Author Topic: TUT0401 Quiz 4  (Read 511 times)

Yuying Chen

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TUT0401 Quiz 4
« on: October 18, 2019, 01:44:42 PM »
$9y^{\prime\prime}+6y^{\prime}+y=0\\$
$\text{We assume that $y=e^{rt}$ is a solution of this equation. Then the characteristic equation is }\\$
$9r^2+6r+1=0\\$
$\quad(3r+1)^2=0\\$
$\qquad\qquad r=-\frac{1}{3},-\frac{1}{3}\\$
$\text{Since the equation has two equal(repeated) roots.}\\$
$\text{Therefore, the general solution:}\\$
$y=c_1e^{-\frac{1}{3}t}+c_2te^{-\frac{1}{3}t}$