Author Topic: TUT0702 Quiz3  (Read 330 times)

Zhangxinbei

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TUT0702 Quiz3
« on: October 11, 2019, 01:51:34 PM »
Find the differential equation has the general solution of :
  y = C1e^(-t/2) + C2e^-2t

Solution:
       The general solution has the form y = C1e^r1t + C2e^r2t
       we have r1 = -1/2.  r2 = -2
       (r + 1/2)(r+2) = 0
       r^2 + 5/2r + 1 = 0
       The differential equation has the from of y'' + p(t)y' +q(t)y = g(t)
       Therefore,
       the differential equation which has the general solution of has the general solution of y = C1e^(-t/2) + C2e^-2t is:
       Y'' + 5/2Y' + Y = 0