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 = 1 . 400 + 0 . 2 1 . 400 2 + 1 . 400 1 . 2 1 . 960; x 3 = 1 . 4 + 0 . 2 = 1 . 6 , y 3 = 1 . 960 + 0 . 2 1 . 960 2 + 1 . 960 1 . 4 2 . 789; x 4 = 1 . 6 + 0 . 2 = 1 . 8 , y 4 = 2 . 789 + 0 . 2 2 . 789 2 + 2 . 789 1 . 6 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 = 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 )] , n = 0 , 1 , 2 , . . . . For the case N = 1, we have h = (1 0) / 1 = 1 which gives us t 1 = 0 + 1 = 1 and φ (1) x 1 = 0 + 1 · (1 + 0 · sin 0) = 1 . For the case N = 2, we have
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