Separable 可分離, Homogeneous 齊次, Exact 恰當 Equation

Content 內容:

• Separable Equation 可分離方程式 / 分離變數法

• Homogeneous Equations 齊次方程式

• Exact Equations 恰當方程式

Order and Degree 階與次

Order 階是指 $\frac{dy}{dx}$ 的最高階數， $\frac{dy}{dx}$ 是一階， $\frac{d^2y}{dx^2}$ 是二階

Degree 次是指 $\frac{dy}{dx}$ 最高的冪數， $\frac{dy}{dx}$ 是一次， $\frac{dy}{dx}^2$ 是二次

Solution form 解答形式

依型態可分為：

• Expilcit Solution 顯解

• Implicit Solution 隱解

依意義可分為：

• General Solution 通解

• Particular Solution 特解

• Singular Solution 奇異解

Separable Equation 可分離方程式 / 分離變數法

把變數分離到方程的兩邊

$$F(x, y, y') = 0$$ $$\begin{align} g(y) dy &= f(x) dx \\ \int g(y)dy &= \int f(x)dx \end{align}$$

例題 1：

$$\begin{align} x \frac{dy}{dx} &= 5y \\ x dy &= 5 y dx \\ \frac{1}{y} dy &= \frac{5}{x} dx \\ \int \frac{dy}{y} &= 5 \int \frac{dx}{x} \\ ln y &= 5 ln x + C \\ y &= Cx^5 \end{align}$$

例題 2：

$$\begin{align} y' &= ay \\ \frac{dy}{y} &= a dx \\ ln y &= ax + C \\ y &= Ce^{ax} \end{align}$$

例題 3：

$$\begin{align} y' + ay + b &= 0 \text{, where } a \not= 0 \\ dy &= -(ay + b) dx \\ \frac{dy}{y + \frac{b}{a}} &= -a dx \\ ln(y + \frac{b}{a}) &= -a x + C \\ y &= Ce^{-ax} - \frac{b}{a} \end{align}$$

Homogeneous Equations 齊次方程式

若 $f(\lambda x, \lambda y) = \lambda^m f(x, y)$ ，則稱 $f(x, y)$ 為 $m$ 次齊次

若一階微分方程式可以以下形式表示：

$$M(x, y) dx + N(x, y) dy = 0$$

當中 $M(x, y)$ $N(x, y)$ 為同次齊次，則稱為 Homogeneous Differential Equaltions 齊次微分方程式

例題 1:

Prove $f(x, y) = x^3 + 10 x^2 y + 3xy^2 + y^3$ is homogeneous of degree 3 三次齊次函數

$$\begin{align} &f(\lambda x, \lambda y) \\ &= (\lambda x)^3 + 10(\lambda x)^2(\lambda y) + 3 (\lambda x) (\lambda y)^2 + (\lambda y)^3 \\ &= \lambda^3 x^3 + 10 \lambda^3 x^2 y + 3 \lambda^3 x y^2 + \lambda^3 y^3 \\ &= \lambda^3 f(x, y) \end{align}$$

若果微分方程不能以 可分離方程式 / 分離變數法 求解

但本身為齊次微分方程式，即 $M(x, y) dx + N(x, y) dy = 0$

這種形式的方程可經過代換成為可分離，做法如下：

$dx$ 前面的函數 $M(x, y)$ 較簡單， $\text{Let } x = vy, dx = v dy + y dv$

$dy$ 前面的函數 $N(x, y)$ 較簡單， $\text{Let } y = ux, dy = u dx + x du$

例題 2:

求解 $y dx = (2x + y) dy$

$$\text{Let }x = vy, dx = vdy + y dv, v = \frac{x}{y}$$ $$\begin{align} y(vdy + y dv) &= (2vy + y) dy \\ vdy + ydv &= (2v+1) dy \\ y dv &= (v + 1) dy \\ \frac{dv}{v + 1} &= \frac{dy}{y} \\ ln(v + 1) &= ln y + C\\ v + 1 &= Cy\\ \frac{x}{y} + 1 &= Cy\\ y + x &= C y^2 \end{align}$$

Exact Equations 恰當方程式

對於 $M(x, y) dx + N(x, y) dy = 0$

若 $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ ，稱為 Exact 恰當

則存在 Potential Function 位勢函數 $F(x, y)$，當中：

$$\partial F = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy$$ $$\begin{align} \frac{\partial F}{\partial x} &= M & \frac{\partial F}{\partial y} &= N \\ F &= \int M dx + k(y) & F &= \int N dy + k(x) \\ \frac{\partial F}{\partial y} &= \frac{\partial}{\partial y} [\int M dx + k(y)] & \frac{\partial F}{\partial x} &= \frac{\partial}{\partial x} [\int N dy + k(x)] \\ N &= \frac{\partial}{\partial y}[ \int M dx + k(y)] & M &= \frac{\partial}{\partial x}[ \int N dy + k(x)] \end{align}$$

