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Akuntansi Manajemen - Tugas 2
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TUGAS 2 PERSAMAAN DIFFERENSIAL BIASA
DISUSUN OLEH :
KELOMPOK 3 HABIB MUNZAMIL
: 1514040043
RAHMA LAYLI LAYLI PUTRI
: 1514040049
WILA APRILIA SAFITRI
: 1514040052
SUTAN SUTAN HAIDAR
: 1514040046
PUTRI FARI
: 15140400 !1
EKA PEBRI FITRIANI
: 1514040041
DOSEN PEMBIMBING : E"#$%& A'($)$*&+ A'($)$*&+ S,S&+ M, S-
URUSAN TADRIS MATEMATIKA .B/ FAKULTAS TARBIYAH DAN KEGURUAN UNIERSITAS ISLAM NEGERI .UIN/ IMAM BONOL PADANG 1439 H 201! M
S$ H$$$* 46 .G*$/
2.
( xy + 2 x + y + 2 ) dx + ( x 2+ 2 x ) dy =0 P*7'$&$*:
( x + 1 ) ( y +2 ) dx + ( x 2+ 2 x ) dy =0 x + 1
⇔
⇔
( x 2+ 2 x )
dx +
1
( y +2 )
dy =0
∫ ( xx++21 x ) dx +∫ ( y 1+2 ) dy =∫ 0 2
misal u=( x
2
+ 2 x )
du =2 x + 2 dx
¿ 2 ( x + 1 ) 1 du= ( x + 1 ) d x 2 1 2
∫ u du = 12 ln u +c = 12 ln| x +2 x|+c
2
missal u= y + 2 du =1 dy du =d y ⇔
⇔
∫ ( xx++21 x ) dx +∫ ( y 1+2 ) dy =∫ 0 2
1 2 ln | x + 2 x|+ ln | y + 2|= ln c (dikalikan dengan 2) 2
⇔
ln ( x
2
+ 2 x ) + ln | y + 2|2= ln c
⇔
ln| x
2
+ 2 x|∙| y + 2| = ln c
⇔
e ln | x
+2 x|∙ | y + 2|
2
2
2
= eln c
| x + 2 x|| y + 2| =c
⇔
4.
2
2
Csc y dy + sec x dy=0 P*7'$&$* :
keduaruas dikalikandengan
1
csc y sec x
Sec x dy + csc y dx =0 ¿ sec x dy + csc ydx
(
1
csc y sec x
) ( =0
1
csc y sec x
)
sec x csc y dy + dx =0 csc y sec x csc y sec x 1
1
dy +
dx =0 csc y secx sin y dy + cos x dx =0 sin y dy + cos x dx = 0 cos y − sin x =c
∫
6.
∫
∫
( e v + 1 ) cos u du + ev ( sin u +1 ) dv =0 Penyelesaian :
( e v + 1 ) cos u du + ev ( sin u + 1 ) dv =0 ⇔
⇔
⇔
⇔
1
cos u
e
v
v
( e + 1 ) ( sin u + 1 ) v ( e v + 1 ) cos u e ( sin u + 1 ) du + v dv =0 ( e v + 1 ) ( sin u + 1 ) ( e + 1 ) ( sin u + 1 ) ( sin u + 1 )
du +
v
( ev + 1 )
cos u
dv =0 e
v
∫ ( sin u +1 ) du +∫ ( e + 1) dv =∫ 0 v
misal u=( sin u + 1 ) du =cos u dv du =cos u dv misal w =( e + 1 ) dw = e v dv v dw = e dv du dw + = 0 ⇔ u w ⇔ ln u + ln w = ln c v
∫
∫
+ 1|¿ ln c ln|sin u + 1||e + 1|= ln c | u + ||e + | e =e c |sin u + 1||e v + 1|=c
ln|sin u + 1|+ ln |e v
⇔ ⇔
ln sin
1
v
1
ln
⇔
( x + y ) dx − x dy =0 P*7'$&$* :
v du + ( u 3
( e v +1 ) ( sin u + 1 )
(( e + 1 ) cos u du + e ( sin u + 1 ) dv = 0 )=0
⇔
10.
1
v
∫
8.
(
dikalikan
3
−u v 2 ) dv = 0
P*7'$&$* :
v
)
2
3
du uv − u = 3 dv v
¿
uv
2
3
−
3
v
u v
3
()
u u ¿ − v v
3
⋯ ⋯ ⋯ ⋯ ( 1)
misal u= wv w=
u v
du d = wv dv dv
¿v
dw + w ⋯ ⋯ ⋯ ⋯ ( 2 ) dv
substitusu persamaan (2 ) ke persamaan ( 1) v
dw + w =w −w 3 dv
v
dw = w − w3− w dv
v
dw =−w3 dv
−dw w
3
dv v
= dikalikan (−1 )
dw dv + =0 3 v w
∫ dww +∫ dvv =∫ 0 3
−1 2w
−1 w
2
2
+ 2 ln v =c
−1
() u v
2
−v2 u
2
ln v
+ ln v =c dikalikan 2
2
+ 2 ln v = c
+ ln v2 =c +c =
v
2
u
2
v
2
=u 2 ( ln v 2+ c )
( 2 s 2+ 2 st + t 2 ) ds + ( s 2 + 2 st + t 2 ) dt =0
12.
14.
( √ x + y + √ x − y ) dx + ( √ x − y − √ x + y ) dy =0
16.
8cos y dx + csc xdy =0, y
18.
2
2
( ) π
12
= π 4
( x + 3 y ) dx − 2 xy dy =0, y ( 2 )=6 2
2
Penyelesaian :
( 3 x 2+ 9 xy +5 y 2 ) dx −( 6 x 2 +4 xy ) dy =0, y (2 )=−6
20.
22. Solve each of the following by two methods (see exercise 21 (a: a. ( x + 2 y ) dx + ( 2 x − y ) dy = 0 Penyelesaian : b.
( 3 x − y ) dx −( x + y ) dy =0 Penyelesaian :
24. !. Prove that if m dx +n dy = 0 is a homogeneous euation! then the
"hange of
varia#el x = uy transform this euation into a se$ara#el euation in the varia#el u and x . #. %se the result of (a) to solve the euatoin of e&m$le 2.12 of te&t. ". %se the result of (a) to solve the euatoin of e&m$le 2.1' of te&t 26. (a) use the method of e&er"ise 2 to solve e&er"ise 8 (#) use the method of e&er"ise 2 to solve e&er"ise
