Narendra
Ara Dui
23/522807 /TK/57752
Example
Tugas
dengan
a = s,
bil ,
cu
=
-
c =
↓ (23
o
2
,
+ = 0
-
10(2)
2,
-
X
2
subs
(general)
/implicit Solution)
c,
=
3
Example H
r2 + (R
r2
-
v
=
y)
Debit
#r
d
t
=
-
Q
:
0
=
.
A
=
-
=
=
32
=
-1
=
20
-
3
solution
general
he
-101 + 11
Steth (particular
=
=
2Ry
air
,
-
yu
menggunakan
rumus
geran jatuh
bebus
AN
Azz
Egy
=
(A1) permlan
,
310)
R2
=
helvar
Eigh
=
C,
1 -10C
R2 - (R-y)2
=
hecepatan
integrasi
=
3
=
batas
CO %-
d = o,
make
(yc2 10)dc (5 + )d +
Suc-10dc S5-+ d +
+
St- 1 +
#c 10
-
,
↓
mala
mencari
3
2
=
t
R-Y
week
.
d
=>
princip
hontinuitys
=
(Av) lubang
t*d2 Vege
-d
)
jy-2Ry dy d
y
=
R
,
+
=
T
=
>
y
=
Sy-2Ryt dy )
dig a
=
0
=
= y - 1Ry)
d
=
T=
(1)
R
=
0 , 04
d 3
=
maka
T
.
16-3
walto
=
sy
m
g
=
9,8
m/s
m
sampai
eairan
habis
:
a
: you
e
I
(2) diameter lubang dibrat
zhali
,
berupa
wahtunya ?
T-
=
mala
Tz =
To
=
10 , 8
Tz
=
2, 7
selon
=
Example I
,
v
=,
he
subs
v (t) =
say Stat
general solution
*
(particular)
Vo e,
=
In (v1 =
-
At
=
,
uxd x
S(D-2)= dD
SuxdX
(D-2) D
D
-
2
=
=
2 +
=
--
=
>
vI + )
=
= v(t) =
se-
7
.
%
2
-
D(D =
(1) (D-2)5 dD =
=
2x2 +
=
V(d)
= c (general)
=
2)
3
.
,
Example 3
uX(D-2
3 (D-
=
↑
+ 4
et e"
VH) =
Jiha v(0)
X 10
(z x)3
8
(= (2)3
+
1 ,
85
2
*
+
2
+
+
=
=
(2 x)3
+
2
=
=
= +
22
2 +C
=
(
=
( (2) (generall
=
=
D(x)
=
2 +
(3x
+
43
(particular)
+
(2) Singular solution
ux(-2)
=
0
=
yx(D
D = 2
-
2)
(singular solution)
D(X)
menuniuthan
Tapi
Letila
D
=
bagaimana laju transmisi
2
,
Sistem
panet
seiring perubahan
mencapai headaan stabil
hemaceton.
(laju horstun) dan tidall dipengarhi
x
(hemacetan)
Ara Dwi Narendra
23/522807/TK/57752
+
Tugas
+
=
..
misal
v
=
=
C = v+
)
I
X
↳
(2)
+ du
+ v
It
=
(1)
he
Fr
+
v
=
:
N
E
sin" (v)
v
=
C
·
((to)
maka ,
=
Sin((n
+ +
k)
=
t
+
,
0
,
o =
to
Sin
(In to + k)
=
In to + k
0
((t)
Sin
-
sin
=
(In
Sin
+
=
General
to > 0
k
jadi
k)
+
=
=
-
+T
Subs
(2)
Yu-S
(2)
he
to +
k)
0
In to
(In + -Into
-
=
(In) Particular
(1)
T
v
=
h
:
2x
=
-
Y +
=
-
2x
=
2
p(x)
Cari
integrating Factor
=
Sdx
-
hedua
ruas
=
In X
*
=
e
=
I-
(In ++ k)
Sin
v =
Kalikan
k
=
·
r(x)
+
E Sin((n
.....