而 $F = C$ 是微分方程式的解

例題 1:

解 $xy' + y + 4 = 0$

改寫為 $M(x, y) dx + N(x, y) dy = 0$ 的形式

$$\begin{align} (y + 4) dx + x dy &= 0 \\ M(x, y) &= y + 4 \\ N(x, y) &= x \end{align}$$ $$\text{Since } \frac{\partial M}{\partial y} = 1 = \frac{\partial N}{\partial x}, \text{It is exact}$$

Method 1 (Use $M(x, y)$ first, then $N(x, y)$ ):

$$\begin{align} F(x, y) &= \int M(x, y) dx + k(y) \\ &= \int (y + 4) dx + k(y) \\ &= x (y + 4) + k(y)\\ \text{Since } \frac{\partial F}{\partial y} &= N \\ \frac{\partial }{\partial y} [x(y + 4) + k(y)] &= x\\ x + k'(y) &= x \\ k'(y) &= 0 \\ k(y) &= C \\ F(x, y) &= x(y + 4) + C\\ \text{Solution is } F(x, y) &= C \\ x(y + 4) &= C \end{align}$$

Method 2 (Use $N(x, y)$ first, then $M(x, y)$ ):

$$\begin{align} F(x, y) &= \int N(x, y) dy + k(x) \\ &= \int x dy + k(x) \\ &= x y + k(x)\\ \text{Since } \frac{\partial F}{\partial x} &= M \\ \frac{\partial }{\partial x} [x y + k(x)] &= y + 4\\ y + k'(x) &= y + 4 \\ k'(x) &= 4 \\ k(x) &= 4x + C \\ F(x, y) &= xy + 4x + C\\ \text{Solution is } F(x, y) &= C \\ xy + 4x &= C \\ x(y + 4) &= C \end{align}$$

有另一個比較方便的公式

$$\int^{x}_{a} M(x, y) dx + \int^{y}_{b} N(a, y) dy = C$$

注意是 $N(a, y)$ 而不是 $N(x, y)$

以下公式推導：

$$\begin{align} F(x, y) &= \int^{x}_{a} M(x, y) dx + k(y) \\ \frac{\partial F}{\partial y} &= \frac{\partial}{\partial y} [ \int^{x}_{a} M(x, y) dx ] + k'(y) \\ \frac{\partial F}{\partial y} &= N \\ N(x, y) &= \int^{x}_{a} \frac{\partial M(x, y)}{\partial y} dx + k'(y) \\ \frac{\partial M}{\partial y} &= \frac{\partial N}{\partial x} \\ N(x, y) &= \int^{x}_{a} \frac{\partial N(x, y)}{\partial x} dx + k'(y) \\ N(x, y) &= N(x, y) - N(a, y) + k'(y) \\ k'(y) &= N(a, y) \\ k(y) &= \int^y_b N(a, y) dy \end{align}$$ $$\begin{align} F(x, y) &= \int^{x}_{a} M(x, y) dx + k(y) \\ &= \int^{x}_{a} M(x, y) dx + \int^y_b N(a, y) dy \end{align}$$

例題 2:

解 $(2x^3+3y)dx+(3x+y-1)dy=0$

$$\frac{\partial M}{\partial y} = 3 = \frac{\partial N}{\partial x}, \text{It is exact}$$ $$\begin{align} \int^{x}_{a} M(x, y) dx + \int^{y}_{b} N(a, y) dy &= C \\ \int^{x}_{a} (2x^3 + 3y) dx + \int^{y}_{b} (3a + y - 1) dy &= C \\ \left[ \frac{1}{2} x^4 + 3xy \right]^x_a + \left[ 3ay + \frac{1}{2}y^2 - y \right]^y_b &= C \\ [\frac{1}{2} x^4 + 3xy - \frac{1}{2}a^4 - 3ay] \ + \ & \\ [3ay + \frac{1}{2} y^2 - y - 3ab - \frac{1}{2} b^2 + b] &= C \\ \frac{1}{2} x^4 + 3xy + \frac{1}{2} y^2 - y &= C \\ x^4 + 6xy + y^2 - 2y &= C \end{align}$$

例題 3:

解上面做過的 $xy' + y + 4 = 0$

$$(y + 4) dx + x dy = 0$$ $$\frac{\partial M}{\partial y} = 1 = \frac{\partial N}{\partial x}, \text{It is exact}$$ $$\begin{align} \int^x_a (y + 4) dx + \int^y_b a dy &= C \\ \left[ x(y+4) \right]^x_a + \left[ ay \right]^y_b &= C \\ x(y+4) &= C \end{align}$$