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24_pdfsam_math 54 differential equation solutions odd

# 24_pdfsam_math 54 differential equation solutions odd -...

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Chapter 1 Plugging these values into the recursive equations above yields t 2 = 0 . 25 + 0 . 25 = 0 . 5 and x 2 = 0 . 25 + 0 . 25 · [1 + 0 . 25 · sin(0 . 0625)] = 0 . 503904 . Continuing in this way gives t 3 = 0 . 75 and x 3 = 0 . 503904 + 0 . 25 · [1 + 0 . 5 · sin(0 . 251952)] = 0 . 785066 , t 4 = 1 . 00 and φ (1) x 4 = 1 . 13920 . For N = 8, we have h = 1 / 8 = 0 . 125 . Thus, the recursive formulas become t n +1 = t n + 0 . 125 and x n +1 = x n + 0 . 125 · [1 + t n sin( t n x n )] . Using these formulas and starting with t 0 = 0 and x 0 = 0, we can fill in Table 1-A. From this we see that φ (1) x 8 = 1 . 19157, which is rounded to five decimal places. Table 1–A : Euler’s method approximations for the solution of x = 1+ t sin( tx ),
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