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Mathematics
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Solve the differential equation
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\[2t^2y''+4ty'-2y=0\] Given , t<0
I know this is a cauchy-Euler equation and my guess is y(t)=(-t)^r but i'm not sure where to go from there.
So \[y(t) = (-t)^r\] \[y'(t)=r(-t)^{(r-1)}\] \[y''(t)=r(r-1)(-t)^{r-2}\]
Substituting: \[2t^2(r(r-1)(-t)^{r-2})+4t(r(-r)^{r-1})-2(-t)^r=0\]
How do I continue? >.< .
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Nvm, got it.
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