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Messages - Keyue Xie

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Quiz-3 / Re: Q3 TUT 0601
« on: October 12, 2018, 06:53:18 PM »
$$y=c_1e^{2t} + c_2e^{-3t}$$
$$r_1 = 2, r_2 = -3$$
$$(r-2)(r+3) = 0$$
$$r^2+ r - 6 = 0$$
$$y''(t) + y'(t) - 6y(t) = 0$$

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Quiz-3 / Re: Q3 TUT 0301
« 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
$$r^2+\frac{5}{2}r+1 = 0$$
$$2r^2 + 5r +2 = 0$$
$$2y''(t) + 5y'(t) + 2y(t) = 0$$

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