Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 9.2 DIRECTION FIELDS AND EULER’S METHOD ¤ 385
(c) (i) For h =0.4: (exact value) − (approximate value) =e 0.4 − 1.4 ≈ 0.0918
(ii) For h =0.2: (exact value) − (approximate value) =e 0.4 − 1.44 ≈ 0.0518
(iii) For h =0.1: (exact value) − (approximate value) =e 0.4 − 1.4641 ≈ 0.0277
Each time the step size is halved, the error estimate also appears to be halved (approximately).
21. h =0.5, x 0 =1, y 0 =0,andF (x, y) =y − 2x.
Note that x 1 = x 0 + h =1+0.5 =1.5, x 2 =2,andx 3 =2.5.
y 1 = y 0 + hF (x 0 ,y 0 )=0+0.5F (1, 0) = 0.5[0 − 2(1)] = −1.
y 2 = y 1 + hF (x 1 ,y 1 )=−1+0.5F (1.5, −1) = −1+0.5[−1 − 2(1.5)] = −3.
y 3 = y 2 + hF (x 2 ,y 2 )=−3+0.5F (2, −3) = −3+0.5[−3 − 2(2)] = −6.5.
y 4 = y 3 + hF (x 3 ,y 3 )=−6.5+0.5F (2.5, −6.5) = −6.5+0.5[−6.5 − 2(2.5)] = −12.25.
23. h =0.1, x 0 =0, y 0 =1,andF (x, y) =y + xy.
Note that x 1 = x 0 + h =0+0.1 =0.1, x 2 =0.2, x 3 =0.3,andx 4 =0.4.
y 1 = y 0 + hF (x 0,y 0)=1+0.1F (0, 1) = 1 + 0.1[1 + (0)(1)] = 1.1.
y 2 = y 1 + hF (x 1,y 1)=1.1+0.1F (0.1, 1.1) = 1.1+0.1[1.1+(0.1)(1.1)] = 1.221.
y 3 = y 2 + hF (x 2 ,y 2 )=1.221 + 0.1F (0.2, 1.221) = 1.221 + 0.1[1.221 + (0.2)(1.221)] = 1.36752.
y 4 = y 3 + hF (x 3 ,y 3 )=1.36752 + 0.1F (0.3, 1.36752) = 1.36752 + 0.1[1.36752 + (0.3)(1.36752)]
=1.5452976.
y 5 = y 4 + hF (x 4 ,y 4 )=1.5452976 + 0.1F (0.4, 1.5452976)
Thus, y(0.5) ≈ 1.7616.
=1.5452976 + 0.1[1.5452976 + (0.4)(1.5452976)] = 1.761639264.
25. (a) dy/dx +3x 2 y =6x 2 ⇒ y 0 =6x 2 − 3x 2 y. Store this expression in Y 1 and use the following simple program to
evaluate y(1) for each part, using H = h =1and N =1for part (i), H =0.1 and N =10for part (ii), and so forth.
h → H: 0 → X: 3 → Y:
For(I,1,N):Y+ H × Y 1 → Y: X + H → X:
End(loop):
Display Y. [To see all iterations, include this statement in the loop.]
(i) H =1,N=1 ⇒ y(1) = 3
(ii) H =0.1,N=10 ⇒ y(1) ≈ 2.3928
(iii) H =0.01,N= 100 ⇒ y(1) ≈ 2.3701
(iv) H =0.001, N = 1000 ⇒ y(1) ≈ 2.3681
(b) y =2+e −x3 ⇒ y 0 = −3x 2 e −x3
LHS = y 0 +3x 2 y = −3x 2 e −x3 +3x 2 2+e −x3 = −3x 2 e −x3 +6x 2 +3x 2 e −x3 =6x 2 =RHS
y(0) = 2 + e −0 =2+1=3