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### Messages - Yu Qi Huang

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##### Quiz-4 / Quiz 4 TUT 0702
« on: October 18, 2019, 02:12:16 PM »
Find the general solution for y'' - 6y' + 9y = 0.

Solution:
changing it into characteristic equation it becomes:
r^2 - 6r + 9 = 0
(r-3)^2 = 0
r = 3

Hence y(t) = c_1e^(3t) + c_2te^(3t)

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##### Quiz-3 / TUT 0702 Quiz 3
« on: October 11, 2019, 02:05:02 PM »
Verify that y_1(t) and y_2(t) are solutions of y'' - 2y' + y = 0 where  y_1(t) = e^t and y_2(t) = te^t. Do they constitute a fundamental set?

Solution:
Transforming into characteristic equation the equation becomes:
r^2 - 2r + 1 = 0
(r - 1)^2 = 0
r = 1
Hence y(t) = c_1*e^t + c_2*te^t

w = e^t * (e^t  + te^t) - (e^t * te^t)
= (e^t)^2 + e^t(te^t) - (e^t)(te^t)
= (e^t)^2
≠ 0

Hence they do constitute a fundamental set.

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##### Quiz-2 / Re: TUT 0702 QUIZ2
« on: October 04, 2019, 05:56:01 PM »
Hey there,

I believe you made a mistake there in your solution as it should be 1/2·x^2 − 1/2·y^(−2) + ln|y| = c
rather than 1/2·x^2 − 1/2·y^(−2) - ln|y| = c
since the derivative of y^(-1) is ln|y| not -ln|y|.

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