big(\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}\Big)-\text{\textcircled 2}
\end{cases}$$

if $I=I(x)$
由 $\text{\textcircled 1}\Rightarrow\underbrace{\dfrac{dI}{I}}_{\because\text{ x only}}=\underbrace{\Big(\dfrac{\tfrac{\partial M}{\partial y}-\tfrac{\partial N}{\partial x}}{N}\Big)}_{\therefore\ f(x)}\ dx$, $I(x)=\exp\big(\displaystyle\int f(x)\ dx\big)$

if $I=I(y)$
由 $\text{\textcircled 2}\Rightarrow\underbrace{\dfrac{dI}{I}}_{\because\text{ y only}}=\underbrace{\Big(\dfrac{\tfrac{\partial M}{\partial y}-\tfrac{\partial N}{\partial x}}{-M}\Big)}_{\therefore\ f(x)}\ dy$, $I(y)=\exp\big(\displaystyle\int f(y)\ dy\big)$

#### 結論

$A_n$ ODE $M(x,\ y)dx+N(x, y)dy=0$ 為非正合
but 存在 $I(x,\ y)$ 使得 $I(x,\ y)M(x,\ y)dx+I(x,\ y)N(x, y)dy=0$ 為正合

check:

|檢查條件|積分因子|
|:---:|:---:|
|$\text{if }\dfrac{\tfrac{\partial M}{\partial y}-\tfrac{\partial N}{\partial x}}{N} = f(x)\text{ only}$|$I(x)=\exp\big(\displaystyle\int f(x)\ dx\big)$|
|$\text{if }\dfrac{\tfrac{\partial M}{\partial y}-\tfrac{\partial N}{\partial x}}{-M} = f(y)\text{ only}$|$I(y)=\exp\big(\displaystyle\int f(y)\ dy\big)$

#### 解題方法

##### 第一步

若 ODE $M(x,\ y)dx+N(x, y)dy=0$ 為非正合 $\Big(\dfrac{\partial M}{\partial y}\ne\dfrac{\partial N}{\partial x}\Big)$

$\text{check if }\begin{cases}
\dfrac{\tfrac{\partial M}{\partial y}-\tfrac{\partial N}{\partial x}}{N} = f(x)\text{ only}\\\\
\dfrac{\tfrac{\partial M}{\partial y}-\tfrac{\partial N}{\partial x}}{-M} = f(y)\text{ only}
\end{cases}\xrightarrow{\text{then\ \ }}\begin{cases}
I(x)=\exp\big(\displaystyle\int f(x)\ dx\big)\\\\
I(y)=\exp\big(\displaystyle\int f(y)\ dy\big)
\end{cases}
$
只會有一種成立，如果兩個都沒成立，就要另找方法

##### 第二步

將 $I$ 乘回原 ODE $\Rightarrow I(x,\ y)M(x,\ y)dx+I(x,\ y)N(x, y)dy=0$ 為正合
則另 ODE 之解為 $\phi(x,\ y)=c$ ，且 $\phi$ 滿足$$\begin{cases}
\Big(\dfrac{\partial\phi}{\partial x}\Big)_y=I(x,\ y)M(x,\ y)\\\\
\Big(\dfrac{\partial\phi}{\partial y}\Big)_x=I(x,\ y)N(x, y)
\end{cases}\Rightarrow\begin{cases}
\phi(x,\ y)=\displaystyle\int^xI(x,\ y)M(x,\ y)\ dx+g(y)\\\\
\phi(x,\ y)=\displaystyle\int^yI(x,\ y)N(x,\ y)\ dy+h(x)
\end{cases}$$
比較上兩式之聯集即可求得 $\phi(x,\ y)$

