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Outline
Persamaan Diferensial I
Drs. Dafik, M.Sc, Ph.D.
Universitas Jember
Fakultas Keguruan dan Ilmu PendidikanUniversitas Jember
PS Pend. Matematika
Dafik Persamaan Diferensial I
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Outline
Outline
1 PDB Linier Order SatuPDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
2 APLIKASI PDB ORDER SATU
Dafik Persamaan Diferensial I
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Outline
Outline
1 PDB Linier Order SatuPDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
2 APLIKASI PDB ORDER SATU
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
Outline
1 PDB Linier Order SatuPDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
2 APLIKASI PDB ORDER SATU
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
Bentuk Umum
Persamaan Umum
dydx
= f (x , y)
dy = f (x , y)dx
M(x , y)dx + N(x , y)dy = 0
Jumlah Diferensial
dF (x , y) =∂F (x , y)
∂xdx +
∂F (x , y)
∂ydy
M(x , y) =∂F (x , y)
∂x; N(x , y) =
∂F (x , y)
∂y
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
Bentuk Umum
Persamaan Umum
dydx
= f (x , y)
dy = f (x , y)dx
M(x , y)dx + N(x , y)dy = 0
Jumlah Diferensial
dF (x , y) =∂F (x , y)
∂xdx +
∂F (x , y)
∂ydy
M(x , y) =∂F (x , y)
∂x; N(x , y) =
∂F (x , y)
∂y
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
PDB Eksak
Sifat
Suatu PDB disebut PDB Eksak bila memenuhi sifat berikut:
∂M(x , y)
∂y=
∂N(x , y)
∂x
Contoh Soal
Apakah PDB berikut Eksak:
(3x2 + 4xy)dx + (2x2 + 2y)dy = 0
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
PDB Eksak
Sifat
Suatu PDB disebut PDB Eksak bila memenuhi sifat berikut:
∂M(x , y)
∂y=
∂N(x , y)
∂x
Contoh Soal
Apakah PDB berikut Eksak:
(3x2 + 4xy)dx + (2x2 + 2y)dy = 0
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
Pembahasan
Jawab
(3x2 + 4xy)dx + (2x2 + 2y)dy = 0
M(x , y)dx + N(x , y)dy = 0∂M(x , y)
∂y= 4x
∂N(x , y)
∂x= 4x
Sehingga PDB ini adalah Eksak. Kerjakan tutorial 2, nomor 1.
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
Solusi PDB Eksak
Solusi Umum
(3x2 + 4xy)dx + (2x2 + 2y)dy = 0 (1)
dF (x , y) =∂F (x , y)
∂xdx +
∂F (x , y)
∂ydy (2)
∂F (x , y)
∂x= (3x2 + 4xy) (3)
∂F (x , y) = (3x2 + 4xy)∂x (4)∂F (x , y)
∂y= (2x2 + 2y) (5)
∂F (x , y) = (2x2 + 2y)∂y (6)
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
Solusi PDB Eksak
Integralkan (6) thd x kemudian turunkan thd y
F (x , y) =
∫M(x , y)∂x + φ(y) =
∫(3x2 + 4xy)∂x + φ(y) (7)
∂F (x , y)
∂y= 2x2 +
dφ(y)
dy(8)
Kombinasikan (7) dan (10)
(2x2 + 2y) = 2x2 +dφ(y)
dy(9)
dφ(y)
dy= 2y (10)
Zdφ(y) =
Z2ydy −→ φ(y) = y2 + c0 (11)
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
Solusi PDB Eksak
Integralkan (6) thd x kemudian turunkan thd y
F (x , y) =
∫M(x , y)∂x + φ(y) =
∫(3x2 + 4xy)∂x + φ(y) (7)
∂F (x , y)
∂y= 2x2 +
dφ(y)
dy(8)
Kombinasikan (7) dan (10)
(2x2 + 2y) = 2x2 +dφ(y)
dy(9)
dφ(y)
dy= 2y (10)
Zdφ(y) =
Z2ydy −→ φ(y) = y2 + c0 (11)
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
Solusi PDB Eksak
Substitusikan (13) ke (9)
F (x , y) =
∫M(x , y)∂x + φ(y) =
∫(3x2 + 4xy)∂x + φ(y) (12)
F (x , y) =
∫(3x2 + 4xy)∂x + y2 + c0 (13)
F (x , y) = x3 + 2x2y + y2 + c0 (14)
Sehingga solusi umumnya dapat ditulis sbb:x3 + 2x2y + y2 = c.
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
Outline
1 PDB Linier Order SatuPDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
2 APLIKASI PDB ORDER SATU
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
Solusi PDB Linier Order Satu Nonhomogen
Bentuk Umum
dydx
+ P(x)y = Q(x) (15)
dydx
+ P(x)y = Q(x)yn (16)
Simplikasi
(P(x)y −Q(x))dx + dy = 0 (17)
M(x , y) = P(x)y −Q(x) dan N(x , y) = 1 (18)
Ikuti langkah-langkah dalam handout.
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
PDB Linier Order Satu HomogenPDB Linier Order Satu Nonhomogen
Solusi PDB Linier Order Satu Nonhomogen
Bentuk Umum
dydx
+ P(x)y = Q(x) (15)
dydx
+ P(x)y = Q(x)yn (16)
Simplikasi
(P(x)y −Q(x))dx + dy = 0 (17)
M(x , y) = P(x)y −Q(x) dan N(x , y) = 1 (18)
Ikuti langkah-langkah dalam handout.
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
Mekanik
Kecepatan
v =dxdt
(m/dt) ; a =dvdt
(m/dt2) (19)
H. Newton
mg = W
ma = F
mdvdt
= F
mdvdx
dxdt
= F ; mvdvdx
= F
Dafik Persamaan Diferensial I
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PDB Linier Order SatuAPLIKASI PDB ORDER SATU
Mekanik
Kecepatan
v =dxdt
(m/dt) ; a =dvdt
(m/dt2) (19)
H. Newton
mg = W
ma = F
mdvdt
= F
mdvdx
dxdt
= F ; mvdvdx
= F
Dafik Persamaan Diferensial I