Author Topic: Q3-T0701  (Read 785 times)

Victor Ivrii

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Q3-T0701
« on: February 10, 2018, 05:19:20 PM »
Find the general solution of the given differential equation.
$$y'' - 2y' - 2y = 0.
$$

Junjie Zhang

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Re: Q3-T0701
« Reply #1 on: February 10, 2018, 05:54:40 PM »
The characteristic equation is $r^2-2r-2=0$, with root of $r = 1+\sqrt{3}, 1-\sqrt{3}$.
Hence, the general solution should be $y = c_{1}exp(1-\sqrt{3})t+c_{2}(1+\sqrt{3})t$
« Last Edit: February 10, 2018, 05:57:22 PM by Junjie Zhang »

Meng Wu

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Re: Q3-T0701
« Reply #2 on: February 11, 2018, 09:33:51 AM »
$$y''-2y'-2y=0$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$r^2-2r-2=0$$
We use the quadratic formula which is
$$r={-b\pm \sqrt{b^2-4ac}\over 2a}$$
Hence,
$$\cases{r_1={1+\sqrt{3}}\\r_2=1-\sqrt{3}}$$
Since the general solution has the form of $$y=c_1e^{r_1t}+c_2e^{r_2t}$$
Therefore, the general solution of the given differential equation is
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$