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129_pdfsam_math 54 differential equation solutions odd

# 129_pdfsam_math 54 differential equation solutions odd -...

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Exercises 3.4 x ( t ) = 98 . 1 t + 981 e t/ 10 981 . To find out when the object will hit the ground, we solve x ( t ) = 1000 for t . Therefore, we have 1000 = 98 . 1 t + 981 e t/ 10 981 98 . 1 t + 981 e t/ 10 = 1981 . In this equation, if we ignore the term 981 e t/ 10 we will find that t 20 . 2. But this means that we have ignored the term similar to 981 e 2 132 . 8 which we see is to large to ignore. Therefore, we must try to approximate t . We will use Newton’s method on the equation f ( t ) = 98 . 1 t + 981 e t/ 10 1981 = 0 . (If we can find a root to this equation, we will have found the t we want.) Newton’s method generates a sequence of approximations given by the formula t n +1 = t n f ( t n ) f ( t n ) . Since f ( t ) = 98 . 1 98 . 1 e t/ 10 = 98 . 1 ( 1 e t/ 10 ) , the recursive equation above becomes t n +1 = t n t n + 10 e t n / 10 (1981 / 98 . 1) 1 e t n / 10 . (3.10) To start the process, let
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