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21_pdfsam_math 54 differential equation solutions odd

# 21_pdfsam_math 54 differential equation solutions odd - 1 Â...

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Exercises 1.4 where C = ± C 2 is any nonzero constant. The value C =0g ives y 0( for x 6 =0) ,wh ichis , clearly, also a solution to the given equation. EXERCISES 1.4: The Approximation Method of Euler, page 28 1. In this initial value problem, f ( x, y )= x/y , x 0 =0 ,and y 0 = 1. Thus, with h =0 . 1, the recursive formulas (2) and (3) on page 25 of the text become x n +1 = x n + h = x n +0 . 1 , y n +1 = y n + hf ( x n ,y n )= y n +0 . 1 · ± x n y n ² ,n =0 , 1 ,.... We set n = 0 in these formulas and obtain x 1 = x 0 +0 . 1=0+0 . 1=0 . 1 , y 1 = y 0 +0 . 1 · ± x 0 y 0 ² = 1+0 . 1 · ± 0 1 ² = 1 . Putting n = 1 in the recursive formulas yields x 2 = x 1 +0 . 1=0 . 1+0 . 1=0 . 2 , y 2 = y 1 +0 . 1 · ± x 1 y 1 ² = 1+0 . 1 · ± 0 . 1 1 ² = 1 . 01 . Continuing in the same manner, we Fnd for n =2 ,3 ,and4 : x 3 =0 . 2+0 . 1=0 . 3 ,y
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Unformatted text preview: 1 Â· Â± . 2 âˆ’ 1 . 01 Â² = âˆ’ 1 . 02980 ; x 4 = 0 . 3 + 0 . 1 = 0 . 4 , y 4 = âˆ’ 1 . 02980 + 0 . 1 Â· Â± . 3 âˆ’ 1 . 02980 Â² = âˆ’ 1 . 05893 ; x 5 = 0 . 4 + 0 . 1 = 0 . 5 , y 5 = âˆ’ 1 . 05893 + 0 . 1 Â· Â± . 4 âˆ’ 1 . 05893 Â² = âˆ’ 1 . 09671 , where we have rounded oÂ± all answers to Fve decimal places. 2. In this problem, x = 0, y = 4, h = 0 . 1, and f ( x, y ) = âˆ’ x/y . Thus, the recursive formulas given in equations (2) and (3) on page 25 of the text become x n +1 = x n + h = x n + 0 . 1 , 17...
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