8/19/2019 Komputasi Sistem Fisis 1

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PR-I

Nama : Nadya Amalia

NIM : 20213042

Mata Kuliah : Komputasi Sistem Fisis (FI5005)

Dosen : Dr.rer.nat. Linus Ampang Pasasa

1. Hitung integral dari ODE ௗ௬ௗ௫  = ݔ2− 

ଷ + 12ݔ

ଶ − ݔ20 + 8.5 dari ݔ = 0 ke ݔ = 4, step size 0.5,

syarat awal ݔ = 0→ ݕ = 1 dengan metode:

a. Euler

b. Heun

c. Raltson

d. Solusi analitik

2.  ௗ௬ௗ௫ ݔݕ = 

ଶ − ݕ1.2

ݔ = 0→ ݔ (0) = 1, dengan menggunakan ℎݕ ,2 =  = 0.5 dan  ℎ  = 0.25

a. Euler

b. Heun

c. Solusi analitik

3.  ௗమ௬ௗ௧మ  − ݐ + ݕ = 0ݕ

ʹ (0) = 0ݕ ,2 = (0)

ݐ

= 0→ ݐ = 4,  ℎ  = 0.1

Gunakan metode Euler dan Heun, plot keduanya dan hitung errornya masing-masing.

4.  ௗ௬ௗ௧ ݅݊ݏݕ = 

ଶݐ

ݕ

ʹ (0) = 0ݕ ,2 = (0)

ݐ = 0→ ݐ (0) = 1, ℎݕ ,3 =  = 0.1

Gunakan metode Heun dan Raltson, plot keduanya dan hitung errornya masing-masing.

PENYELESAIAN

EulerFirst order differential equation

ାଵݕ + ݕ =  ℎݕʹ Second order differential equation

݀ݕ

݀ݐ

=ݒ 

= ଵ(ݐ, ,ݕ (ݒ

ଶݕ

݀ݐ

ଶ   =݀ݒ

݀ݐ

= ଶ(ݐ,ݕ, (ݒାଵݕ + ݕ =  ℎݕᇱ   ାଵݕ → + ݕ =  ℎ ଵ(ݐ,ݕ , (ݒାଵݒ + ݒ =  ℎݒᇱ   ାଵݒ → + ݒ =  ℎ ଶ(ݐ,ݕ, (ݒ

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Heun

First order differential equation

ାଵݕ  + ݕ =  ଵ +  ଶ,  =  = ଵଶ

ାଵݕ ) + ଵଶݕ =  ଵ + ଶ)

ଵ =  ℎ ݔ)

ݕ,

)

ଶ  =  ℎ ݇ߚ + ݕ,ℎߙ + ݔ) ଵ), ߙܾ ܾߚ = = ଵଶ ߙ, ߚ = = 1ଶ  =  ℎ + ݔ) ℎ, ( + ଵݕ

Second order differential equation

݀ݕ

݀ݐ

=ݒ 

= ଵ(ݐ, ,ݕ (ݒ

ଶݕ

݀ݐ

ଶ   =݀ݒ

݀ݐ

= ଶ(ݐ,ݕ, (ݒ

ାଵݕ  + ଵଶ൫ݕ =  ଵ௬ + ଶ௬൯

ଵ

௬ =  ℎ

ଵ(ݐ

,ݕ

,ݒ

)

ଶ௬  =  ℎ ଵ(ݐ + ℎ,ݕ + ଵ௬,ݒ + ଵ௩)

ାଵݒ ) + ଵଶݒ =  ଵ௩ + ଶ௩)

ଵ௩  =  ℎ ଶ(ݐ,ݕ, (ݒଶ௩  =  ℎ ଶ(ݐ + ℎ,ݕ + ଵ

௬( + ଵ௩ݒ,

Raltson

ାଵݕ  + ݕ =  ଵ +  ଶ,  = ଶଷ , = 1 − =

 ଵଷ

ାଵݕ  + ଵଷݕ =  ଵ + ଶଷ ଶ

ଵ  =  ℎ ݔ) (ݕ,ଶ  =  ℎ ݇ߚ + ݕ,ℎߙ + ݔ) ଵ), ߙܾ ܾߚ = =

 ଵଶ ߙ, ߚ = =

 ଵଶ :

