Sep 21, 2009 #1 S Sicar Newbie level 2 Joined Sep 20, 2009 Messages 2 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,281 Activity points 1,290 First time taking this course, concept is lost on me. Problem seems simple enough. x" + 4x = 0 x(0) = 1 x(0)'=0 given initial values .. I just don't know where to start with it. It needs to be done without using LaPlace transforms?
First time taking this course, concept is lost on me. Problem seems simple enough. x" + 4x = 0 x(0) = 1 x(0)'=0 given initial values .. I just don't know where to start with it. It needs to be done without using LaPlace transforms?
Sep 22, 2009 #2 _Eduardo_ Full Member level 5 Joined Aug 31, 2009 Messages 295 Helped 118 Reputation 238 Reaction score 103 Trophy points 1,323 Location Argentina Activity points 2,909 x" + 4x = 0 x(0) = 1 x(0)'=0 given initial values .. You must know some theory of differential equations: https://en.wikipedia.org/wiki/Linear_differential_equation Then, the characteristic equation is z^2+4 = 0 --> z = +/- 2j (j imaginary unit) The general solution is: x(t) = A*exp(j 2t) + B*exp(-j 2t) = C*cos(2t) + D*sin(2t) But x(0) = 1 and x(0)'=0 --> C=1, D=0 --> x(t) = cos(2t)
x" + 4x = 0 x(0) = 1 x(0)'=0 given initial values .. You must know some theory of differential equations: https://en.wikipedia.org/wiki/Linear_differential_equation Then, the characteristic equation is z^2+4 = 0 --> z = +/- 2j (j imaginary unit) The general solution is: x(t) = A*exp(j 2t) + B*exp(-j 2t) = C*cos(2t) + D*sin(2t) But x(0) = 1 and x(0)'=0 --> C=1, D=0 --> x(t) = cos(2t)
Sep 22, 2009 #3 S Sicar Newbie level 2 Joined Sep 20, 2009 Messages 2 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,281 Activity points 1,290 Thanks man. Very good explanation!