I need help with problem 5 because the Current answer I got is I = e-40t( 2cos(30t) + (242/3)sin(30t) ) Which is different from the answer shown below

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Linear Second-Order Equations

I need help with problem 5 

because the Current answer I got is

I = e-40t( 2cos(30t) + (242/3)sin(30t) )

Which is different from the answer shown below 

6.3.5 (p. 295) I = -e-40t (2 cos 30t - 86 sin 30t)
Transcribed Image Text:6.3.5 (p. 295) I = -e-40t (2 cos 30t - 86 sin 30t)
6.3 Exercises
In Exercises 1-5 find the current in the RLC circuit, assuming that E (t) = 0 for t > 0.
1. R = 3 ohms; L = .1 henrys; C = .01 farads; 2o = 0 coulombs; Io = 2 amperes.
L = .05 henrys; C = .01 farads'; Qo = 2 coulombs; Io = -2 amperes.
2.
3.
L = .1 henrys; C = .01 farads; 20
=
2 coulombs; Io = 0 amperes.
4.
L = .1 henrys; C = .004 farads'; Qo
5.
L = .05 henrys; C = .008 farads; Qo = -1 coulombs; Io = 2 amperes.
R = 2 ohms;
R = 2 ohms;
R = 6 ohms;
R = 4 ohms;
= 3 coulombs; Io = -10 amperes.
Transcribed Image Text:6.3 Exercises In Exercises 1-5 find the current in the RLC circuit, assuming that E (t) = 0 for t > 0. 1. R = 3 ohms; L = .1 henrys; C = .01 farads; 2o = 0 coulombs; Io = 2 amperes. L = .05 henrys; C = .01 farads'; Qo = 2 coulombs; Io = -2 amperes. 2. 3. L = .1 henrys; C = .01 farads; 20 = 2 coulombs; Io = 0 amperes. 4. L = .1 henrys; C = .004 farads'; Qo 5. L = .05 henrys; C = .008 farads; Qo = -1 coulombs; Io = 2 amperes. R = 2 ohms; R = 2 ohms; R = 6 ohms; R = 4 ohms; = 3 coulombs; Io = -10 amperes.
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