ଶଷ =

 ଷସ

ଶ  =  ℎ ቀݔ + ଷସℎ,ݕ + ଷସ ଵቁAnalitik

1.  ௗ௬ௗ௫  = ݔ2− 

ଷ + 12ݔଶ  − ݔ20 + 8.5Û ݀ݕ ଶݔଷ + 12ݔ2−) =  − ݔ20 + 8.5)݀ݔÛ  න݀ݕ = න(−2ݔଷ + 12ݔଶ  − ݔ20 + 8.5)݀ݔÛ  ݕ =  −

ଵଶݔ

ସ + 4ݔଷ  − ݔଶ + 8.5ݔ10 + ݕ

(0) = 0

Û    −ଵଶ(0)

ସ + 4(0)ଷ  − 10(0)ଶ + 8.5(0) +  = 1

Û   = 1

ݕ

(ݔ

) =  −ଵଶݔ

ସ + 4ݔ

ଷ − ݔଶ + 8.5ݔ10 + 1

2.  ௗ௬ௗ௫ ݔݕ = 

ଶ − ݕ1.2

Û 

݀ݕ

݀ݔ

ଶݔ)ݕ =  − 1.2)

Û 

݀ݕ

ݕ

= (ݔ

ଶ − 1.2)݀ݔ

Û  න݀ݕݕ

= න(ݔଶ  − 1.2)݀ݔÛ  lnݕ = ଵଷݔଷ  − ݔ1.2 + Û  ݕ = exp [

ଵଷݔ

ଷ − ݔ1.2 + ]

(0) = 1ݕ

Û  exp ቂଵଷ(0)ଷ  − 1.2(0) + ቃ = 1Û c = 0

ݕ

= exp [ଵଷݔ

ଷ − [ݔ1.2

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3.  ௗమ௬ௗ௧మ  − ݐ ݕ + = 0

Û 

ଶݕ

݀ݐ

ଶ  + ݕ ݐ =ݕ

ᇱᇱ + ݕ = 0Û ଶ + 1 = 0ݎ 

Û  ݎ = ±

ݕ

=௦ݕ + ݕ 

ݕ  = ଵ cos ݐ + ଶ sin ݐ௦ݕ  =  +ݐ  ʹ ௦ݕ  = ʹʹ ௦ݕ  = 0Û   0 +   +ݐ ݐ =Û   = 1,  = 0

ݕ

=௦ݕ + ݕ 

Û  ݕ = ଵ cos ݐ + ଶ sin ݐ + ݐݕ

(0) = 2

Û   ଵ cos(0) + ଶ sin(0) + (0) = 2Û   ଵ  = 2ݕ

ʹ  =  − ଵ sin ݐ + ଶ cos ݐ + 1

ݕ

ʹ (0) = 0

Û    − ଵ sin(0) + ଶ cos(0) + 1 = 0Û   ଶ cos(0) + 1 = 0Û   ଶ  =  −1Û  ݕ  = 2 cos ݐ − sin ݐ + ݐ

4.  ௗ௬ௗ௧ ݅݊ݏݕ = 

ଶݐ

Û 

݀ݕ

ݕ

݅݊ݏ = ଶݐ݀ݐ

Û  න݀ݕݕ

= න ݅݊ݏ ଶݐ݀ݐÛ  lnݕ =

 ௧ଶ−

ଵସ sin2ݐ + 

Û  ݕ = exp[ ௧ଶ−

ଵସ sin2ݐ + ]

ݕ

(0) = 1

Û    exp[()ଶ   −

ଵସ sin2(0) + ] = 1

Û   = 0

ݕ

= exp[ ௧ଶ −

ଵସ sin2ݐ]