> Example
>
> Find $6xy\ dx+(4y+9x^2)dy=0$
>
> 另 $\begin{cases}
M(x,\ y)=6xy\\\\
N(x,\ y)=4y+9x^2
\end{cases}\Rightarrow{\text{判斷是否為正合}}\begin{cases}
\dfrac{\partial M}{\partial y}=6x\\\\
\dfrac{\partial N}{\partial x}=18x
\end{cases}\Rightarrow\dfrac{\partial M}{\partial y}\ne\dfrac{\partial N}{\partial x}$
>
> 故原 ODE 為非正合，then 檢查是否存在積分因子 $I$ 能使原 ODE 變為正合
>
> $\dfrac{\tfrac{\partial M}{\partial y}-\tfrac{\partial N}{\partial x}}{-M}=\dfrac{-12x}{-6xy}=\dfrac{2}{y}=f(y)\text{ only}$
>
> 故存在積分因子 $I(y)=\displaystyle\int^y\Big(f(y)\ dy\Big)=\exp\Big[\displaystyle\int\dfrac{2}{y}dy\Big]=\exp\Big[\ln y^2\Big]=y^2$
>
> 將 $I(y)$ 乘回原 ODE，可得 $6xy^3\ dx+(4y^3+9x^2y^2)\ dy=0$ 為正合 equation
>
> 故令 ODE 的解 $\phi(x, y)=c$，且 $\phi(x,\ y)$ 滿足 $$\begin{cases}
\Big(\dfrac{\partial\phi}{\partial x}\Big)_y=I(x,\ y)M(x,\ y)=6xy^3\\\\
\Big(\dfrac{\partial\phi}{\partial y}\Big)_x=I(x,\ y)N(x, y)=4y^2+9x^2y^2
\end{cases}$$
> 將上式作偏積分可得 $\begin{cases}
\phi(x,\ y)=\displaystyle\int^x6xy^3\ dx=3x^2y^3+g(y)\\\\
\phi(x,\ y)=\displaystyle\int^y(4y^3+9x^2y^2)\ dy=y^4+3x^2y^3+h(x)
\end{cases}$
>
> 比較上兩式可得 $\phi(x,\ y)=3x^2y^3+y^4$ (即 $g(y)=y^4$, $h(x)=0$
> 代入 $\phi(x,\ y)=c$，可得 ODE 之通解為 $3x^2y^3+y^4=c$

### 線性行微分方程式[^9] (Linear Differential Equation)

[^9]: 即修正要正合

#### 定義

A list-order ODE: $y'(x)+P(x)y=Q(x)$

* when $Q(x)=0$，it is called `homogeneous`、齊性方程式
* when $Q(x)\ne0$，it is called `non-homogeneous`、非齊性方程式

$\Rightarrow \dfrac{dy}{dx}+P(x)y=Q(x)$

$\Rightarrow dy+\big[P(x)y-Q(x)\big]dx=0$

$\Rightarrow \underbrace{1}_Ndy+\big[\underbrace{P(x)y-Q(x)}_M\big]dx=0$

$\Rightarrow{\text{判斷是否為正合}}$ 令 $\begin{cases}
M(x,\ y)=P(x)y-Q(x)\\
N(x,\ y)=1
\end{cases}$

則 $\begin{cases}
\dfrac{\partial M}{\partial y}=\dfrac{\partial }{\partial y}\big[P(x)y-Q(x)\big]=p(x) \\\\
\dfrac{\partial N}{\partial x}=\dfrac{\partial (1)}{\partial x}=0
\end{cases}\Rightarrow\dfrac{\partial M}{\partial y}\ne\dfrac{\partial N}{\partial x}$, non-exact

, then check if an integration factor exists, 將原本之非正合 ODE 轉為正合 ODE

We observe $\dfrac{\tfrac{\partial M}{\partial y}-\tfrac{\partial N}{\partial x}}{N}=p(x)$ only, 故 ODE 存在積分因子 $I(x)=\exp\Big[\displaystyle\int p(x)\ dx\Big]$

則原 ODE 可化成 $I(x)y'(x)+I(x)P(x)y=I(x)Q(x)$ 為正合[^9]

[^9]: 如果將 $I$ 乘回改完後 ODE：$I\ dy+\Big[IPy-IQ\Big]=0$ 為正合，求解較複雜

$\Rightarrow (Iy)'=IQ$
> 證明
> $(Iy)'=I'y+y'I$
> $\because I=\exp\Big(\displaystyle\int P(x)\ dx\Big)$
> $I'=PI$
> $(Iy)'=PIy+y'I$