·= =
·
In +
=
=
misal
-
Subs
Week 4
x
=
rex)
da
-
& [ - v]
=
-E
Sv] S-Edx
=
*
-
4
v
make T
-
=
( - Ex + (x)
General solution
O
y= 2
UX
dyney
-16y
-
:
I
4Y
d -ne
I
=
:
-16y
y
·
=
Cosh uX
=
u sin
dy-16y
y
·
15-Ibe ~
:
=
sink
=
ux
=o
16 Coshux-16 losh x
=
·
tank
~
ux
A Ucosh uxy
dy-1y
= cost
=
=
sin a
1
Sinh ux-16 sinhux
16
=o
he
uX
de =4 sechu
↓y
=
-32
dy-16 y
=
I
tank
-32
ux
sechu
tanhux sechux
-
I thex on
X
y
·
eY
=
= ex
-
-
y=2
·
~
Heh-16e
=
a)
- 29-3y
"
+
y
·
=
due
= 22x
&
2 dy-by
+
2
+ze-ze =
enX
=
-
d
e
Ye + (e)-ze
X
=
3X
-
·
Y
= ze
dy
-
3x
de
dy
be
a 2dy-by
+
-
·
y
ge + 2
=
-
Y(x)
2) y(0)
=
or
(xe*
=
=
+
*
C2
-
C,
(2
2
=
-
-
3(z = 8
3(z
=
y(z
=
(z
=
-
=
=
Y(x)
-
( 2)
=
-
ze"
X
(2e 3X/
, (0)
0
C + (2 = 0
-
-
Solution
↓
C
3
=He
-ye d
=
6) General
=
be
4x
d+ -3 =+
C1
ge
e
=
↳
=
2/
=
-
-
ze
=
34
f
2
-
2
-
Ara Dui
Narendra
23/52280 +/Tk/57752
Tugas
Week 5
# Persamaan
ranghaian RLC Calam AC
Persumaan
:
dengan
L Ra +
+
·
:
·
L = Induktansi (H)
resistanci (ohm)
R
·
=
C
=
Kapasitansi (Furad)
# Kasus
9) misal
H
R
,
220 1
=
C =
,
0, 1
menjadi 220t)
Persumaan
asumsi
1
1 =
+
9(t)
berbentra
solusi
make characteristic
equation
r2 + 220r + 10
=
0
+
+
F
10
·
9 (0)
e
:
-
219 , 9
in 20
=
-
act)
=
v
q(t)
9 (d)
=
C , e"
2, e
value
Initial
=
-219
,
Distinct
Real
Roots
9
Solution :
=
+
+
+
(e
0 , 03t
problem
r2t
-
t
(2e
219 , 95t
F
:
1
9(0) = ( , e + (2e
,
rt
=
-
General
1 C
gl-
·
:
=
=
1
+
(
z =
...
17
=
been
a
10)
=
0
Subs
-
0 , 05(1
-
(2)
he
11)
0
(2)
-
.
-
219
+ 0
, 03(2
05
93(2
,
-
219
(2
Specific
Solution
6)
Misal
=
R
Persuman
=
:
0
95/2
,
C
0
=
=
1
0 , 000227 =
0
, 999 + +3
↑
:
0 .03 +
-
100Ko
L
,
=
219 , 95 +
F
0 , 00022te
+
0
, 999773e
10 H
,
C =
4 UF
,
10 1000 +
+
10
-
0 , 000227
=
-
act)
=
d 100000dat
+
+
23
.
9(0)
=
=
12
das o
0
100 a-
characteristic equation :
108
,5
lor" + 100 000 r + 2
.
=
0
General
)ert
=
(c
,
+
9(t) =
((
,
+ (2 +) e
value
problem
a(t)
Initial
9(0)
solution :
=
(2 +
-
3000 +
:
1
-500eo
(2
-
5000
(1)
=
0
(2 = 5000
Unique
Solution :
9 (H) =
(1
+ 3000 +
12
-
5000 +
=
repeated
real
root,
2) misal
R
=
100k
Persamaan menjadi
H +
1
L = 100k
H
,
,
2=
2
/F
,
9(0)
=
0
,
d
:
look ok +To
Eat
Characteristic
r2 + r
d
+
equation
+2
=
+
zac
:
0
· =
General
Solution
C, e
q(t)
=
q(t)
=
*
Ge
Initial
9(0)
value
cos(Et) + (ze
*
sin
(E + )
-
:
o
=
4+ 0
q(+ )
:
(os(B +) + (zett sin (B + )
**
=
0
-
40
+ 12
=
Ge
sin
(E + )
= Ecos
d
=
