MAT244--2019F > Quiz-3

TUT0702 Quiz3

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**Zhangxinbei**:

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

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