Solving an Initial Value Problem: 9r^2-12r+4=0

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SUMMARY

The discussion focuses on solving the initial value problem represented by the differential equation 9y'' - 12y' + 4y = 0 with initial conditions y(0) = 2 and y'(0) = -1. The characteristic equation 9r^2 - 12r + 4 = 0 yields a double root at r = 2/3, leading to the general solution y(t) = c_1 e^(2/3 t) + c_2 e^(2/3 t). To account for the double root, the second linearly independent solution is t e^(2/3 t). The correct general solution incorporates both solutions, ensuring the initial conditions can be applied correctly.

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Solve the given initial value problem.
9y"-12y'+4y=0 y(0)=2 y'(0)=-1

9r^2-12r +4=0
(3r-2)^2
so r=2/3
so my general solution would be y(t)=c_1e^2/3t +c_2e^2/3t
Whenever I try to use the initial conditions I can't get them to work. One disappears.
Is my general solution even right?
 
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You only have one linearly independent solution there. When you get a double root like that, the other solution is given by, in this case, t e^{2/3 t} [/tex]. You can verify this by plugging it into the DE.
 
I get it.. thanks a lot!
 

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