-
E
Unique
9(t)
Solution :
:
*
e
sin(E)
-
in
=
2
Narendra
Ara Dui
23/52280 + /TK/5 + 752
Tugas
# Example
+
dy-u
=>
Week 6
Se-bos2x
by :
y(
,
=
2
d)
,
Yc(x)
dy-Ydzy-
=
(r2
*
Asumsi Solusi
:
y = e
y
y"
re
=
=
+y
)
3)erx
=
0
H
r2 yr + 3
r2 erX
=
yr +
-
0
=
(r 1) (r 3)
↳ 1 Vr 3
=
-
-
make
0
,
y((x)
=
=
yc(X) = C , e
" "
Ge
=
+ (2
eix
1eX
+
=> Yp(x)
Yp(X)
=
Yp'(x)
=
Yp"(x)
=
At #Blos2X + C
-
Sin 2x
#e"- zB Sin 2x + 2C los 2x
*
Ae - YBcos2X- 42
Sin 2x
Subs :
*
Ae
=
=
YB los 2x
-
UC
-
(A + YA + 3A)e
(8A)e
8A
"
(
+
-
5
=
A=3
2x-11-Ae-2B sin
Sin
x
B
( uB
+
-
-
B
-
B
-
-
-
-
b)
=
E
-
-
8
-
C
c
65/
5e* +
=>
4
+ 8B +
2x
=
3()Sin 2x
3
"
-
8 (os2x
0
=
8B = C
B= ↳
make , Yp(x) =
-
C + &B) sin
8B
8
b(8B) =
los2x) + 3 (Ae"+ Blos2x + (sin 2x)
81 + 3B)(os2x + (
81)(os2x + (
-
2c
2x +
=
0(8)
+
losx
sin
=
2x
Y(x)
Y(X)
Y(X)
Yc(x)
=
(,
=
Y(0) = 2
X
e
=
Yp(X)
+
+
( , e + (20
( + (2 +
Y'(0)
=
3
=
Fe +
2e3 +
3 f
+
( , e + 3(ze"
2 + 3(z
( + 3(z
ze
+
-
3
=
+
3 +
=
+
1
=
3
3 1
-
+
bos(0)
2()(1 + (2
Ge
-
cos2x
*
-
+
=
by Sin 10)
(General
sin2x
b sino
3 t
2
+
-
2
-
1
s
10)
=
3
Solution a
(1 + (2
=
2
-
3
t
-
# = 7)
(3 +
-
maka
,
2
Y(X) =
-
2
41
+
2
=
-
3 5
-
Zo
=
Le + e + +Cox +
Sin2x
(unique Solution e
# Exercise
-6
Y(x)
=>
Yc(x)
=
+
gy
+
yp(x)
ne
=
+
3cx
+
g)erx
=
x
yo
,
misal : y
+
gy-
rX
=
y
=
y"
=
(v2
e
re
+ Y
-
=>
Gr
+
0
↓
r2erX
r2 br
+g
=
0
(r -3) (r 3)
=
-
Maka ,
0
y((x)
Yp(X)
y
↓
He
-
6
+
gy
ne
=
3cx
+
2
-
-
-
exponential
sinusoidal
polinomial
Y
Yp(x)
yp' (x)
Yp"(X)
=
=
=
Ae
Y
Subs :
Ae"-6Ae" + gAe"
Aex
A
-
6A + gA
A
Ae
maha
,
Yp (x)
,
=
C,e
*"
+
y , (x) = ( , e3
r = 3
·
-
Y((x)
dy-by
=
:
=
eY
=
4
=
1
=
Yex
(2Xe
+
rX
3X
(2xe
·
X2
Yp(x)
By + (x
=
Yp'(x)
=
Yp"(x)
subs :
+ D
2B
2Bx + C
=
G /2BX + c)
-
2B
-
12BX
-
-
=
B
·
S
,
Ypz(x)
+ gBx2 + g(x
6
gBx2 + (94 12B)X + 2B
2B
gB
make
g(BX + (x + D)
+
=
9) c
1
+
g
+x2 + +
X
=
C-
-
-
+ gD
6 + 9)
12B = 0
9) =
12 B
c
&B
=
= (g)
c
=
c
Y
CS2
=
x2
=
x2
2B
2
B
-
-
x2
=
61
+ gD
=
0
E(z)B gD
+
6B
=
9D
(+3)
=
9D
R
D
=
=
0
27D
=
Yp =?
Isulit
Karena tidah bisa
tetapi
Karena
Y(x) ahan
=>
diselesailan dalum
bentrh
Yp(x) besifat saling independent
valid
tetap
,
mana
sinusoidal)
dengan mengabaihan bagian (5 CSCX)
.
Y(x)
Y(x)
=
Y(0)
=
( , e3x +
(2xe + e + +x
+ +
x
(General
Solution)
2
(0) = (1 +
1
+
= 3e
z
+
=
(
(= )
2
=
-
3(2xe + e
+
+
20
=
34
+
3(z)
+
( +
(2
1
=
+u
=
3
25
=
-
27/
Y (x)
=
z3cxe
+
e
+
x+
(unique Solution
,