Rangkuman Kuis Y Deret Euler dari e"*

i2= 1

zio itis"
-

bi i FI
=

2 a
=
+

iY
(X,4) 2 x
= +

Re(2) a
= - -

(m(2) b
.......
=

an
X
2 x
=

yi
+ +
(x,y)
2
121 x
=

y2
+

- :tan" (E)
7
I X
arg (2)
-yi(X, -4)
=

·penjumlahan atav selisch

2, 2 (X, Xz) =i(y, yz)

=(1 2 it ...) +i(0 8...
+ = -
+ - +
+

a2, b2z (ax, dxz) = i(ay

+

byz)
= +
+

2
perkallan
gi0=COSO+ i sin O

2, 22:(X,X2-4,Y)+:(X,Y x2Y,(
+

3 Dembagian jadi

2 22:
2Z Notes:Itcerminan cos (0) 1

2 t...
= -

2 +
=

1221 YkonduktF
Notes
Bentuk Kutub/piler dari belangan kompleks
sin (8):0
-1it...
2 121((010+ic1nO)
=
0 121 =
sin (x+B):sinx cos cOSX
+

sin
121 ei0 reiO (X-B):Sin COSB-cOSX sin
sin x
2 = =

z r = cis O Sin (2x):sInx COSx+COSSinx

:2 SInx COSA

(eig:eit (x B):COSACOSB-Sinx sin

cos +

COS (&-B) COSCOSx+sind (in

=

leig."Ocono (22):COscos Sinxsind

-

cos

:cOS - sin
=COS 20 + i s in CO
cos (0):(0) x (OSx + sinx <inx

1:COSX (
+ in x
2 x =
i7
+ +
z r =
(IS0
COS =1 -
Sin2 x
z" (r ci) p) =

"
2 xs(n0)
=

5
COS (2x):COS &-sin &
27 sin
Akar- akar dari suatu
persamaan bilangan komplexs =71-sin2 -

2 c, CE Riel, komplexs -1-2 Sin X

JIPG
=

~PC (90)
p(1s(5) = 3 r
P
= =

n0 5 =
Ir pn =

n =0 2k =

4 1
=0 =

Ok
I=
=

akar-akar:2k p5
=
(s(8x) x 0,1,2,
=

.
.
.,
(n-1)

=r cis (OK)
 Materi Orde Dua Persamaan Diferencial
y f(x)
=

Acosta
Y:sin (f(x)) ->

re
-* e
PDorde dua homogen Y:2
f(x) -

(x) +
X). i
d
+

encas.
:a
y a
=
->

ay" by
+
+
cy 0
=

a, b, c konstant a

karakteristik ax +bx c 0
persamaan
+ =

y en(f(x))
:is I
= -

2
D b = -
4aC

11K9D 0 akarnya Sama

=

Materi Orde Satu

x12 =
-
b b2
=
-
4aC
PD biasa orde satu tax
homogen
24
y +f(x)y r(x) =

x12:
-t
x1 x) x
1 solusi
= =
x =

:e-"[Serax c] +

solusi umum
** n ( f(x)
=
dx
y (c +(X) e
=

C,dan 22 Konstanta

2D b2
=

4a) >0 berbeda

akarnya
-

x,2 - b1b2 = 4aC

x:
ac x2=-b-br-4ac

20

solusi umum

+cre*
* * **
y c,e
=

3 Sika DC0 akarnya kompleks

Nic
b1
-
=

24

Xi2 1 =

i
solusi
**
y e
=

(C,cOS(B) + i c sin())
4 4(x,y)
=
solusi

dari u=4(X,4) (mdx k(y)

diferential total adalah u
=
+

=((OS(x y) k(y)
1Ydx+Ye
du:
+ +

=(in(x y) +
k(y)
+

du 0 =>u ?
=

Jika
=

Sdu S0 dx
&4 cos(x y)
dex)
=

N
+ +
=
=

u =
C dy

u(x,y) G
=

c0S(x y)

d4)
3y2 2y
+

U 0 c0)(x+)
JIka
+
=
+ +

orde satu homogene

dudx+du
maka 0 =PDB
dy
=

342
d14):
+
y

m(x,y)dx + N(x,y) dy 0
=
(dK(y):( (34 2
24) dy
+

3)y'dy
44 M(xy
d4 N(X k(Y) Sydy
=
=
c
+

k(Y)
3.1y3 c.1y2
=

c
+
+

:
N:lae =y3 yz +
c
+

y u (
=
N dy 1(x) +

:)(34
2
2y
+ + c0S (x + y) dy

=y3 y2 f(x)
=A
(x + y)

