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8.2 (2). Consider the linear equation Y(x) = XY (2) + (1 - 1) cos(x) - (1+) sin(x), quady (0) = 1 The true solution is Y (1)

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1 2 - 3 - 4 - 5- 6 - clear x(1)=0; y(1)=1;lambda=5; h=0.0625; m=10/h; E for i=1:m x(i+1)=x (i) +h; y(i+1)=y(i)+h* (lambda*y(i(a) x = -1, h = 0.5 Y(x) = -1 Y(x) + (1-(-12) cosx - (1 + (-1)) sinn xo=o. or, y(x) = -Y(«) + 2 cos x • Y (o)=1 Y=1, 08, F(% Euler Method clear all clc; format short for k=1:3 lambda=-1 h=1/2^k f=@(X,Y)lambda*Y+(1-lambda)*cos(x)- (1+lambda)*sin(x);7.5000 1.5590 1.2846 8.0000 1.1261 0.8439 8.5000 0.4176 0.1965 9.0000 -0.3932 -0.4990 9.5000 1.1077 -1.0723 10.0000 -1.5510 -8.7500 -0.0750 -0.1561 9.0000 -0.4467 -0.4990 9.2500 -0.7906 -0.8109 9.5000 -1.0853 -1.0723 9.7500 -1.3126 -1.2671 10.0000 -14.5000 7.4346 -1.1883 5.0000 12.1294 -0.6753 5.5000 19.1531 0.0031 6.0000 29.4352 0.6808 6.5000 44.4322 1.1917 7.0000 66.43327.2500 77.4487 1.3910 7.5000 96.3993 1.2846 7.7500 120.0301 1.0984 8.0000 149.5404 0.8439 8.2500 186.4308 0.5369 8.5000 232.53.7500 0.7333 -1.3921 3.8750 0.9678 -1.4123 4.0000 1.2562 -1.4104 4.1250 1.6024 -1.3866 4.2500 2.0108 -1.3411 4.3750 2.4859 -9.8750 665.0709 -1.3355 10.0000 748.3136 -1.3831

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