153_pdfsam_math 54 differential equation solutions odd

# 153_pdfsam_math 54 differential equation solutions odd -...

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Exercises 3.6 Table 3–E : Euler’s method approximations to the solution of y = 20 y , y (0) = 1, on [0 , 1] with h = 0 . 1 , 0 . 2 , and 0 . 025. x n n y n n ( h = 0 . 2) ( 2) . y n y n ( h = 0 . 1) h = 0 y n ( h = 0 . 025) . h = 0 0.1 1 0.062500 0.2 3 1 0.003906 0.3 1 0.000244 0.4 9 1 0.000015 0.5 1 0.000001 0.6 27 1 0.000000 0.7 1 0.000000 0.8 81 1 0.000000 0.9 1 0.000000 1.0 243 1 0.000000 19. In this problem, the variables are t and p . With suggested values of parameters, the initial value problem (13) becomes dp dt = 3 p p r , p (0) = 1 . Therefore, f ( t, p ) = 3 p p r and, with h = 0 . 25, functions F and G in improved Euler’s method subroutine have the form F = f ( t, p ) = 3 p p r ; G = f ( t + 0 . 25 , p + 0 . 25 F ) = 3[ p + 0 . 25 F ] [ p + 0 . 25 F ] r = 3 [ p + 0 . 25 (3 p p r )] [ p + 0 . 25 (3 p
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Unformatted text preview: . 25 (3 p − p r )] r . The results of computations are shown in Table 3-F. These results indicate that the limiting populations for r = 1 . 5, r = 2, and r = 3 are p ∞ = 9, p ∞ = 3, and p ∞ = √ 3, respectively. Since the right-hand side of the given logistic equation, f ( t, p ) = 3 p − p r , does not depend on t , we conclude that this equation is autonomous. Therefore, its equilibrium solutions (if any) 149...
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