Posted by Carly on Friday, July 29, 2011 at 12:25am.
Integral of dt = Integral of (1/4)y^-6 dy
t = -1/(20 y^-5) + C
0 = 1/20*243 +C
t = -1/(20 y^-5) -1/4860
1/(20 y^-5) = -(1/4860) - t
y^5 = (1/20)/[(-1/4860) - t]
= 243/[-1 -4860 t]
y = -3*[1 +4860 t]^(1/5)
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