兩邊積分 $\Rightarrow Iy=IQ\ dx+c$

> Example
> Find $y'+y \tan x=\sin2x$, $y(0)=1$[^10]
> $\begin{cases}
> P(x)= \tan x \\
> Q(x)=\sin2x
> \end{cases}$
>
> $\therefore$ 1st-order, linear ODE
> 存在積分因子 $I=\exp\Big[\displaystyle\int P(x)\ dx\Big]$
>
> $=\exp\Big[\displaystyle\int \tan x\ dx\Big]$
>
> $=\exp\Big[\displaystyle\int \dfrac{\sin x}{\cos x}\ dx\Big]$
>
> $=\exp\Big[\displaystyle\int \dfrac{-d(\cos x)}{\cos x}\Big]=\exp\Big[-\ln \cos x\Big]=\dfrac{1}{\cos x}$
>
> $(Iy)'=IQ$
>
> $\Rightarrow Iy=\displaystyle\int IQ\ dx=\displaystyle\int \dfrac{1}{\cos x}\sin 2x\ dx$
>
> $\displaystyle\int \dfrac{2\sin x\cos x}{\cos x}\ dx=\displaystyle\int 2\sin x\ dx=-2\cos x+c$
> $\therefore y=-2\cos^2 x+c\cos x$ (general solution)
>
> $\because y(0)=1$ 代入上式 $\Rightarrow c=3$ (general solution)

故 ODE 之解為

$y=-2\cos^2x+3\cos x$ (particular solution 特解)

[^10]: Initial-value Problem (IVP)

### Bermoullis' Equation[^11]

#### 定義

$y'+P(x)y=Q(x)y^n,\ n\ne 1$, 1st-order non-linear, ODE

[^11]: 工程上 (e.g. 流體力學) 常見之微分方程式

#### 如何解

(non-linear $\rightarrow$ linear)

$y'+P(x)y=Q(x)y^n$
$\Rightarrow y^{-n}y'+P(x)y^{1-n}=Q(x)$ ($\Rightarrow{改成}y'+Py=Q$)

變數變換法，令 $u=y^{1-n}$，以 $u$ 取代 $y$ 的角色，為了將 ODE 變為 Linear

代入原式可得 $\dfrac{du}{dx}+(1-n)P(x)u=(1-n)Q(x)$ (u 之 1st-order linear ODE)

令積分因子 $I=\exp\Big[\displaystyle\int(1-n)P(x)\ dx\Big]$
$\Rightarrow Iu=\displaystyle\int I(1-n)Q(x)\ dx+c$，即可求解

> Example
>
> Find $xy'+2y=xy^3$
>
> 原式可改為 $y'+\dfrac{2}{x}y=y^2$ (Bernonllis equation $y'+P(x)y=Q(x)y^n$
>
> $\Rightarrow y^{-3}y'+\dfrac{2}{x}y^{-2}=1$ (希望可改成 $y'P(x)y=Q(x)$)
>
> 令 $u=y^{-2} (為了將 ODE 轉為 linear)$
> $\Rightarrow u'=-2y^{-3}y'$
> 代入原式可得：$-\dfrac{u'}{2}+\dfrac{2}{x}u=1$
> $u'-\dfrac{4}{x}u=-2$ (u 之 1st-order linear ODE)
> $\begin{cases}
P(x)=-\dfrac{4}{x}\\
Q(x)=-2
\end{cases}$
> 積分因子 $I=\exp\Big[\displaystyle\int P(x)dx\Big]=\exp\Big[\displaystyle\int-\dfrac{4}{x}dx\Big]=\dfrac{1}{x^4}$
>
> $Iu=\displaystyle\int IQ(x)dx=\displaystyle\int\dfrac{1}{x^4}(-2)dx=\dfrac{2}{3}\dfrac{1}{x^3}+c$
>
> 即 $u=\dfrac{2}{3}x+cx^4=y^{-2}$ (general solution)