Author Topic: TUT0502 Quiz4  (Read 533 times)

Xinqiao Li

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TUT0502 Quiz4
« on: October 18, 2019, 09:59:31 PM »
Find the general solution of the given differential equation:
$$9y'' + 6y' + y = 0$$
Solve:

The characteristic polynomial is given by $9r^2 + 6r + 1 = 0$

Factor it we obtain $(3r + 1)(3r + 1)$

Then $r_1 = -\frac{1}{3}$ and $r_2 = -\frac{1}{3}$

We observe repeated roots here.

Then the general solution of the given differential equation is $y(t) = c_1e^{-\frac{1}{3} t} + c_2te^{-\frac{1}{3}t}$