Download - kalkulus 3
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1) (x2 + y2 + 2x) dx + xy dy = 0 ……………………………(a)
x2 dx + y2 dx + 2x dx + xy dy = 0
M = x2 + y2 +2x
= 2y
N = xy = y
=
( ) =
dx
∫ = ∫
ln F(x) = ln x + C
ln F(x) – ln x = C
ln
= C
= eC = C
F(x) = C x ; jika C=1 F(x) = x …………………… (b)
Persamaan (a) dikali persamaan (b)
x [(x2 + y2 + 2x) dx + xy dy ] = 0
(x3 + xy2 + 2 x2 ) dx + x2y dy = 0
M = x3 + xy2 + 2 x2 = 2xy
N = x2
y = 2xy
=
( -
)
Eksak
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Solusi Eksak :
M = x3 + xy2 + 2x2 =
δu = (x3 + xy2 + 2x2)
∫ = ∫ dx + y2 ∫ dx + 2∫ dx
u = x4 + x2y2 + x3 + C (y) ..(c)
N = =
= 0 + 2 x2y + 0 + C’ (y)
= x2y + C’ (y)
N = x2y x2y + C’ (y) = x2y
C’ (y) = x2y - x2y
C’ (y) = 0
∫ = ∫
C (y) = C ……………………… (d)
Masukkan persamaan (d) ke persamaan (c) didapatkan Solusi Umum
u = x4 +
x2y2 +
x3 + C
Solusi Khusus:
u = x4 +
x2y2 +
x3 + C
jika x = 6 ; y = 9 ; u = 0
0 = (6)4 +
(6)2(9)2 +
(6)3 + C
0 = 324 + 1458 +144 + C
0 = 1926 + C
C = -1926
u = x4 +
x2y2 +
x3 – 1926
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2). x2 dx – 2xy dy + y2 dx = 0
(x2 + y2) dx – 2xy dy = 0 …………………………… (a)
M = x2 + y2
= 2y
N = -2xy = -2y
=
( ) =
dx
∫ = ∫
dx
∫ = ∫
dx
ln F(x) = -2 ln x + C
ln F(x) = -ln x2 + C
ln F(x) + ln x2 = C
ln F(x). x2 = C
F(x). x2 = eC = C F(x) =
; jika C = 1 F(x) =
…… (b)
Kalikan persamaan (a) dan (b) :
(x2 + y2) dx -
(2xy) dy = 0
dx +
dx - dy = 0
1
dx +
dx - dy = 0
=
( -
)
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M = 1 +
=
N = -
=
Solusi Eksak :
M =
= δu = (1 +
) dx
∫ = ∫ + y2 ∫ dx
u = x -
+ C (y) …………………. (c)
N = =
= 0 - + C’ (y)
= - + C’ (y)
N = -
-
+ C’ (y) = -
C’ (y) = +
C’ (y) = 0
∫ = ∫
C (y) = C ………………………. (d)
Masukkan persamaan (d) ke persamaan (c) didapatkan solusi umum
u = x -
+ C
Eksak
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Solusi Khusus :
u = x -
+ C
jika x = 6 ; y = 9 ; u = 0
0 = 6
+ C
0 =
+ C
0 = - + C
C =
C = 7,5
u = x -
+ 7,5
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3). 2x2 dy = 2x dx – 6 dx
2x2 dy - 2x dx + 6 dx = 0 …………………………….. (a)
M = -2x + 6
= 0
N = 2x2 = 4x
=
( 0 – 4x ) =
dx
∫ = ∫
ln F(x) = -2 ln x + C
ln F(x) = -ln x2 + C
ln F(x) + ln x
2
= C
ln F(x). x2 = C
F(x). x2 = eC = C F(x) =
jika C = 1 F(x) =
……………. (b)
Persamaan (a) dikali persamaan (b) :
(2x2 dy – 2x dx + 6 dx) = 0
dy dx +
dx = 0 2 dy
dx +
dx = 0
=
( -
)
Tidak Eksak
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M = +
= 0
N = = 0
Solusi Eksak :
M = = -
δu = dx +
dx
∫ = - ∫
dx + 6 ∫ dx
U = -2 ln x + (-6) + c(y)
U = -2 ln x - + c(y)
N = =
= 0 - 0 + c(y)
N = 2 → 2+ cꞌ(y) = 2
∫ = ∫
C(y) = 2y + C
c(y) = 2y + C
didapatkan Solusi Umum yaitu : U = -2 ln x - + 2y + C
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Solusi Khusus :
X = 6 dan y = 9
U = -2 ln x -
+ 2y + C
0 = -2 ln 6 - + 2(9) + C
0 = 3,6 – 1 + 18 + C
C = - 12,6
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2) x2 dx – 2 xy dy + y2 dx = 0
(x2+y2) dx – 2xy dy = 0 ………………. (a)
M =
→ x2+y2 = 2y
N = → -2xy = -2y
=
(2y + 2y) dx
=
dx
∫
=
∫
dx
Ln f(x) = -2 ln x + c
Ln f(x) = -ln x2 + c
Ln F(x) + ln x2 = c
Ln (f(x) . x2) = c
F(x) . x2 = e2
F(x) =
Misalkan C = 1
F(x) =
…………… (b)
Kalikan persamaan (b) ke persamaan (a)
( x
2
+ y2
) dx -
( 2xy) dy = 0
+
dx - dy = 0
dx - dy = 0
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Solusi eksak :
M =
→ =
N = → =
Turunkan
menjadi -2yx-1 = 2yx-2
M =
∫ = ∫
∫
= y2
∫
= y2.
=
U =
+ c (y) …………… (c)
N =
= (
) + c (y) diturunkan
=
+ Cꞌ(y)
∫ = ∫
C = c (y) ………….(d)
Dari persamaan (c) dan (d) didapatkan sousi umum (SU)
U =
= c (y)
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U =
+ c (su)
Jika x = 1 dan y = 9
0 =
+ C
=
+ C
0 = 81 + C
C = - 81
Jadi didapatkan Solusi Khusus:
U = – 81