169. 附錄六： Hyperbolic Function
比較：

173. 附錄七 Linear DE 解法的步驟 (參照講義 page 151)
Step 1: Find the general solution (i.e., the complementary function ) of the associated homogeneous DE
(Sections 4-2, 4-3, 4-7)
Step 2: Find the particular solution
(Sections 4-4, 4-5, 4-6)
Step 3: Combine the complementary function and the particular solution
Extra Step: Consider the initial (or boundary) conditions

174. 4-3 Homogeneous Linear Equations with Constant Coefficients
本節使用 auxiliary equation 的方法來解 homogeneous DE
KK: [             ]
限制條件
限制條件: (1) homogeneous
(2) linear
(3) constant coefficients
a0, a1, a2, …. , an are constants
(the simplest case of the higher order DEs)

175. 4-3-2 解法 解法核心： Suppose that the solutions has the form of emx解法
解法核心：
Suppose that the solutions has the form of emx
Example: y''(x) 3 y'(x) + 2 y(x) = 0
Set y(x) = emx, m2 emx  3m emx + 2 emx = 0
m2  3m + 2 = solve m
可以直接把 n 次微分用 mn 取代，變成一個多項式
這個多項式被稱為 auxiliary equation

176.  解法流程
Step 1-1
auxiliary function
Step 1-1
Find n roots , m1, m2, m3, …., mn
(If m1, m2, m3, …., mn are distinct)
Step 1-2
n linearly independent solutions
(有三個 Cases)
Step 1-3
Complementary function

177. 4-3-3 Three Cases for Roots (2nd Order DE)
solutions
Case 1 m1  m2, m1, m2 are real
(其實 m1, m2 不必限制為 real)

178. Case 2 m1 = m2 (m1 and m2 are of course real)
First solution:
Second solution: using the method of “Reduction of Order”

179. Case 3 m1  m2 , m1 and m2 are conjugate and complex
Solution:
Another form:
set c1 = C1 + C2 and c2 = jC1 − jC2
c1 and c2 are some constant

180. Example 1 (text page 134)
(a)
2m2 − 5m − 3 = 0, m1 = −1/2, m2 = 3
(b)
m2 − 10m + 25 = 0, m1 = 5, m2 = 5
(c)
m2 + 4m + 7 = 0,

181. 4-3-4 Three + 1 Cases for Roots (Higher Order DE)
For higher order case
auxiliary function:
roots: m1, m2, m3, …., mn
(1) If mp  mq for p = 1, 2, …, n and p  q
(也就是這個多項式在 mq 的地方只有一個根)
then is a solution of the DE.
(2) If the multiplicities of mq is k (當這個多項式在 mq 的地方有 k 個根),
are the solutions of the DE.
重覆次數

182. (3) If both  + j and  − j are the roots of the auxiliary function,
then
are the solutions of the DE.
(4) If the multiplicities of  + j is k and the multiplicities of  − j is also k，then

183. Note: If  + j is a root of a real coefficient polynomial,
then  − j is also a root of the polynomial.
a0, a1, a2, …. , an are real

184. Example 3 (text page 135)
Solve
Step 1-1
m1 = 1, m2 = m3 = 2
Step 1-2
3 independent solutions:
Step 1-3
general solution:

185. Example 4 (text page 136)
Solve
Step 1-1
four roots: i, i, i, i
Step 1-2
4 independent solutions:
Step 1-3
general solution:

186. How to Find the Roots
(1) Formulas
Solutions:
(太複雜了)

187. (2) Observing
例如：1 是否為 root 看係數和是否為 0
又如：
factor: 1,3
factor: 1,2,4
possible roots: 1, 2, 4, 1/3, 2/3, 4/3
test for each possible root find that 1/3 is indeed a root

188. (3) Solving the roots of a polynomial by software
Maple
Mathematica (by the commands of Nsolve and FindRoot)
Matlab ( by the command of roots)

189. 4-3-6 本節需注意的地方 (1) 注意重根和 conjugate complex roots 的情形
本節需注意的地方
(1) 注意重根和 conjugate complex roots 的情形
(2) 寫解答時，要將 “General solution” 寫出來
(3) 因式分解要熟練
(4) 本節的方法，也適用於 1st order 的情形

190. 練習題
Sec. 4-1: 3, 7, 8, 10, 13, 20, 26, 29, 33, 36
Sec. 4-2: 2, 4, 9, 13, 14, 16, 18, 19
Sec. 4-3: 7, 16, 20, 24, 28, 33, 39, 41, 52, 54, 56, 59, 61, 63

191. 4-4 Undetermined Coefficients – Superposition Approach
This section introduces some method of “guessing” the particular solution.
方法適用條件
(1)
(2)
Suitable for linear and constant coefficient DE.
(3) g(x), g'(x), g'' (x), g'''(x), g(4)(x), g(5)(x), ………contain finite number of terms.

