ページ3:

•Non-exact equation
加一個卡進去:
161 F 1 £ F can be F(x), Fly), F(x,y)
If FP (x, y) dx + FQ (x,y) dy.
non-exact.
= 0
is exact equation
Jp (x,y)
JQ(x,y)
JX
J(FQ)
=
J (FP)
dy
dx
Fy P - FPy
+
=
FxQFQ₂
Find F(x) →
0+ FPy
= FxQ + FQx
對F(g) 反之亦然
Fx =
Q [FPy - F Qx ]
F [
Q
dy
-
JQ
-]
(+ * Jx - 1 + 143 1
dx
F Jx
=
Inf = SRdx
dy
·R (x)
JQ J J x
Jdx
F
= Ce
SRix)dx

ページ4:

homogeneous linear ODE
y' + pix) y = r(x) = 0
dy
dx
dy
+
· P(x)y = 0
y = -pwy
S dy = S-p(x)dx
y₁ = ce
-Spix) dx
Nonhomogeneous linear ODE
y' + P(x) y = x(x) & 0
dy
dx = -(p(x)y - r(x)) x 1
Idy = - (pix)y-rux)) dx
上一页的中
Q
(p(x)y -rwx)) dx + bdy = 0

ページ5:

SPW)dx
FW) = e
=
R(x) + (Py- Qx) · P(x)
N
hix)
e" (pwy - rw) dx + edy.
du = Ny
ry
hw
u. S ( 1 ) dy = se dy
=
=0
e "xy + k (x) → Ju = M
> y = e^ Se "r(x) d x + ce"
dx
6 直接代公式
πt' @ 13 k(x)

ページ6:

modeling an engineering problem
Step 1. Find independent variable t or X,
Y₁ z
Step 2. Find dependent variable:
target
Step 3. Find the model (governing equation)
控制方程式
Ex. Torricellis's Law
ho
V(t)
= 225 cm
An area of hole
AT: cross-section area of tank
`AM = 0.5³TV [cm³]
A Tank = 100² πv [cm³]
Step 1. t
Step 2 h(t) = ?
Step 3
用
Bernoulli's equation
mass balance
:
V(+) = 0.6 √2gh(t)
out
=
in
比較高
V(+) AH ot
=
hit)
hittot)
(h(t)-h(t+ot)) At

ページ7:

√ (+) x-
AH
A1
h(t)-h(trot)
h(t+ot)-h(t)
=
=> -lim
ot
st
V(t) AH
Jhle
x
A1
dt.
It Bernoulli's equation
Jhle
dt
= 0.6 √2gh(t) x k
Jodh(t)=
0.6 k√zg dt
dh
=
(1/1) h² = 0.6 kt√³g + C
0.6 k√29.
(sh (t)
=
( = x ( 0.6 k + √ zg + c ) ) ²
b t
2
h (o) = 225 = (-2)², c=30
h(t)
=
(-0.3 k t√2g + 15 ) ²
2
but whent t∞h won't → ∞
So, tc
15
0.3k√29

ページ8:

Ex Mixing
Jw
dt
=
-input-output
(ratex Concentration).
W(+)
ric₁ - r, w(t)
Wlt)

ページ9:

Ex.
money in bank
複利
Step 1. t
Continuous compounding n=00
Step 2. y(t) = ?
Step 3.
y (trot) -y (t) = ay⋅ ot
·y (trot)-y(t)
ot
lim y (trot) -y(t)
otro
ot
dy
dt
=
=
==
ay
ay
ay
Sjdy Sade
=
lny at +c.
y = e
Y
at+c₁
=
t time
y(t): money in bank.
y (o) = Yo
a rate
monthly α = 3%
t-1
daily a
at
C₂eaz
10 ) = C ₂ e° = C₂ = Yo
.. y = y₁ eat
3
=
= 0.1%
30
t = 30

ページ10:

:
nonlinear Bernoulli
equation
a
y' + py = qy², a+ Oorl
•When a Oor 1 linear ODE
=
set yka=u,對u微分伐回原式
+
(-a) | u = (1-a) q
→
·y' + p(x) y = x(x) 1st linear ODE of u
-h
u = eth Se" rix dx + ceh
h = Spw) dx = 1 (1-α)p Jx
8(x)
=
(1-9)9
解完後
記得代回 y = u²解y

ページ11:

