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MAT244--2019F
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tut 0202 quiz4
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Topic: tut 0202 quiz4 (Read 4186 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
$$ set \ r^2-2r+6=0$$
$$r=\frac{2\pm\sqrt{(-2)^2-4*6}}{2}$$
$$r=\frac{2\pm2i\sqrt{5}}{2}$$
$$r=1\pm\sqrt{5}i$$
$$y_c(t)=c_1e^tCos(\sqrt{5}t)+c_2e^tSin(\sqrt{5}t)$$
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Last Edit: October 18, 2019, 02:07:46 PM by rj127
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Toronto Math Forum
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MAT244--2019F
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tut 0202 quiz4