AEM_3e_Chapter_06

# The solution of the corresponding linear system gives

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The solution of the corresponding linear system gives 6. We identify P ( x ) = 5, Q ( x ) = 0, f ( x ) = 4 x , and h = (2 1) / 6 = 0 . 1667. Then the finite difference equation is 1 . 4167 y i +1 2 y i + 0 . 5833 y i 1 = 0 . 2778(4 x i ) . The solution of the corresponding linear system gives 9. We identify P ( x ) = 1 x , Q ( x ) = x , f ( x ) = x , and h = (1 0) / 10 = 0 . 1. Then the finite difference equation is [1 + 0 . 05(1 x i )] y i +1 + [ 2 + 0 . 01 x i ] y i + [1 0 . 05(1 x i )] y i 1 = 0 . 01 x i . The solution of the corresponding linear system gives 12. We identify P ( r ) = 2 /r , Q ( r ) = 0, f ( r ) = 0, and h = (4 1) / 6 = 0 . 5. Then the finite difference equation is 1 + 0 . 5 r i u i +1 2 u i + 1 0 . 5 r i u i 1 = 0 . The solution of the corresponding linear system gives 99

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x n Euler h 0.1 Euler h 0.05 Imp. Euler h 0.1 Imp. Euler h 0.05 RK4 h 0.1 RK4 h 0.05 0.50 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.55 0.5500 0.5512 0.5512 0.60 0.6000 0.6024 0.6048 0.6049 0.6049 0.6049 0.65 0.6573 0.6609 0.6610 0.70 0.7095 0.7144 0.7191 0.7193 0.7194 0.7194 0.75 0.7739 0.7800 0.7801 0.80 0.8283 0.8356 0.8427 0.8430 0.8431 0.8431 0.85 0.8996 0.9082 0.9083 0.90 0.9559 0.9657 0.9752 0.9755 0.9757 0.9757 0.95 1.0340 1.0451 1.0452 1.00 1.0921 1.1044 1.1163 1.1168 1.1169 1.1169 x n y n 0.0 2.00000000 init.cond. 0.1 1.65620000 RK4 0.2 1.41097281 RK4 0.3 1.24645047 RK4 0.4 1.14796764 ABM CHAPTER 6 REVIEW EXERCISES CHAPTER 6 REVIEW EXERCISES 3. 6. The first predictor is y 3 = 1 . 14822731. 100
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