192. 4-4-2 方法 把握一個原則： g(x) 長什麼樣子，particular solution 就應該是什麼樣子. 記熟下一頁的規則方法
把握一個原則：
g(x) 長什麼樣子，particular solution 就應該是什麼樣子.
記熟下一頁的規則
(計算時要把 A, B, C, … 這些 unknowns 解出來)

193. Trial Particular Solutions (from text page 143)
g(x)
Form of yp
1 (any constant) A
5x + 7
Ax + B
3x2 – 2
Ax2 + Bx + C
x3 – x + 1
Ax3 + Bx2 + Cx + E
sin4x
Acos4x + Bsin4x
cos4x
e5x
Ae5x
(9x – 2)e5x
(Ax + B)e5x
x2e5x
(Ax2 + Bx + C)e5x
e3xsin4x
Ae3xcos4x + Be3xsin4x
5x2sin4x
(Ax2 + Bx + C)cos4x + (Ex2 + Fx + G)sin4x
xe3xcos4x
(Ax + B)e3xcos4x + (Cx + E)e3xsin4x
It comes from the “form rule”. See page 198.

194. yp = ?
yp = ?
yp = ?

195. 4-4-3 Examples Example 2 (text page 141)
Step 1: find the solution of the associated homogeneous equation
Guess
Step 2: particular solution
A = 6/73, B = 16/73
Step 3: General solution:

196. Example (text page 142)
Step 1: Find the solution of
Step 2: Particular solution
guess
guess

197. Particular solution
Step 3: General solution

198. 方法的解釋
Form Rule: yp should be a linear combination of g(x), g'(x),
g'' (x), g'''(x), g(4)(x), g(5)(x), …………….
Why? 如此一來，在比較係數時才不會出現多餘的項

199. When g(x) = xn
When g(x) = cos kx
When g(x) = exp(kx)

200. When g(x) = xnexp(kx) :
會發現 g(x) 不管多少次微分，永遠只出現

201. 4-4-5 Glitch of the method:
Example 4
(text page 142)
Particular solution guessed by Form Rule:
(no solution)
Why?

202. Glitch condition 1: The particular solution we guess belongs to the complementary function.
For Example 4
Complementary function
解決方法：再乘一個 x

203. Example 7
(text page 144)
From Form Rule, the particular solution is Aex
如果乘一個 x 不夠，則再乘一個 x

204. Example 8 (text page 145)
Step 1
注意： sinx, cosx 都要 乘上 x
Step 2
Step 3
Step Solving c1 and c2 by initial conditions
(最後才解 IVP)

205. Example 11 (text page 146)
From Form Rule
yp 只要有一部分和 yc 相同就作修正 修正
乘上 x
乘上 x3
If we choose
沒有 1, x2ex 兩項，不能比較係數，無解

206. If we choose
沒有 x2ex 這一項，不能比較係數，無解
If we choose
A = 1/6, B = 1/3, C = 3, E = 12

207. Glitch condition 2: g(x), g'(x), g'' (x), g'''(x), g(4)(x), g(5)(x), ……………
contain infinite number of terms.
If g(x) = ln x
If g(x) = exp(x2) :

208. 本節需要注意的地方
(1) 記住 Table 4.1 的 particular solution 的假設方法 (其實和 “form rule” 有相密切的關聯)
(2) 注意 “glitch condition”
另外，“同一類” 的 term 要乘上相同的東西 (參考 Example )
(3) 所以要先算 complementary function，再算 particular solution
(4) 同樣的方法，也可以用在 1st order 的情形
(5) 本方法只適用於 linear, constant coefficient DE

209. 4-5 Undetermined Coefficients – Annihilator Approach
For a linear DE:
Annihilator Operator:
能夠「殲滅」 g(x) 的 operator
方法適用條件
(1) Linear， (2) Constant coefficients
(3) g(x), g'(x), g'' (x), g'''(x), g(4)(x), g(5)(x), ………contain finite number of terms.

210. 4-5-2 Find the Annihilator
Example 1: (text page 150)
annihilator: D4
annihilator: D + 3

211. annihilator: (D − 2)2
(D − 2)2 = D2 − 4D + 4
註：當 coefficient 為 constants 時，function of D 的計算方式和 function of x 的計算方式相同
(x − 2)2 = x2 − 4x + 4
 (D − 2)2 = D2 − 4D + 4

212. General rule 1: If
then the annihilator is
注意： annihilator 和 a0, a1, …… , an 無關
只和 , n 有關

213. General rule 2: If
b1  0 or b2  0
then the annihilator is
Example 2: (text page 151)
annihilator
Example 5: (text page 154)
annihilator
Example 6: (text page 155)
annihilator

214. General rule 3:
If g(x) = g1(x) + g2(x) + …… + gk(x)
Lh[gh(x)] = 0 but Lh[gm(x)]  0 if m  h,
then the annihilator of g(x) is the product of Lh (h = 1 ~ k)
Proof:
(因為 L1, L2 為 linear DE with constant coefficient,
L1L2 = L2L1 )

