Undergraduate
Mathematics & Statistics
工數ch01.一階ODE複習筆記
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103
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內容含名詞解釋、ODE種類、解題方式、一些簡單的例題示範
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資料來源：YouTube頻道—姚賀騰教授 （其工數影片）、上課筆記
ノートテキスト
ページ1:
Chol. ODE - Introduction . 分類 ODE - e.g.: Y = Y₁t) [ PDE - ** eg.: u = u(x,t) ·基本名詞- (1) order: func的最高階权 (3) linear 5+ £J >R Q [(2) degree : 滿足: degree 為1次 2 無互乘項 © #124% func, 1.9.: = func -,e.g.: ⇒ ex: 1. (y²+(y+y' => 24% ODE 2. y" + 3y² + 4y = cosy > = 1 > == 94% ODE 3. Y" + y² + 4y = 2xY> = 1244 ODE 4. 8°4 +84 = 0 → = √143 144 PDE 5. dy = √1+y . 解的种類 → (x)=1+y⇒一階2次非線性ODE (1) general solution: e.g. C. cosx+ Casin x (2) particular solution: e.g.: 204x+ STUX (3) singular solution:通解中無法表示的解 y= x² 包絡線→異解 C=1 C=2 C=-2 C = -1 任一條→特解
ページ2:
⇒ ex:y=x²一本(C為任意常報)確認是否可滿足本(y)-y' + y = 0 Sol: ½ 4' λ 7 à 1² - (x + (cx- & c²) =0 C = 1 → 1 = x - 4 7 4 →通解 @ 4, 157 7 4 + 4x² - 2x² + X² = 0 → *14 DE的問題分類 ·IVP e.g: "+y=0, y(0)=y(0)=1(給一个自變數值) -BVP e.g.: y" +4=0. yo)=1y(1)=2(給两个以上自受权值) 各類1-order DDE 作題: ① 分離 ②齊次②函報 正合工 (1)分離妾权 ①y' = f(x) fz(y) e.g. y' = y ex yzdy = exdx-y=ex+c M₁ (x) (M = (y) dx + N₁ (X) N₂ (y) dy = 0 e.g. (4+1)dx = y secx dy sex dx = dy 51+0x42x dx = = ln Y+1] +Cz → →x+sin2x = = Quly²+11+ C (2) 齋性ODE (不用判斷dy/k的次方) @y' = f(x,y); fix) 0 FR func 1次 e.g. y' y = 0 •R 34 -v dy = vox + xdv y= vdx+xdv dv dx = H+V = V I+ V → lux + C = -luv + → In |X|+C = − lu | × | + → Inlyl - ₤ = c
ページ3:
© M(x, y) dx + N (x,y) dy = 0, M, N to k³R R func e.g. y² dx + (x²- xy ) dy ==>> R + 2³R = >+R =0 Ž V = ¥ dy = vdx+xdv → lulx|+ C = V - lulVI √dx = (1 + √ ] (vdx+xdv) → > lu|x| + C = = = = = lu\/\/1 vdx = (-1+V)x dv (3)函权型 ½ ax+by+C=t, by = t-ax+c, y = t-ax-c e.g. dy = (x+y+1)" dt dt dy dt-dx dx t = x + y + 1, y = t-x-1, dx = d½-1 = t², 11, de = x+c₁, & t = Tame S sede secodo = tan 't + C₂ (4) 正合ODE Odp= 20 b , dy = 3x · dx + 3x dy = f(x,y) dx + N(x, y) dy ; dt - adx b →tant = x+c 代入原ODE >X+y+1 = tan (x+C) 判別式:談-M(x,y)、謎=N(X,Y)→判斷正合⇒ 4=4 ex: (y-x + 1) dx - (y + x+5) dy = 0, £ 4 ( • ) = 4 Sol: M = Y-X+1, N = Y+X+5 > \ = 1 = 34 = exact xy-≤x²+x-≤y²+5y = c y10)代入:-8+20=C = 12 JX ↑同類型 M dy > xy-≤ x²+X - ≤ y²+5y=12 #
ページ4:
| Mdx +Ndy =0 11 14
積分因子(I)
ay
x = f(x)
N
$x = full
exp (ffiy)ody )
②判别式不相等,乘上積分因子(I)使之 exact
exp(Sfwdx)
Note:
(11) 積分因子非唯一,可有0多个
(2) 若工為積分因子,工仍是
(3) 四大常見積分因子如左表格
= f(x+y)
exp ( ) f(x+y) d (x+y))
x = flxy)
exp (fixy) d (ay))
am
-M
N
-=
N-M
YN-XM
N
ex:ydx+ (2x+5+5my)dy =0
2 1 = 7+[ = y
ex:(3y²+x+1)dx+2y(x+1)dy=0)y(0)=1
20
4y
→303y²(x+1)dx+(x+1)dx+2y(x+1)3
y+(x+1)+(x+1)+=
个
= C *
y
3.xy² + y)-ycosy + smy = C xx
1
7/1
0
STRY
-Dosy
-Siny
2: 号
1° by ₤24
→ I=(x+1)+
4
=
2y(x+1) X+1
(5) 線性 DDE
☆ly' + P(xxy = Q(x) =>
{
Q(x) 70 →
Qx) = 0 → homogeneous
monhomogeneous
; L = exp ( ) puodx]
→ Iy = fIQdx+c
ex: y+ xy=3x) y (1) = 2
Sol: | I=X
2.xy=S3xdx+C
→xy = xX²+C
3° C = 1
4:xy=x+1#
ページ5:
(6)線性化一階 ODE Bernoulli's equ y' + P(x) = Q(x) yim), nto or 1, 求解:同ìyhin可為分权)→令yu→化成以U為未知权的 一階線性 ODE ex: y-yy = Sal: 1 y² -ƒ'= > u=4", u' = -4³y' 3 +u' + 4 = ②函权微分型 >= = → xu= ≤ x²+c →X41 == x²+ & ())+ P(x)v(y)=Q10⇒合U=(y(x)),則u=y + P(x) v(y) = Q(x) = U = v(Y (x)), 1) u'= → u' + p(x) u = Q(x) ex: (>Coshy + 3x) dx + (x Stuhy) dy = 0 7 3° 1 = x² · Sinhy y' =0. x²u = -x³ + C > sinhy y' + Coshy = -3 x²-cosh x = -x² + C * 2° ½ coshy = u u= y' sinhy 41+4=3 Riccati equ y=p(x)y² + Q(x)y+R(x)=>夾雜線性&柏努利 观察一特解yp(x). → 令 y=yp(x)+uwxx代入,可化或以人為自变权, U為因受权的linear ODE ex: Y₁ = ½³ y ² - 1 / 4 + 1 ; y (1) = 3 x4²-u' = = = (4x²+2×4+1)-\*-* Sol: 1'. & Y = X 4' ^ PRODE → ∞ = 1 = 1 − 1 + 1 → → y=x為ODE之一解 2 1/2 y = x + 1 = (or X+u) y' = 1-u' 3.I-wu= = (x+a)-(x+ù)+1 (u=(x+a)-文(x+ù) u² = 1½ + ½ 12 → u² + ¼ u = - x ³ ³ I = X x4 = - In\x1+C ✗ > 1 x = -lu | x + ≤1 44
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