131_pdfsam_math 54 differential equation solutions odd

131_pdfsam_math 54 differential equation solutions odd - 2...

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Exercises 3.4 This is a linear equation. Solving yields d ( e 2 t/ 5 v 1 ) = e 2 t/ 5 gdt v 1 ( t )= 5 g 2 + C 1 e 2 t/ 5 ; 0= v 1 (0) = 5 g 2 + C 1 e 0 = 5 g 2 + C 1 C 1 = 5 g 2 v 1 ( t )= 5 g 2 ( 1 e 2 t/ 5 ) x 1 ( t )= t Z 0 v 1 ( s ) ds = 5 g 2 ± s + 5 2 e 2 s/ 5 ²³ ³ ³ ³ s = t s =0 = 5 g 2 ± t + 5 2 e 2 t/ 5 5 2 ² . To fnd the time t when the chute opens, we solve 20 = v 1 ( t ) 20 = 5 g 2 ( 1 e 2 t / 5 ) t = 5 2 ln ± 1 8 g ² 4 . 225 (sec) . By this time the parachutist has Fallen x 1 ( t )= 5 g 2 ± t + 5 2 e 2 t / 5 5 2 ² 5 g 2 ± 4 . 225 + 5 2 e 2 · 4 . 225 / 5 5 2 ² 53 . 62 (m) , and so she is 2000 53 . 62 = 1946 . 38 m above the ground. Setting the second equation, we For convenience reset the time t .D en o t in gb y x
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Unformatted text preview: 2 ( t ) := x 2 ( t ) her velocity, we have 75 dv 2 dt = 75 g 90 v 2 , v 2 (0) = v 1 ( t ) = 20 , x 2 (0) = 0 . Solving, we get v 2 ( t ) = 5 g 6 + C 2 e 6 t/ 5 ; 20 = v 2 (0) = 5 g 6 + C 2 C 2 = 20 5 g 6 v 2 ( t ) = 5 g 6 + 20 5 g 6 e 6 t/ 5 x 2 ( t ) = t Z v 2 ( s ) ds = 5 g 6 s 5 6 20 5 g 6 e 6 s/ 5 s = t s =0 = 5 g 6 t + 5 6 20 5 g 6 ( 1 e 6 t/ 5 ) . 127...
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