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160_pdfsam_math 54 differential equation solutions odd

# 160_pdfsam_math 54 differential equation solutions odd -...

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Chapter 3 k 2 = hf ( x + h/ 2 , y + k 1 / 2) = 0 . 25[2(1 + ( 1) / 2) 6] = 1 . 25 ; k 3 = hf ( x + h/ 2 , y + k 2 / 2) = 0 . 25[2(1 + ( 1 . 25) / 2) 6] = 1 . 3125 ; k 4 = hf ( x + h, y + k 3 ) = 0 . 25[2(1 + ( 1 . 3125)) 6] = 1 . 65625 . Step 4 then yields x = 0 + 0 . 25 = 0 . 25 , y = 1 + 1 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) = 1 + 1 6 ( 1 2 · 1 . 25 2 · 1 . 3125 1 . 65625) ≈ − 0 . 29688 . Now we go back to Step 3 and recalculate k j ’s for new values of x and y . k 1 = 0 . 25[2( 0 . 29688) 6] = 1 . 64844 ; k 2 = 0 . 25[2( 0 . 29688 + ( 1 . 64844) / 2) 6] = 2 . 06055 ; k 3 = 0 . 25[2( 0 . 29688 + ( 2 . 06055) / 2) 6] = 2 . 16358 ; k 4 = 0 . 25[2( 0 . 29688 + ( 2 . 16358)) 6] = 2 . 73022 ; x = 0 . 25 + 0 . 25 = 0 . 5 , y = 0 . 29688 + 1 6 ( 1 . 64844 2 · 2 . 06055 2 · 2 . 16358 2 . 73022) ≈ − 2 . 43470 . We repeat the cycle two more times: k 1 = 0 . 25[2( 2 . 43470) 6] = 2 . 71735 ; k 2 = 0 . 25[2( 2 .
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Unformatted text preview: 43470 + ( − 2 . 71735) / 2) − 6] = − 3 . 39670 ; k 3 = 0 . 25[2( − 2 . 43470 + ( − 3 . 39670) / 2) − 6] = − 3 . 56652 ; k 4 = 0 . 25[2( − 2 . 43470 + ( − 3 . 56652)) − 6] = − 4 . 50060 ; x = 0 . 5 + 0 . 25 = 0 . 75 , y = − 2 . 43470 + 1 6 ( − 2 . 71735 − 2 · 3 . 39670 − 2 · 3 . 56652 − 4 . 50060) ≈ − 5 . 95876 and k 1 = 0 . 25[2( − 5 . 95876) − 6] = − 4 . 47938 ; k 2 = 0 . 25[2( − 5 . 95876 + ( − 4 . 47938) / 2) − 6] = − 5 . 59922 ; 156...
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