re
+
sin
+
+

a= (OS(x y) + +

dx
suatu PDB m(x,y)dx+N(x,y) dy 0
=
adalah

PD
m:
eksal
jika
cos(x+Y)

dicose
+
in

Solusi PDB eksak

u (mdx
=
k
+
(y) Atay U:/ Ndy 12x)
+

e(x) c
=

Soal
-

1. (OS(x y)dx+ (3y2 cy

+
+ + cos (x+y)) dy 0
=

Jadi,
2. xy' y 4 0
+ +
=

u(x,y) y3 y2
=
+ xn(x y)
+
+ +
c

3. 2xsin (34) dx + 3x c0s (34) dy 0

=
solusi

4. Ydx +exydy 0
=

U(x,y) 3in(x y)
=
+
y3 y2
+ +
c
+

apakah PDB datas eksak dan tentukan solusing a

3 2xsin (34) dx + 3x c0s (34) dy 0
=

Jawab m(x,y) 2
=
3 in
x (34)
-

=(OS(x y)dx+ (3y2 cy +

+ +cos (x+y)) dy 0
=
N (x,y) =
3x COS (34)

d3
m(x,y)dx N(x,y)dy:0 d (2xsin ( dN_
AM:
+

m (x,y) COS
= (x y) +

N(x,y):(342 cy + + cos (x +y)) dy d dx

M:d(COS(X+y)) :2x COS 6x (OS (34)

dNhereyou
(34). 3
=

dy dy =
6x COS (34)

-Sin(X+y)

Am AN= PDB
=
eKa

=0 0+
+
(-sin (x+ y)) -
=-sin(x+Y)

am:dN
Jadi PDB
eke
Maka e
solusi Materi
u (mdx
=

k(y)
+

y
=

f(x)
:Sax

-*
sin (34) dx + (y)
=x * 3in (24) K(Y) +

PDorde homogen
e
dua

=
Xu1012
atani de dy
ay"
a, b, c

persamaan
D b =
by
+

2 -
4aC
+
cy
konstant a

karakteristik
0
=

ax +bx +
c 0
=

K(Y) C
=

11K9D 0 akarnya Sama

=

solusiu x
=
sin (34) + C x12 =
-
b b2
=
-
4aC

24

Cara lain x12:

-t
x1 x) x
1
= =
x =

2xsin (34) dx + 3x c0s (34) dy 0

=

2xsinysdY= ocame e
solusi umum
**
y (c +(X) e
=

C,dan 22 Konstanta

2) Idx= 3)
dy
-

2D b2
=

4a) >0 berbeda

akarnya
-

misal u= sin(y) x,2 - b1b2 = 4aC

a=3CO) (34)
du=3 COS (34) dy
x:
ac x2=-b-br-4ac

20

solusi umum
maka,
+cre*
* * **

oax:theene
y c,e
=

3 Sika DC0 akarnya kompleks

Nic
b1
=
=

zenix=-)I
24
du
Xi2 1 =

i
2 n1X1: -2n141 d
+
solusi
**

en x
=
- In 131 (34) 1 d
+
Y = e (C,cOS(B) +
icsin(p))
enx2 en +
1 sin (34)1 d
=

en Contoh
12 se -
Tentukan solusi umum dan PDB orde homogen

(34):e*
2
x S in 1). Y" +34' 2y + 0
=

x
:
Sin (34):C* 27.Y" y'
+
4y
+
0
=

u x2 in
=

(34) +
e 0
=
3). 2Y" 3y' 4y + + 0
=
Jawab
-

1Y" 34' +
2y
+
0
=

2
x 3 x
+
2X
+
0
=

x1 -
=
3 32.4.1.2
+
X2 =
-
3-32.4.1.2
2. 1 2. 1

-- I -
-
2

Solusi umum
x - ax

y C,e
=
(ze
+

- X 2X
=- (ie
-

y - 2(ze
+4(ze- 2*
*
*
=Ce
y

maka,
*

34' cie +4 +3)-ce 2)

-
"
y +

2y
+ = - ze +

2x)
* -

2(C,e (e
+

3e *-6e *+ie 2
*
=Cie 4(2-
+ +
 Soal Cek
- -
"
1.y +y
=0
y 2. =

20s(Ex) 2(n (Ex) +

i

c, sin (Ex)+ i. (2 cOS (Ex)

