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PERSAMAAN DIFERENSIALPertemuan 5, PD Eksak
Nikenasih [email protected]
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1
β’ Solusi PD Eksak
β’ Metode Standar
β’ Metode Grouping
2β’ Solusi PD Eksak - Metode Alternatif
PRESENTATION PARTS
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Solusi PD Eksak
Diberikan PD eksakπ π₯, π¦ ππ₯ + π π₯, π¦ ππ¦ = 0.
terdapat πΉ(π₯, π¦) sedemikian sehinggaππΉ(π₯, π¦)
ππ₯= π π₯, π¦ dan
ππΉ(π₯, π¦)
ππ¦= π π₯, π¦
AkibatnyaππΉ(π₯, π¦)
ππ₯ππ₯ +
ππΉ(π₯, π¦)
ππ¦ππ¦ = 0.
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Jadi,ππΉ π₯, π¦ = 0
Sehingga diperoleh solusi umumnyaπΉ π₯, π¦ = πΆ
Pertanyaannya...Which πΉ π₯, π¦ ??
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Diketahui PD bersifat eksak maka
ππΉ(π₯, π¦)
ππ₯= π π₯, π¦ πΌ
ππΉ(π₯, π¦)
ππ¦= π π₯, π¦ (πΌπΌ)
Dari persamaan I maka diperoleh
πΉ π₯, π¦ = π π₯, π¦ ππ₯ + π(π¦)
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Dari persamaan II,
π
ππ¦ π(π₯, π¦) ππ₯ + π(π¦) = π(π₯, π¦)
Jadi,
π π¦ = π π₯, π¦ β π
ππ¦π π₯, π¦ ππ₯ ππ¦
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Jadi,
πΉ π₯, π¦ = π π₯, π¦ ππ₯ + π(π¦)
dengan
π π¦ = π π₯, π¦ β π
ππ¦π π₯, π¦ ππ₯ ππ¦
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Contoh : Tentukan solusi umum dari
Jawab :
Latihan :
3π₯2 + 4π₯π¦ ππ₯ + 2π₯2 + 2π¦ ππ¦ = 0
(2π₯ cos π¦ + 3π₯2π¦) ππ₯ + π₯3 β π₯2 sin π¦ β π¦ ππ¦ = 0
π₯3 + 2π₯2π¦ + π¦2 = π
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Metode Grouping
Syarat :Persamaan Diferensial merupakan jumlahan daripersamaan diferensial eksak.
Contoh :
1. 3π₯2 + 4π₯π¦ ππ₯ + 2π₯2 + 2π¦ ππ¦ = 0
2. (2π₯ cos π¦ + 3π₯2π¦) ππ₯ + π₯3 β π₯2 sin π¦ β π¦ ππ¦ = 0
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Metode Alternatif
Jika
π π₯, π¦ ππ₯ = π π₯, π¦ + π1 π₯ + β1(π¦)
Dan
π π₯, π¦ ππ₯ = π π₯, π¦ + π2 π₯ + β2(π¦)
makaπΉ π₯, π¦ = π π₯, π¦ + π1 π₯ + β2(π¦)
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Contoh :
1. π₯2 + π¦2 ππ₯ + 2π₯π¦ + cos π¦ ππ¦ = 0
2. 1 β 2π₯π¦ ππ₯ + 4π¦3 β π₯2 ππ¦ = 0
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Referensi :
1. Ross, L. Differential Equations. John Wiley & Sons.
2. Oswaldo GonzΒ΄alez-Gaxiola1 and S. HernΒ΄andezLinares. An Alternative Method to Solve ExactDifferential Equations. International MathematicalForum, 5, 2010, no. 54, 2689 - 2697
3. http://math.stackexchange.com/questions/215324/proof-for-exact-differential-equations-shortcutdiakses tanggal 5 oktober 2016
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THANK YOU
Nikenasih [email protected]
Karangmalang Sleman Yogyakarta
Exercise 5.1
Solve the following equations.
1. (3π₯ + 2π¦) ππ₯ + (2π₯ + π¦)ππ¦ = 0
2. (π¦2 + 3) ππ₯ + (2π₯π¦ β 4)ππ¦ = 0
3. (6π₯π¦ + 2π¦2 β 5)ππ₯ β (π₯3 + π¦) ππ¦ = 0
4. (π2 + 1) cos π ππ + 2π sin π ππ = 0
5. (π¦ sec2 π₯ + sec π₯ tan π₯) ππ₯ + (tan π₯ + 2π¦)ππ¦ = 0
Exercises 5.2
1. Consider the differential equation
(4π₯ + 3π¦2) ππ₯ + 2π₯π¦ ππ¦ = 0
a. Show that this equation is not exact
b. Find the integratiiing factor of the form π₯π, where n is a positive integer.
c. Multiply the given equation through by the integrating factor found in (b) and solve
the resulting exact equation.
2. Consider the differential equation
(π¦2 + 2π₯π¦)ππ₯ β π₯2 ππ¦ = 0
a. Show that this equation is not exact.
b. Multiply the given equatttion through by π¦π, where n is an integer and then
determine n so that π¦π is an integrating factor of the given equation.
c. Multiply the given equation through by the integrating factor found in (b) and solve
the resulting exact equation.
d. Show that y = 0 is a solution of the original nonexact equation but is not a solution
of the essentially equivalent exact equation found in step (c).
e. Graph several integral curves of the original equation, including all those whose
equations are (or can be writen) in some special form.