### Author Topic: TUT0702 Quiz3  (Read 760 times)

#### Zhangxinbei

• Jr. Member
• Posts: 8
• Karma: 0
##### 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