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PERSAMAAN DIFERENSIAL (DIFFERENTIAL EQUATION)

PERSAMAAN DIFERENSIAL (DIFFERENTIAL EQUATION). metode euler metode runge-kutta. Persamaan Diferensial. Persamaan paling penting dalam bidang rekayasa, paling bisa menjelaskan apa yang terjadi dalam sistem fisik. Menghitung jarak terhadap waktu dengan kecepatan tertentu, 50 misalnya.

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PERSAMAAN DIFERENSIAL (DIFFERENTIAL EQUATION)

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  1. PERSAMAAN DIFERENSIAL(DIFFERENTIAL EQUATION) metode euler metode runge-kutta

  2. Persamaan Diferensial • Persamaan paling penting dalam bidang rekayasa, paling bisa menjelaskan apa yang terjadi dalam sistem fisik. • Menghitung jarak terhadap waktu dengan kecepatan tertentu, 50 misalnya.

  3. Rate equations

  4. Persamaan Diferensial • Solusinya, secara analitik dengan integral, • C adalah konstanta integrasi • Artinya, solusi analitis tersebut terdiri dari banyak ‘alternatif’ • C hanya bisa dicari jika mengetahui nilai x dan t. Sehingga, untuk contoh di atas, jika x(0) = (x saat t=0) = 0, maka C = 0

  5. Klasifikasi Persamaan Diferensial Persamaan yang mengandung turunan dari satu atau lebih variabel tak bebas, terhadap satu atau lebih variabel bebas. • Dibedakan menurut: • Tipe (ordiner/biasa atau parsial) • Orde (ditentukan oleh turunan tertinggi yang ada • Liniarity (linier atau non-linier)

  6. PDO Pers.dif. Ordiner = pers. yg mengandung sejumlah tertentu turunan ordiner dari satu atau lebih variabel tak bebas terhadap satu variabel bebas. y(t) = variabel tak bebas t = variabel bebas dan turunan y(t) Pers di atas: ordiner, orde dua, linier

  7. PDO • Dinyatakan dalam 1 peubah dalam menurunkan suatu fungsi • Contoh:

  8. Partial Differential Equation • Jika dinyatakan dalam lebih dari 1 peubah, disebut sebagai persamaan diferensial parsial • Pers.dif. Parsial mengandung sejumlah tertentu turunan dari paling tidak satu variabel tak bebas terhadap lebih dari satu variabel bebas. • Banyak ditemui dalam persamaan transfer polutan (adveksi, dispersi, diffusi)

  9. PDO Ordiner, linier, orde 3 Ordiner, linier, orde 2 Ordiner, non linier, orde 1

  10. Solusi persamaan diferensial • Secara analitik, mencari solusi persamaan diferensial adalah dengan mencari fungsi integral nya. • Contoh, untuk fungsi pertumbuhan secara eksponensial, persamaan umum:

  11. Rate equations

  12. But what you really want to know is… the sizes of the boxes (or state variables) and how they change through time That is, you want to know: the state equations There are two basic ways of finding the state equations for the state variables based on your known rate equations: 1) Analytical integration 2) Numerical integration

  13. Suatu kultur bakteria tumbuh dengan kecepatan yang proporsional dengan jumlah bakteria yang ada pada setiap waktu. Diketahui bahwa jumlah bakteri bertambah menjadi dua kali lipat setiap 5 jam. Jika kultur tersebut berjumlah satu unit pada saat t = 0, berapa kira-kira jumlah bakteri setelah satu jam?

  14. Solusi persamaan diferensial • Jumlah bakteri menjadi dua kali lipat setiap 5 jam, maka k = (ln 2)/5 • Jika P0 = 1 unit, maka setelah satu jam…

  15. The Analytical Solution of the Rate Equation is the State Equation Rate equation (dsolve in Maple) State equation

  16. There are very few models in ecology that can be solved analytically.

  17. Solusi Numerik • Numerical integration • Eulers • Runge-Kutta

  18. Numerical integration makes use of this relationship: Which you’ve seen before… • Relationship between continuous and discrete time models • *You used this relationship in Lab 1 to program the • logistic rate equation in Visual Basic:

  19. Fundamental Approach of Numerical Integration y = f(t), unknown yt+t, unknown yt+t, estimated y , known yt, known t, specified t

  20. Calculate dN/dt*1 at Nt Add it to Nt to estimate Nt+ t Nt+ t becomes the new Nt Calculte dN/dt * 1 at new Nt Use dN/dt to estimate next Nt+t Repeat these steps to estimate the state function over your desired time length (here 30 years) Euler’s Method: yt+ t≈ yt + dy/dtt

  21. Example of Numerical Integration point to estimate Analytical solution to dy/dt Y0 = 10  t = 0.5

  22. Euler’s Method: yt+ t≈ yt + dy/dtt analytical y(t+ t) m1 = dy/dt at yt m1 = 6*10-.007*(10)2 y = m1*t yest=yt + y y estimated y(t+ t) y yt = 10  t = 0.5

  23. Runge-Kutta Example point to estimate Problem: estimate the slope to calculate y y  t = 0.5

  24. Runge-Kutta Example Weighted average of > 1 slope ½ Δt Δt t estimated yt+Δt Unknown point to estimate, yt+Δt estimated yt+Δt estimated yt+Δt yt  t = 0.5

  25. Runge-Kutta, 4th order Uses the derivative, dy/dt, to calculate 4 slopes (m1…m4) within Δt: These 4 slopes are used to calculate a weighted slope of the state function between t and t + Δt, which is used to estimate yt+ Δt:

  26. Step 1: Evaluate slope at current value of state variable. y0 = 10 m1 = dy/dt at y0 m1 = 6*10-.007*(10)2 m1 = 59.3 y m1=slope 1 y0

  27. Step 2: A) Calculate y1at t +t/2 using m1. B) Evaluate slope at y1. A) y1 = y0 + m1* t /2 y1 = 24.82 B) m2 = dy/dt at y1 m2 = 6*24.8-.007*(24.8)2 m2 = 144.63 m2=slope 2 y1  t = 0.5/2

  28. Step 3: Calculate y2 at t +t/2 using k2. Evaluate slope at y2. y2 = y0 + k2* t /2 y2 = 46.2 k3 = slope 3 k3 = dy/dt at y2 k3 = 6*46.2-.007*(46.2)2 k3 = 263.0 y2  t = 0.5/2

  29. Step 4: Calculate y3 at t +t using k3. Evaluate slope at y3. y3 = y0 + k3* t y3 =141.5 y3 k4 = slope 4 k4 = dy/dt at y3 k4 = 6*141.0-.007*(141.0)2 k4 = 706.9 y2  t = 0.5

  30. Now you have 4 calculations of the slope of the state equation between t and t+Δt m4 = slope 4 m3 = slope3 m2 = slope 2 m1 = slope 1  t = 0.5

  31. Step 5: Calculate weighted slope. Use weighted slope to estimate y at t +t weighted slope = true value weighted slope estimated value  t = 0.5

  32. Conclusions Analytical • 4th order Runge-Kutta offers substantial improvement over Eulers. • Both techniques provide estimates, not “true” values. • The accuracy of the estimate depends on the size of the step used in the algorithm. Runge-Kutta Eulers

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