Author Topic: tut 0202 quiz4  (Read 476 times)

Ruojing Chen

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tut 0202 quiz4
« on: October 18, 2019, 02:02:34 PM »
Find the general solution for equation y''-2y'+6y=0

$$ set \ r^2-2r+6=0$$

$$r=\frac{2\pm\sqrt{(-2)^2-4*6}}{2}$$

$$r=\frac{2\pm2i\sqrt{5}}{2}$$

$$r=1\pm\sqrt{5}i$$

$$y_c(t)=c_1e^tCos(\sqrt{5}t)+c_2e^tSin(\sqrt{5}t)$$
« Last Edit: October 18, 2019, 02:07:46 PM by rj127 »