### Author Topic: quiz3 tut0502  (Read 389 times)

#### yueyangyu

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##### quiz3 tut0502
« on: October 11, 2019, 02:00:02 PM »
Find the general solution of the given differential equation.
$$y''-2y'-2y=0$$

Assume that $$y=e^{rt}$$ and it follows that r must be a root of characteristic equation
$$r^2-2r-2=0$$
$$r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Then,
$$r_1=1+\sqrt{3} \quad r_2=1-\sqrt{3}$$

Therefore, the general solution of the given differential equation is:
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$