161_pdfsam_math 54 differential equation solutions odd

# 161_pdfsam_math 54 - x 1 − 77734375 y Hence in Step 4 we have x = x 0 25 y = y 1 6 k 1 2 k 2 2 k 3 k 4 Using the initial conditions x = 0 and y =

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Exercises 3.7 k 3 =0 . 25[2( 5 . 95876 + ( 5 . 59922) / 2) 6] = 5 . 87918 ; k 4 =0 . 25[2( 5 . 95876 + ( 5 . 87918)) 6] = 7 . 41895 ; x =0 . 75 + 0 . 25 = 1 . 00 , y = 5 . 95876 + 1 6 ( 4 . 47938 2 · 5 . 59922 2 · 5 . 87918 7 . 41895) ≈− 11 . 7679 . Thus φ (1) ≈− 11 . 7679 . The actual solution, φ ( x )=3 2 e 2 x , evaluated at x =1,gives φ (1) = 3 2 e 2(1) =3 2 e 2 ≈− 11 . 7781 . 9. For this problem we will use the 4th order Runge-Kutta subroutine with f ( x, y )= x +1 y . Using the step size of h =0 . 25, the number of steps needed is N = 4 to approximate the solution at x =1 . ForStep3wehave k 1 = hf ( x, y )=0 . 25( x +1 y ) , k 2 = hf ± x + h 2 ,y + k 1 2 ² =0 . 25(0 . 875 x +1 0 . 875 y ) , k 3 = hf ± x + h 2 ,y + k 2 2 ² =0 . 25(0 . 890625 x +1 0 . 890625 y ) , k 4 = hf ( x + h, y + k 3 )=0 . 25(0 .
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Unformatted text preview: x + 1 − . 77734375 y ) . Hence, in Step 4 we have x = x + 0 . 25 , y = y + 1 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) . Using the initial conditions x = 0 and y = 1, c = 1, and N = 4 for Step 3 we obtain k 1 = 0 . 25(0 + 1 − 1) = 0 , k 2 = 0 . 25(0 . 875(0) + 1 − . 875(1)) = 0 . 03125 , k 3 = 0 . 25(0 . 890625(0) + 1 − . 890625(1)) ≈ . 0273438 , k 4 = 0 . 25(0 . 77734375(0) + 1 − . 77734375(1)) ≈ . 0556641 . Thus, Step 4 gives x = 0 + 0 . 25 = 0 . 25 , 157...
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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