MAT244--2018F > Quiz-3
Q3 TUT 0301
(1/1)
Victor Ivrii:
Find a differential equation whose general solution is
$$y=c_1e^{-t/2}+c_2e^{-2t}.$$
Yunqi(Yuki) Huang:
in the attchement
Keyue Xie:
$$
y=c_1e^{\frac{-1}2t} + c_2e^{(-2t)}
$$
$$
r_1=\frac{-1}2, r_2=-2
$$
$$
(r+\frac{1}{2})(r+2) = 0
$$
$$
r^2+ 2r+\frac{1}2r +1 = 0
$$
Therefore
$$
r^2+\frac{5}{2}r+1 = 0
$$
$$
2r^2 + 5r +2 = 0
$$
$$
2y''(t) + 5y'(t) + 2y(t) = 0
$$
Yunqi(Yuki) Huang:
$$
r=-\frac{1}{2}or-2
$$
$$
(r+\frac{1}{2})(r+2) = 0
$$
$$
r^2+ 2r+\frac{1}2r +1 = 0
$$
$$
r^2+\frac{5}{2}r+1 = 0
$$
$$
2r^2 + 5r +2 = 0
$$
Therefore
$$
2y'' + 5y' + 2y = 0
$$
Victor Ivrii:
Yuki, only ASCII file names!
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