292_pdfsam_math 54 differential equation solutions odd

# 292_pdfsam_math 54 differential equation solutions odd - 3...

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Chapter 5 18. For starting values we take t 0 =0 , x 0 , 1 = 10, and x 0 , 2 = 15, which are determined by the initial conditions. Here h =0 . 1, and f 1 ( t, x 1 ,x 2 )= (0 . 1) x 1 x 2 , f 2 ( t, x 1 ,x 2 )= x 1 . Now, using the defnitions o± t n , x i ; n , k i, 1 , k i, 2 , k i, 3 ,and k i, 4 on page 258 o± the text, we have k 1 , 1 = hf 1 ( t n ,x 1; n ,x 2; n )= h (0 . 1) x 1; n x 2; n , k 2 , 1 = hf 2 ( t n ,x 1; n ,x 2; n )= hx 1; n , k 1 , 2 = hf 1 ± t n + h 2 ,x 1; n + k 1 , 1 2 ,x 2; n + k 2 , 1 2 ² = h (0 . 1) ± x 1; n + k 1 , 1 2 ²± x 2; n + k 2 , 1 2 ² , k 2 , 2 = hf 2 ± t n + h 2 ,x 1; n + k 1 , 1 2 ,x 2; n + k 2 , 1 2 ² = h ± x 1; n + k 1 , 1 2 ² , k 1 , 3 = hf 1 ± t n + h 2 ,x 1; n + k 1 , 2 2 ,x 2; n + k 2 , 2 2 ² = h (0 . 1) ± x 1; n + k 1 , 2 2 ²± x 2; n + k 2 , 2 2 ² , k 2 , 3 = hf 2 ± t n + h 2 ,x 1; n + k 1 , 2 2 ,x 2; n + k 2 , 2 2 ² = h ± x 1; n + k 1 , 2 2 ² , k 1 , 4 = hf 1 ( t n + h, x 1; n + k 1 , 3 ,x 2; n + k 2 , 3 )= h (0 . 1) ( x 1; n + k 1 , 3 )( x 2; n + k 2 , 3 ) , k 2 , 4 = hf 2 ( t n + h, x 1; n + k 1 , 3 ,x 2; n + k 2 , 3 )= h ( x 1; n + k 1 , 3 ) . Using these values, we fnd t n +1 = t n + h = t n +0 . 1 , x 1; n +1 = x 1; n + 1 6 ( k 1 , 1 +2 k 1 , 2 +2 k 1 , 3 + k 1 , 4 ) , x 2; n +1 = x 2; n + 1 6 ( k 2 , 1 +2 k 2 , 2 +2 k
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Unformatted text preview: , 3 + k 2 , 4 ) . In Table 5-B we give approximate values ±or t n , x 1; n , and x 2; n . From Table 5-B we see that the strength o± the guerrilla troops, x 1 , approaches zero, there±ore with the combat e²ectiveness coeﬃcients o± 0 . 1 ±or guerrilla troops and 1 ±or conventional troops the conventional troops win. 19. See the answer in the text. 288...
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