215. Similarly, :
Therefore,

216. Example 7 (text page 154)
annihilator: D3
annihilator: D − 5
annihilator: (D − 2)3
annihilator of g(x): D3 (D − 2)3 (D − 5)

217. 4-5-3 Using the Annihilator to Find the Particular Solution
Step 2-1
Find the annihilator L1 of g(x)
Step 2-2
如果原來的 linear & constant coefficient DE 是
那麼將 DE 變成如下的型態：
(homogeneous linear & constant coefficient DE)
註： If
then

218. Step 2-3
Use the method in Section 4-3 to find the solution of
Step 2-4
Find the particular solution.
The particular solution yp is a solution of
but not a solution of
(Proof):
Since , if g(x)  0, should be nonzero.
Moreover,
Step 2-5
Solve the unknowns

219. solutions of
particular solution yp
solutions of
particular solution yp  solutions of
 solutions of
本節核心概念

220. 4-5-4 Examples Example 3 (text page 152)
Step 1: Complementary function
(solution of the associated homogeneous function)
Step 2-1: Annihilation: D3
Step 2-2:
Step 2-3:
auxiliary function
roots: m1 = m2 = m3 = 0, m4 = −1, m5 = −2
移除和 complementary function 相同的部分
Solution for :

221. Step 2-4: particular solution

222. Example 4 (text page 153)
Step 1: Complementary function
From auxiliary function, m2 − 3m = 0, roots: 0, 3
Step 2-1: Find the annihilator
D − annihilate but cannot annihilate
(D2 + 1) annihilate but cannot annihilate
(D − 3)(D2 + 1) is the annihilator of
Step 2-2:

223. Step 2-3:
auxiliary function:
易犯錯的地方
solution of :
Step 2-4: particular solution
代回原式
並比較係數
Step 2-5:
Step 3: general solution

224. 4-5-5 本節要注意的地方 (1) 所以要先算 complementary function，再算 particular solution
本節要注意的地方
(1) 所以要先算 complementary function，再算 particular solution
(2) 若有兩個以上的 annihilator，選其中較簡單的即可
(3) 計算 auxiliary function 時有時容易犯錯
(4) 的解和 的解不一樣。
(5) 這方法，只適用於 constant coefficient linear DE
(因為，還需借助 auxiliary function)

225. The thing that can be done by the annihilator approach can always be done by the “guessing” method in Section 4-4, too.

226. 4-6 Variation of Parameters
方法的限制
The method can solve the particular solution for any linear DE
(1) May not have constant coefficients
(2) g(x) may not be of the special forms

227. 4-6-2 Case of the 2nd order linear DE
associated homogeneous equation:
Suppose that the solution of the associated homogeneous equation is
Then the particular solution is assumed as:
(方法的基本精神)

228. 代入原式後，總是可以簡化 代入
zero
zero

229. 簡化
進一步簡化： 假設
聯立方程式

230. where
| |: determinant
可以和 1st order case (page 58) 相比較

231. 4-6-3 Process for the 2nd Order Case
Step 2-1 變成 standard form
Step 2-2
Step 2-3
Step 2-4
Step 2-5

232. 4-6-4 Examples Example 1 (text page 159) Step 1: solution of Step 2-2:

233. Step 2-4:
Step 2-5:
Step 3:

234. Example 2 (text page 159)
Step 1: solution of
Step 2-1: standard form:
Step 2-2:
Step 2-3:
Step 2-4:
(未完待續) 注意 算法

235. Step 2-5:
Step 3:
Note: 課本 Interval (0, /6) 改為(0, /3)
Example 3 (text page 160)
Note: 沒有 analytic 的解
所以直接表示成 (複習 page 45)

236. 4-6-5 Case of the Higher Order Linear DE
Solution of the associated homogeneous equation:
The particular solution is assumed as:

238. Wk: replace the kth column of W by
For example, when n = 3,

239. 4-6-6 Process of the Higher Order Case
Step 2-1 變成 standard form
Step 2-2 Calculate W, W1, W2, …., Wn (see page 237)
Step ………
Step …….
Step 2-5

240. 4-6-7 本節需注意的地方 養成先解 associated homogeneous equation 的習慣 記熟幾個重要公式
本節需注意的地方
養成先解 associated homogeneous equation 的習慣
記熟幾個重要公式
這裡 | | 指的是 determinant
(4) 算出 u1(x) 和 u2(x) 後別忘了作積分
(5) f(x) = g(x)/an(x) (和 1st order 的情形一樣，使用 standard form)
(6) 計算 u1'(x) 和 u2'(x) 的積分時，+ c 可忽略
因為 我們的目的是算particular solution yp
yp 是任何一個能滿足原式的解
(7) 這方法解的範圍，不包含 an(x) = 0 的地方
特別要小心