23_pdfsam_math 54 differential equation solutions odd

# 23_pdfsam_math 54 differential equation solutions odd -...

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Exercises 1.4 x 2 =1 . 2+0 . 2=1 . 4 ,y 2 . 400 + 0 . 2 ± 1 . 400 2 +1 . 400 1 . 2 ² 1 . 960; x 3 . 4+0 . . 6 3 . 960 + 0 . 2 ± 1 . 960 2 . 960 1 . 4 ² 2 . 789; x 4 . 6+0 . . 8 4 =2 . 789 + 0 . 2 ± 2 . 789 2 +2 . 789 1 . 6 ² 4 . 110 . 7. For this problem notice that the independent variable is t and the dependent variable is x . Hence, the recursive formulas given in equations (2) and (3) on page 25 of the text become t n +1 = t n + h and φ ( t n +1 ) x n +1 = x n + hf ( t n ,x n ) ,n =0 , 1 , 2 ,... . For this problem, f ( t, x )=1+ t sin( tx ), t 0 ,and x 0 = 0. Thus the second recursive formula above becomes x n +1 = x n + h [1 + t n sin( t n x n )] , 1 , 2 For the case N ,wehave h =(1 0) / 1=1wh ichgivesus t 1 =0+1=1 and φ (1) x 1 =0+1 · (1 + 0 · sin 0) = 1 . N h / 2=0 . 5 . Thus we have t 1 =0+0 . 5=0 . 5 1 . 5 · (1 + 0 · sin 0) = 0
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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 Berkeley.

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