141_pdfsam_math 54 differential equation solutions odd

# 141_pdfsam_math 54 differential equation solutions odd -...

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Exercises 3.5 (e) Using the equation v e = 2 gR for the escape velocity and converting meters to kilometers we have v e = p 2 = ± 2 · 9 . 81 m / sec 2 · (1 km / 1000 m)(6370 km) 11 . 18 km / sec . (f) Similarly to (e), we Fnd v e = p 2( g/ 6) R = p 2(9 . 81 / 6)(1 / 1000)(1738) = 2 . 38 (km / sec) . EXERCISES 3.5: Electrical Circuits, page 122 1. In this problem, R =5Ω , L =0 . 05 H, and the voltage function is given by E ( t ) = 5 cos 120 t V. Substituting these data into a general solution (3) to the Kirchho±’s equation (2) yields I ( t )= e Rt/L ²Z e Rt/L E ( t ) L dt + K ³ = e 5 t/ 0 . 05 e 5 t/ 0 . 05 5 cos 120 t 0 . 05 dt + K ³ = e 100 t ² 100 Z e 100 t cos 120 tdt + K ³ . Using the integral tables, we evaluate the integral in the right-hand side and obtain I ( t e 100 t ´ 100 e 100 t (100 cos 120 t + 120 sin 120
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