Author Topic: Q3 TUT 0301  (Read 979 times)

Victor Ivrii

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Q3 TUT 0301
« on: October 12, 2018, 06:04:24 PM »
Find a differential equation whose general solution is 
$$y=c_1e^{-t/2}+c_2e^{-2t}.$$

Yunqi(Yuki) Huang

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Re: Q3 TUT 0301
« Reply #1 on: October 12, 2018, 06:16:38 PM »
in the attchement

Keyue Xie

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Re: Q3 TUT 0301
« Reply #2 on: October 12, 2018, 06:34:41 PM »
$$
y=c_1e^{\frac{-1}2t} + c_2e^{(-2t)}
$$
$$
r_1=\frac{-1}2, r_2=-2
$$
$$
(r+\frac{1}{2})(r+2) = 0
$$
$$
r^2+ 2r+\frac{1}2r +1 = 0
$$
Therefore
$$
r^2+\frac{5}{2}r+1 = 0
$$
$$
2r^2 + 5r +2 = 0
$$
$$
2y''(t) + 5y'(t) + 2y(t) = 0
$$
« Last Edit: October 12, 2018, 06:55:18 PM by Keyue Xie »

Yunqi(Yuki) Huang

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Re: Q3 TUT 0301
« Reply #3 on: October 12, 2018, 06:49:55 PM »
$$
r=-\frac{1}{2}or-2
$$
$$
(r+\frac{1}{2})(r+2) = 0
$$
$$
r^2+ 2r+\frac{1}2r +1 = 0
$$
$$
r^2+\frac{5}{2}r+1 = 0
$$
$$
2r^2 + 5r +2 = 0
$$
Therefore
$$
2y'' + 5y' + 2y = 0
$$

Victor Ivrii

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Re: Q3 TUT 0301
« Reply #4 on: October 12, 2018, 07:44:17 PM »
Yuki, only ASCII file names!