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Find the general solution of \(y'''-3y''+3y'-y=0\)
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From this I should have, \(r^3-3r^2+3r-1=0\) Right? Factor that to get.... \((r-1)^3=0\) \(r=1\)
Now... where would I go from here? \(y=c_1e^{t}+c_2te^{t}+c_3t^2e^{t}\) Is this correct?
yup
Coolio, well I have to go now unfortunately... Have a great day all!
you too
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