Toronto Math Forum

MAT244--2018F => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on October 12, 2018, 06:04:24 PM

Title: Q3 TUT 0301
Post by: Victor Ivrii on October 12, 2018, 06:04:24 PM
Find a differential equation whose general solution is
$$y=c_1e^{-t/2}+c_2e^{-2t}.$$
Title: Re: Q3 TUT 0301
Post by: Yunqi(Yuki) Huang on October 12, 2018, 06:16:38 PM
in the attchement
Title: Re: Q3 TUT 0301
Post by: Keyue Xie 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
$$r^2+\frac{5}{2}r+1 = 0$$
$$2r^2 + 5r +2 = 0$$
$$2y''(t) + 5y'(t) + 2y(t) = 0$$
Title: Re: Q3 TUT 0301
Post by: Yunqi(Yuki) Huang 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
$$2y'' + 5y' + 2y = 0$$
Title: Re: Q3 TUT 0301
Post by: Victor Ivrii on October 12, 2018, 07:44:17 PM
Yuki, only ASCII file names!