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139_pdfsam_math 54 differential equation solutions odd

# 139_pdfsam_math 54 differential equation solutions odd -...

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Exercises 3.4 Solving this linear equation and using the initial condition, v (A) 1 (0) = 2, we get v (A) 1 ( t )= 65 8 49 8 e 4 t/ 3 , and so x (A) 1 ( t t Z 0 ± 65 8 49 8 e 4 s/ 3 ² ds = 65 8 t 147 32 ( e 4 t/ 3 1 ) . The boat A will have the velocity 5 m/sec at t = t satisfying 65 8 49 8 e 4 t / 3 =5 t = 3ln(25 / 49) 4 0 . 5 (sec) , and it will be x (A) 1 ( t 65 8 t 147 32 ( e 4 t / 3 1 ) 1 . 85 (m) away from the starting point or, equivalently, 500 1 . 85 = 498 . 15 meters away from the Fnish. Similarly to (3.12), resetting the time, we obtain an equation of the motion of the boat A starting from the moment when its velocity reaches 5 m/sec. Denoting by x (A) 2 ( t ) the distance passed by the boat A and by v (A) 2 ( t ) its velocity, we get x (A) 2 (0) = 0, v (A) 2 (0) = 5, and m dv (A) 2 dt = 650 b 2 v (A) 2 dv (A) 2 dt = 65 6 v (A) 2 v (A) 2 ( t 65 6 35 6 e t x (A) 2 ( t t Z 0 ± 65 6 35 6 e s ² ds = 65 6 t + 35 6 ( e t 1 ) . Solving the equation x (A) 2 ( t ) = 498 . 15, we Fnd the time (counting from the moment when the
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