### Author Topic: Quiz 5 Victor's Section Lec5101  (Read 649 times)

#### quxipeng

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• Karma: 0 ##### Quiz 5 Victor's Section Lec5101
« on: October 31, 2019, 11:42:43 AM »
Question: y'' + 4y' + 4y = t^-2 * e^-2t, t > 0

Solutions:
(1) Consider the homogeneous equation y'' + 4y' + 4y = 0 first
(2) Consider its characteristic equation: r^2 + 4r + 4 =0
(r + 2)^2 = 0
r1 = r2 = -2
so y1(t) = e^-2t, y2(t) = te^-2t
thus, yc(t) = C1e^-2t+ C2te^-2t, where C1 and C2 are two arbitrary constants
(3) Apply the method of variation of parameter
y1'(t) = -2e^-2t, y'2(t) = e^-2t - 2te^-2t
Wronskian = y1(t)*y'2(t) - y2(t)*y'1(t) = e^-4t
(4) Compute u1(t) and u2(t)
u1(t) = integral (-te^-2t * t^-2 * e^-2t) / e^-4t dt
= integral -t^-1dt
= - In(t)
u2(t) = integral (e^-2t * t^-2 * e^-2t) / e^-4t dt
= integral t^-2 dt
= - t^-1
(5) Figure out the particular solution
yp(t) = u1(t) * y1(t) + u2(t) * y2(t)
= -In(t) * e^-2t + t*e^-2t * -t^-1
= -e^-2t* In(t) - e^ -2t
(6) Write down the general solution
y(t) = yc(t) + yp(t)
= C1e^-2t+ C2te^-2t -e^-2t* In(t) - e^ -2t
= (C1 - 1)e^-2t+ C2te^-2t - e^-2t* In(t)
=K1*e^-2t+ K2*te^-2t - e^-2t * In(t)     (In here, we substitute C1 - 1, C2 = K1, K2 respectively, because they are both arbitrary constants)
Now we are done !
« Last Edit: October 31, 2019, 01:56:24 PM by quxipeng »