4th order RK ---------------------method.pptx

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1. DEPARTMENT OF SCIENCE AND HUMANITIES Name of the Faculty    : Mr.R.VIMAL KUMAR Subject Name & Code :Statistics and Numerical Methods & MA3251 Branch & Department : B.Tech IT Year & Semester         : I / II Academic Year            :2023-24 16-04-2024 MA3251/B.Tech IT/I Year UG/II SEM/KG-KiTE 1     KGiSL Institute of Technology (Approved by AICTE, New Delhi; Affiliated to Anna University, Chennai) Recognized by UGC, Accredited by NBA (IT) 365, KGiSL Campus, Thudiyalur Road, Saravanampatti , Coimbatore – 641035 .

2. FOURTH ORDER RUNGE KUTTA FOR SOLVING FIRST ORDER EQUATIONS SECOND ORDER R-K METHOD If the initial values of (x,y) for the differentiation equation dy /dx=f(x,y) then the first increment in y namely ∆ y is calculated as 16-04-2024 MA3251/B.Tech IT/I Year UG/II SEM/KG-KiTE 2 /14

3. FOURTH ORDER RUNGE KUTTA FOR SOLVING FIRST ORDER EQUATIONS THIRD ORDER R-K METHOD The algorithm for this method is 16-04-2024 MA3251/B.Tech IT/I Year UG/II SEM/KG-KiTE 3 /14

4. FOURTH ORDER RUNGE KUTTA FOR SOLVING FIRST ORDER EQUATIONS FOURTH ORDER R-K METHOD The algorithm for this method is 16-04-2024 MA3251/B.Tech IT/I Year UG/II SEM/KG-KiTE 4 /14

5. PROBLEMS Consider an ordinary differential equation dy /dx = x 2  + y 2 , y(1) = 1.2. Find y(1.05) using the fourth order Runge-Kutta method . Solution: Given, dy /dx = x 2  + y 2 , y(1) = 1.2 So, f(x, y) = x 2  + y 2 x = 1 and y = 1.2 Also, h = 0.05 16-04-2024 MA3251/B.Tech IT/I Year UG/II SEM/KG-KiTE 5 /14

6. PROBLEMS Let us calculate the values of k 1 , k 2 , k 3  and k 4 . k 1  = hf (x , y ) = (0.05) [x 2  + y 2 ] = (0.05) [(1) 2  + (1.2) 2 ] = (0.05) (1 + 1.44) = (0.05)(2.44 ) = 0.122 16-04-2024 MA3251/B.Tech IT/I Year UG/II SEM/KG-KiTE 6 /14

7. PROBLEMS k 2  = hf [x  + (½)h, y  + (½)k 1 ] = (0.05) [f(1 + 0.025, 1.2 + 0.061)] {since h/2 = 0.05/2 = 0.025 and k 1 /2 = 0.122/2 = 0.061 } = (0.05) [f(1.025, 1.261)] = (0.05) [(1.025) 2  + (1.261) 2 ] = (0.05) (1.051 + 1.590) = (0.05)(2.641) = 0.1320 16-04-2024 MA3251/B.Tech IT/I Year UG/II SEM/KG-KiTE 7 /14

8. PROBLEMS K 3  = hf [x  + (½)h, y  + (½)k 2 ] = (0.05) [f(1 + 0.025, 1.2 + 0.066)] {since h/2 = 0.05/2 = 0.025 and k 2 /2 = 0.132/2 = 0.066 } = (0.05) [f(1.025, 1.266)] = (0.05) [(1.025) 2  + (1.266) 2 ] = (0.05) (1.051 + 1.602) = (0.05)(2.653) = 0.1326 16-04-2024 MA3251/B.Tech IT/I Year UG/II SEM/KG-KiTE 8 /14

9. PROBLEMS k 4  = hf (x  + h, y  + k 3 ) = (0.05) [f(1 + 0.05, 1.2 + 0.1326)] = (0.05) [f(1.05, 1.3326)] = (0.05) [(1.05) 2  + (1.3326) 2 ] = (0.05) (1.1025 + 1.7758) = (0.05)(2.8783) = 0.1439 16-04-2024 MA3251/B.Tech IT/I Year UG/II SEM/KG-KiTE 9 /14

10. PROBLEMS By RK4 method, we have; y 1  = y  + (⅙) (k 1  + 2k 2  + 2k 3  + k 4 ) y 1  = y(1.05) = y  + (⅙) (k 1  + 2k 2  + 2k 3  + k 4 ) By substituting the values of y , k 1 , k 2 , k 3  and k 4 , we get; y(1.05) = 1.2 + (⅙) [0.122 + 2(0.1320) + 2(0.1326) + 0.1439] = 1.2 + (⅙) (0.122 + 0.264 + 0.2652 + 0.1439) = 1.2 + (⅙) (0.7951) = 1.2 + 0.1325 = 1.3325 16-04-2024 MA3251/B.Tech IT/I Year UG/II SEM/KG-KiTE 10 /14