Ex population model
by (t) = Ay (t) - Dylt)
dt
=
cy (t)
y = c₁ect
When t=0, C₁ = y₁
y(t) population
A growth
:
rate
D death rate
→∞
Malthu's Law t+∞ y\)→
Verhust (logistic) equation:
y' - Ay (1-1/2)
k
Carrying capacity
Ay - Ay²
H
harvest rate
· Ay - By
y' + py = 2y"
PA
a
a = 2.
9 = -B - A
h(t)
=
k
·S (1- ^) Pdt = At
It is it k
B
A
-At
u = b + c e²² = y "

ページ12:

y
B + Ce At
=
B
Yo
A
When t-o, y-g. 16 C-11-1
t=0, =
B
y = 12 + (y
A + (y - B) e
For harvesting
-At
=
y' - Ay - By ² - Hy
=
(A-H)y. By "
y' = 0 for equilibrum y = 0 & ATH
:
t
y.
t
A-H
=k
A

ページ13:

Ex. Epidemics
y'lt) = A∙y (1-y)
y(+) population
· Ay Ay'
A
Constant
y infected people
y' - Ay = -Ay"
y' Ay = -By²
y
B
B
A
+
q
I
Ce At
29
:
(1-y) noninfected people
1+ ce
-At
t70
{
when yo=0 ;y-o
0
when pay styl Yo
Stable
J=0 unstable
0ī
→t
· Yo⋅oy (1000) = 0

ページ14:

2nd order ODE
homogeneous and linear
·Y" + P(x)y' + 9(x)y=0
solution: y₁yz y = C₁y, + Czyz
2 →
Yz
Const.
Euler - Cauchy equations y" + ay' + by =0
set y=e^x, y' = ^e^, y". x²e
(x²+an+b) e²
^
=
Three cases
a
2
t
0346
2
nx
=
:0
=
e²
ß
7x
.)
-19-678
1.
0-4670
real roots.
y = c₁₂ e
+ C₂e
-ax
-xx
1.
a²-46=0
double roots
y = c₁e
+ xe
1=0
complex
-9x
Ⅲ、
9-4600
conjugate y = e" (Acs (wx) + B sin (wx))
Yoots
·B = ±iw

ページ15:

Ex. Mass - Spring system.
equilibrium
position
k. spring constant
t time
m mass
y(+) displacement from y=0.
Step 1: t
Step 2. y(t) = ?
Step 3 ↑ +ΣF = F₂ =
ky
(Fs - by
= ma = my" + my" = my"
Im is constant.
· m² = 0)
"
my" + ky - 0
=

ページ16:

Ex. damped mass-spring system
5
Fd
-mm-
oil
- y = 0
my" + cy" - ky
y'
"
"
t
+
t
k
C: drag coefficient (damping)
Fd: -cy'
=
↑ +ΣF - F₁ + Fd
= -
by + (-cy)
:-ky-cy" = my"
= 0
by + y = 0
m
I
@y+
y (+) = c₁ e
m
+ by =0
nit
m
+ C₂ e
士
12t
2
一情
+
2
m
α ±
2
m √√c² 4 km

ページ17:

1. c²-41cm 70 overdamping]
=
J-oce
(α-13)t
(-0-3)t
=
+ C₂ e
C₁e (αip) ==
(-9-ß)t
-
C₂ e
C
Cz
> |-
70
1
-2ßt
It does not exist, no cross
<0
t exists, cross.
Dy (o) = Y₁ < 0,y10) < 0
yo.
⇒ y (0) = Yoco, y' lo) =
t
@ y (o) = yo<o, y' 10) >

ページ18:

1. C²-4 km=0
critical damping
-at
-ast
y = C₁ ea²² + C₂t ea
=
( C₁ + C₁t) e = mt
C²-4 km <0
y= e
=
gt
[ underdamping]
(Acos wt + B sin wt)
A²+B² e
-at
cos (wot-0)
Y(+)
-at
e√A²+B²
=
2 m
·T
penod
272
W = 2TH = 2πf
=-
論
2 m
α ±
D
± 1 / 1 √4/am-c²
ß
2 m
-

ページ19:

IV. harmonic oscillation
C = 0
·y' + my
= 0
B = iWo
natural angular frequency W..
=
y' + w² yo
Yet)
=
m
To = 1
fo
A²+B²
Wo
=
27V
To = 2πuf₂
natural frequency