Toronto Math Forum
MAT244--2018F => MAT244--Tests => Quiz-3 => Topic started by: Victor Ivrii on October 12, 2018, 06:02:55 PM
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Find the general solution of the given differential equation.
$$y''-2y'-2y=0.$$
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In the attachment.
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$$y′′−2y′−2y=0$$
$$r^2−2r−2=0$$
$$r = \frac{−b \pm \sqrt{b^2−4ac}}{2a}$$
$$r_1=1+\sqrt{3}, r_2=1−\sqrt{3}$$
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$
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LOL: Yulin, you typed it to produce in the end png. Why not directly into forum?