"
=
6.44 254
+
0;y(0)
= =
5 4 (0) = -
2,5 y =

3. Y"+0,641 0.094:0 +

y
"
=

-()" ccos(Ex) -

iE)" sin (Ex)

Y10) 2 2 :x'(0) =

0,14;
=
e013x;xe-0,3x
4. Y"+24' 2Y 0
+
= ;

Y(0) 0;
=
x'(0):15, e "cosx, e *sinX 4Y" +
25Y

5.Y" y' +
3,44
+
0
=

(-(2)"cos(Ex)
=
iE)" sin (Ex)) -

6.94" -30y' +
(54 0
=
+
c5 ((2)" cos(Ex) -i()" sin (Ex))
7.84" 24' - -
y 0 = 4(0) 0.2;
;

= y'(0) =
-0.325 =O

Jawab Solusi umum a

-
E* +icce Ex
-

1.4
"
Y'
+
0
= y Cie =

x +1 0
=

(is(x) i(a(s) Ex)

:
+
-

x(x 1) 0
(Ex) isin (Ex)] ( [cos(- +isin) [Bx)]
=[CO)
=
+
-
+ +

x, x, 1 di cos(EX) da i sin
(-2X)
=

0
= - : +

x, Ix2

Tentupan solusi khusus

-

Y(0) 3 -> X 0, y 3
=
=

Solusi
=

umum

Flie** ce"* + + (0) = -2,5 + X 0, =

y =
- 2.5

:.e* (z 2*
-

y C1
= cce
+
*
y C,cos
=

(Ex) + i (2 sin (x)

x 0 = + Y C,COS10) +i C2
= (in 10)

Cek y C,
= y 3 =

-
*
x C,
=

(22
+ C1 3 =

C,sin (Ex)+ i. (2 (Ex)

x
= **= cOS
-

- c,e =

* *
y
-

=
c, e

x 0
= +
y
=
-
I.(,xn (0) is COS (0) +

xex)
x
xy ce ( = i.5.C2 =
y
0 y Y
=

2,5
= -
+
-

i. 2,5(a 2,5
2.4 Y"+254 0
=
=
-

4x2 +
0x +
25 0
= i.22= -
1

2 22: i
4 x
-

25
+
0
=

2
4 x =
- 25
2

knuses, Ex)+
x
= i (- il sin (i
x y 3cos
=

sin (Ex)
1.
-
=

f1.
=

d
i.
=

1iI
X =

x12 2 1
B x 0
=5
=
= =

Solusi umum

Fex(C, cos x+ icsinx)

=e (CICOS[Ex)+i2 sin (Ex))
=C,cos (Ex) + i C Sin
(Ex 1)
PDB Orde dua tidak homogen
2
dy" by' + +
2y f(x)
= x 3x+2
+

0
=

11. d2Xc+...
f(x) do dr X= +
(x 2)(X 1) 0
=
+ + +
=

solusi pastikulir
Yp:fo+f,x fax+. . . fax"
+
+ x2 +
3x 2
+
0
=

Yp' f1
= +
2f, +...+
k.f.XP- (x 2)(x 1)
+
+
0
=

- 1

Yp"=2fat... 2k 1) fx.x
-

X
+

=
- 2 x = - 1
- 2 *

Yo Ce
:
(z e
+

disubstitusikan kex

aYp" bYp'+cYp
*"
do + diX d2 x dx
=
+ +
.
+
..
+

didapat fo, f., fa, ....f

Solusi Umum
y Yo =
+
Yp
to adalah solusi ay"+by'+ cy: 0

2) f(x):
AedX

(1kaax2 bx c 0
+ =
+

i) DI 0

x, dan x2

Ed x2 I d
Sikad, dan

mak a

yp e4x
=

Yp' Bded*
=

Yp"=Bilded*

aYp" byp' eyp +

+

A
=
ed*
didapatnilaiB

Solusi umum

Y Yo =
+
YP
*
to C,e** Ce
: +

ii) O
DF

X, d
=

muka,
**

YP: (B cx) + e

Yp' = Bde** +

Y" =

aYp" bYp'+cYp Aed* +

=

didapatB dan c

d*
iii) D 0
dyp" byp'+ c4p:Ae
=
+

xi x2 d didapatB, C, dan D
= =

Yp (B
= (x
+
+
DX2)c4* solusi umum

yp'= Y Yo= +
Y4

Yp" =