Finding general soln of differential eqn (zero under the root)

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SUMMARY

The differential equation 9y'' + 6y' + y = 0 has a characteristic equation r^2 + 6r + 1. The roots are calculated as r = -3 ± √(36 - 4) / 2, leading to a double root of λ = -1/3. When encountering a double root, the general solution takes the form y(t) = C_1e^{λt} + C_2te^{λt}, which is essential for solving such equations. The confusion regarding zero under the root is clarified by recognizing that it indicates a double root scenario.

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Homework Statement



9y'' + 6y' + y = 0

r^2 + 6r + 1

( -6 +/- root 36 - 36 ) / 18

lamda = -6/18 = -(2/9)
mu = 0

(this is the part I am unsure about, because if i end up with a zero under the root, then does this make the final equation into this:

y = c_1*cos(0*x) + c_2*sin(0*x)

thanks for any help.

Homework Equations





The Attempt at a Solution

 
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First, [itex]-\frac{6}{18}=-\frac{1}{3}\neq-\frac{2}{9}[/tex]. <br /> <br /> Second, when you have a double root, [itex]\lambda[/itex], you look for a solution of the form [itex]y(t)=C_1e^{\lambda t}+C_2 t e^{\lambda t}[/itex][/itex]
 

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