LATIHAN SOAL
PERSAMAAN DIFERENSIAL YANG DAPAT
DIPISAH
1. (1 + 2x2)y. y’ = 2x (1 + y2)
2. x2y dx + (x +1) dy
3. y’ + (y+1) cos x = 0
4. sin x . cos y dx + tan y . cos x dy =
0
5.
6. (x2 + 4) = (y + 2) (x + )
7. =
8. 2xy (4 – y2) dx + ( y – 1)
(x2 + 2) dy = 0
Jawab:
1. (1
+ 2x2)y. y’ = 2x (1 + y2)
Penyelesaian:
(1 + 2x2)y. y’ = 2x (1 +
y2)
(1 + 2x2)y. = 2x (1 + y2)
(1 + 2x2)y. dy = 2x (1 +
y2) dx
(1 + 2x2)y. dy - 2x (1 +
y2) dx = 0
Mis: mis :
u = 1 + y2 u = 1 + 2x2
du = 2y dy du = 4x dx
mis : 2C1= ln C
=
=
= C
= C .
2. x2y dx + (x +1) dy = 0
Penyelesaian :
x2y dx + (x +1) dy = 0
mis : 2C1
= ln C
3. y’ + (y+1) cos x = 0
Penyelesaian:
y’ + (y+1) cos x = 0
mis:
C1 = ln C
4. sin x . cos y dx + tan y . cos x dy =
0
Penyelesaian:
sin x . cos y dx + tan y . cos x dy = 0
Mis: Mis:
u
= cos x u = cos y
. du
du =
= ln cos x
= sec y
mis: C1= ln C
mis: C1= ln C
5.
Penyelesaian:
Mis: mis:
mis: 2C1=ln
C
6. (x2 + 4) = (y + 2) (x + )
Penyelesaian:
(x2 + 4) = (y + 2) (x + )
Misal u= u=
du= du=
Misal C1=ln
C
7.
Penyelesaian:
Untuk:
untuk: Misal: u =2
=C
8. 2xy (4 – y2) dx + ( y – 1)
(x2 + 2) dy = 0
Penyelesaian:
2xy (4 – y2) dx + ( y – 1) (x2 + 2) dy = 0
mis:
4A
= -1...............................................................(1)
A = -
2B
+ 2C = 1 ........................................................(2)
-(A + B – C) = 0
...................................................(3)
Subsitusikan A = - ke persamaan (3)
-(A + B – C) = 0 -
A – B + C = 0
- B + C = A
- B + C = - .................(4)
Dari persamaan (2) dan (4)
2B + 2C = 1 x 1 2B + 2C = 1
- B + C = - x 2 -2B + 2C = -
-
4B =
-
4B =
B =
Subsitusikan B = ke persamaan (2)
2B + 2C = 1
2 + 2C = 1
2C = 1 -
2B + 2C = 1
2 + 2C = 1
2C = 1 -
C =
Dengan mensubsitusikan nilai A, B dan C. Maka :
Jadi,
mis: