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36_pdfsam_math 54 differential equation solutions odd

# 36_pdfsam_math 54 differential equation solutions odd -...

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Chapter 2 If we let t be the variable of integration and replace x 1 by x and y (0) by 1 in the above equation, then we can express the initial value problem as x Z 0 e t 2 dt = 1 3 ± y ( x ) 3 1 ² . Solving for y ( x ) we arrive at y ( x )= 1+3 x Z 0 e t 2 dt 1 / 3 . (2.3) (c) The diFerential equation dy/dx = 1+sin x (1 + y 2 ) separates if we divide by (1 + y 2 ) and multiply by dx . We integrate the separated equation from x =0to x = x 1 and ±nd x 1 Z 0 1+sin xdx = x = x 1 Z x =0 (1 + y 2 ) 1 dy =tan 1 y ( x 1 ) tan 1 y (0) . If we let t be the variable of integration and replace x 1 by x and y (0) by 1 then we can express the solution to the initial value problem by y ( x )=tan x Z 0 1+sin tdt + π 4 . (d) We will use Simpson’s rule (Appendix B) to approximate the de±nite integral found in part (b). (Simpson’s rule is implemented on the website for the text.) Simpson’s rule
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Unformatted text preview: requires an even number of intervals, but we don’t know how many are required to obtain the desired three-place accuracy. Rather than make an error analysis, we will compute the approximate value of y (0 . 5) using 2, 4, 6, . . . intervals for Simpson’s rule until the approximate values for y (0 . 5) change by less than ±ve in the fourth place. ²or n = 2, we divide [0 , . 5] into 4 equal subintervals. Thus each interval will be of length (0 . 5 − 0) / 4 = 1 / 8 = 0 . 125. Therefore, the integral is approximated by . 5 Z e x 2 dx = 1 24 h e + 4 e (0 . 125) 2 + 2 e (0 . 25) 2 + 4 e (0 . 325) 2 + e (0 . 5) 2 i ≈ . 544999003 . 32...
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